\(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx\) [583]

Optimal result

Integrand size = 23, antiderivative size = 220 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {(a-b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 a^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right ) d}+\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 a \sqrt {\tan (c+d x)}}{b^2 d}+\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d} \] Output:

1/2*(a-b)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d+1/2*(a-b 
)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d+2*a^(7/2)*arctan( 
b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(5/2)/(a^2+b^2)/d+1/2*(a+b)*arctanh(2^ 
(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/(a^2+b^2)/d-2*a*tan(d*x+c)^ 
(1/2)/b^2/d+2/3*tan(d*x+c)^(3/2)/b/d
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.01 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-\frac {6 \sqrt {2} (a-b) b^2 \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )}{a^2+b^2}+\frac {24 a^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} \left (a^2+b^2\right )}-\frac {3 \sqrt {2} b^2 (a+b) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )}{a^2+b^2}-24 a \sqrt {\tan (c+d x)}+8 b \tan ^{\frac {3}{2}}(c+d x)}{12 b^2 d} \] Input:

Integrate[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x]),x]
 

Output:

((-6*Sqrt[2]*(a - b)*b^2*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[ 
1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]))/(a^2 + b^2) + (24*a^(7/2)*ArcTan[(Sqrt[b 
]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[b]*(a^2 + b^2)) - (3*Sqrt[2]*b^2*(a 
+ b)*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2] 
*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]))/(a^2 + b^2) - 24*a*Sqrt[Tan[c + d*x] 
] + 8*b*Tan[c + d*x]^(3/2))/(12*b^2*d)
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.16, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.957, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4137, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{7/2}}{a+b \tan (c+d x)}dx\)

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2 \int -\frac {3 \sqrt {\tan (c+d x)} \left (a \tan ^2(c+d x)+b \tan (c+d x)+a\right )}{2 (a+b \tan (c+d x))}dx}{3 b}+\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\int \frac {\sqrt {\tan (c+d x)} \left (a \tan ^2(c+d x)+b \tan (c+d x)+a\right )}{a+b \tan (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\int \frac {\sqrt {\tan (c+d x)} \left (a \tan (c+d x)^2+b \tan (c+d x)+a\right )}{a+b \tan (c+d x)}dx}{b}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 \int -\frac {a^2+\left (a^2-b^2\right ) \tan ^2(c+d x)}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{b}+\frac {2 a \sqrt {\tan (c+d x)}}{b d}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\int \frac {a^2+\left (a^2-b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{b}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\int \frac {a^2+\left (a^2-b^2\right ) \tan (c+d x)^2}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{b}}{b}\)

\(\Big \downarrow \) 4137

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {\int \frac {a b^2-b^3 \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {a^4 \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {\int \frac {a b^2-b^3 \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 \int \frac {b^2 (a-b \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 b^2 \int \frac {a-b \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 b^2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 b^2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 b^2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 b^2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 b^2 \left (\frac {1}{2} (a+b) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 b^2 \left (\frac {1}{2} (a+b) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 b^2 \left (\frac {1}{2} (a+b) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {2 b^2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{b}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^4 \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}+\frac {2 b^2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 a^4 \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {2 b^2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{b}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d}-\frac {\frac {2 a \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 b^2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {2 a^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} d \left (a^2+b^2\right )}}{b}}{b}\)

Input:

Int[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x]),x]
 

Output:

-((-(((2*a^(7/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[b]*(a 
^2 + b^2)*d) + (2*b^2*(((a - b)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/ 
Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + ((a + b)*( 
-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + 
Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)* 
d))/b) + (2*a*Sqrt[Tan[c + d*x]])/(b*d))/b) + (2*Tan[c + d*x]^(3/2))/(3*b* 
d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

rule 4130

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. 
) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && 
 NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4137

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) 
+ (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Sim 
p[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*Simp[a*(A - C) - (A*b - b*C)*T 
an[e + f*x], x], x], x] + Simp[(A*b^2 + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan 
[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{ 
a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && 
 NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+a \sqrt {\tan \left (d x +c \right )}\right )}{b^{2}}+\frac {2 a^{4} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{b^{2} \left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}}{d}\)	\(255\)
default	\(\frac {-\frac {2 \left (-\frac {b \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+a \sqrt {\tan \left (d x +c \right )}\right )}{b^{2}}+\frac {2 a^{4} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{b^{2} \left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}}{d}\)	\(255\)

Input:

int(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2/b^2*(-1/3*b*tan(d*x+c)^(3/2)+a*tan(d*x+c)^(1/2))+2/b^2*a^4/(a^2+b^ 
2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))+2/(a^2+b^2)*(1/8*a*2 
^(1/2)*(ln((tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+1)/(tan(d*x+c)-2^(1/2)*tan 
(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2) 
*tan(d*x+c)^(1/2)))-1/8*b*2^(1/2)*(ln((tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2) 
+1)/(tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c) 
^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 766 vs. \(2 (187) = 374\).

Time = 0.19 (sec) , antiderivative size = 1558, normalized size of antiderivative = 7.08 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c)),x, algorithm="fricas")
 

Output:

[1/6*(6*a^3*sqrt(-a/b)*log((2*b*sqrt(-a/b)*sqrt(tan(d*x + c)) + b*tan(d*x 
+ c) - a)/(b*tan(d*x + c) + a)) - 6*sqrt(1/2)*(a^2*b^2 + b^4)*d*sqrt((a^2 
- 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*arctan(-((a^4 + 2*a^2*b^2 + 
b^4)*d^2*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt((a^2 
 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) + 2*sqrt(1/2)*(a^3 + a^2*b 
+ a*b^2 + b^3)*d*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*s 
qrt(tan(d*x + c)))/(a^2 - b^2)) - 6*sqrt(1/2)*(a^2*b^2 + b^4)*d*sqrt((a^2 
- 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*arctan(((a^4 + 2*a^2*b^2 + b 
^4)*d^2*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt((a^2 
- 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 2*sqrt(1/2)*(a^3 + a^2*b + 
 a*b^2 + b^3)*d*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sq 
rt(tan(d*x + c)))/(a^2 - b^2)) + 3*sqrt(1/2)*(a^2*b^2 + b^4)*d*sqrt((a^2 + 
 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(2*sqrt(1/2)*(a^2 + b^2)*d 
*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(tan(d*x + c) 
) + (a + b)*tan(d*x + c) + a + b) - 3*sqrt(1/2)*(a^2*b^2 + b^4)*d*sqrt((a^ 
2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(-2*sqrt(1/2)*(a^2 + b^ 
2)*d*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(tan(d*x 
+ c)) + (a + b)*tan(d*x + c) + a + b) - 4*(3*a^3 + 3*a*b^2 - (a^2*b + b^3) 
*tan(d*x + c))*sqrt(tan(d*x + c)))/((a^2*b^2 + b^4)*d), 1/6*(12*a^3*sqrt(a 
/b)*arctan(b*sqrt(a/b)*sqrt(tan(d*x + c))/a) - 6*sqrt(1/2)*(a^2*b^2 + b...
 

Sympy [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\tan ^{\frac {7}{2}}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \] Input:

integrate(tan(d*x+c)**(7/2)/(a+b*tan(d*x+c)),x)
 

Output:

Integral(tan(c + d*x)**(7/2)/(a + b*tan(c + d*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.92 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {24 \, a^{4} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{2} b^{2} + b^{4}\right )} \sqrt {a b}} + \frac {3 \, {\left (2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a + b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a + b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )}}{a^{2} + b^{2}} + \frac {8 \, {\left (b \tan \left (d x + c\right )^{\frac {3}{2}} - 3 \, a \sqrt {\tan \left (d x + c\right )}\right )}}{b^{2}}}{12 \, d} \] Input:

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c)),x, algorithm="maxima")
 

Output:

1/12*(24*a^4*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^2*b^2 + b^4)*sqrt( 
a*b)) + 3*(2*sqrt(2)*(a - b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x 
+ c)))) + 2*sqrt(2)*(a - b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x 
+ c)))) + sqrt(2)*(a + b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 
1) - sqrt(2)*(a + b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/ 
(a^2 + b^2) + 8*(b*tan(d*x + c)^(3/2) - 3*a*sqrt(tan(d*x + c)))/b^2)/d
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c)),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 3.39 (sec) , antiderivative size = 4927, normalized size of antiderivative = 22.40 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(7/2)/(a + b*tan(c + d*x)),x)
 

Output:

(2*tan(c + d*x)^(3/2))/(3*b*d) - atan((((((1/(b^2*d^2*1i - a^2*d^2*1i + 2* 
a*b*d^2))^(1/2)*(((1/(b^2*d^2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2)*((((32*( 
12*a^2*b^9*d^4 + 24*a^4*b^7*d^4 + 12*a^6*b^5*d^4))/(b^3*d^5) - (16*tan(c + 
 d*x)^(1/2)*(1/(b^2*d^2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2)*(16*b^12*d^4 + 
 16*a^2*b^10*d^4 - 16*a^4*b^8*d^4 - 16*a^6*b^6*d^4))/(b^3*d^4))*(1/(b^2*d^ 
2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2))/2 + (32*tan(c + d*x)^(1/2)*(14*a*b^ 
9*d^2 + 16*a^9*b*d^2 - 4*a^3*b^7*d^2 - 2*a^5*b^5*d^2))/(b^3*d^4)))/2 - (32 
*(4*a^9*d^2 + a*b^8*d^2 + a^3*b^6*d^2 + 16*a^5*b^4*d^2 - 16*a^7*b^2*d^2))/ 
(b^3*d^5)))/2 - (32*tan(c + d*x)^(1/2)*(2*a^8 + b^8))/(b^3*d^4))*(1/(b^2*d 
^2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2)*1i)/2 - ((((1/(b^2*d^2*1i - a^2*d^2 
*1i + 2*a*b*d^2))^(1/2)*(((1/(b^2*d^2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2)* 
((((32*(12*a^2*b^9*d^4 + 24*a^4*b^7*d^4 + 12*a^6*b^5*d^4))/(b^3*d^5) + (16 
*tan(c + d*x)^(1/2)*(1/(b^2*d^2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2)*(16*b^ 
12*d^4 + 16*a^2*b^10*d^4 - 16*a^4*b^8*d^4 - 16*a^6*b^6*d^4))/(b^3*d^4))*(1 
/(b^2*d^2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2))/2 - (32*tan(c + d*x)^(1/2)* 
(14*a*b^9*d^2 + 16*a^9*b*d^2 - 4*a^3*b^7*d^2 - 2*a^5*b^5*d^2))/(b^3*d^4))) 
/2 - (32*(4*a^9*d^2 + a*b^8*d^2 + a^3*b^6*d^2 + 16*a^5*b^4*d^2 - 16*a^7*b^ 
2*d^2))/(b^3*d^5)))/2 + (32*tan(c + d*x)^(1/2)*(2*a^8 + b^8))/(b^3*d^4))*( 
1/(b^2*d^2*1i - a^2*d^2*1i + 2*a*b*d^2))^(1/2)*1i)/2)/(((((1/(b^2*d^2*1i - 
 a^2*d^2*1i + 2*a*b*d^2))^(1/2)*(((1/(b^2*d^2*1i - a^2*d^2*1i + 2*a*b*d...
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{a +\tan \left (d x +c \right ) b}d x \] Input:

int(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c)),x)
 

Output:

int((sqrt(tan(c + d*x))*tan(c + d*x)**3)/(tan(c + d*x)*b + a),x)