\(\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx\) [571]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [F(-2)]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 218 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d} \] Output:

1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/d+1/ 
2*(a+b)*(a^2-4*a*b+b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/d-1/2*( 
a-b)*(a^2+4*a*b+b^2)*arctanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1 
/2)/d+2*b*(3*a^2-b^2)*tan(d*x+c)^(1/2)/d+8/5*a*b^2*tan(d*x+c)^(3/2)/d+2/5* 
b^2*tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.54 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.55 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\frac {-5 \sqrt [4]{-1} (i a+b)^3 \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-5 (-1)^{3/4} (a+i b)^3 \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 b \sqrt {\tan (c+d x)} \left (15 a^2-5 b^2+5 a b \tan (c+d x)+b^2 \tan ^2(c+d x)\right )}{5 d} \] Input:

Integrate[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^3,x]
 

Output:

(-5*(-1)^(1/4)*(I*a + b)^3*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - 5*(-1)^ 
(3/4)*(a + I*b)^3*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 2*b*Sqrt[Tan[c 
+ d*x]]*(15*a^2 - 5*b^2 + 5*a*b*Tan[c + d*x] + b^2*Tan[c + d*x]^2))/(5*d)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.14, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 4049, 27, 3042, 4113, 3042, 4011, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3dx\)

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\tan (c+d x)} \left (12 a b^2 \tan ^2(c+d x)+5 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (5 a^2-3 b^2\right )\right )dx+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \sqrt {\tan (c+d x)} \left (12 a b^2 \tan ^2(c+d x)+5 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (5 a^2-3 b^2\right )\right )dx+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \sqrt {\tan (c+d x)} \left (12 a b^2 \tan (c+d x)^2+5 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (5 a^2-3 b^2\right )\right )dx+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {1}{5} \left (\int \sqrt {\tan (c+d x)} \left (5 a \left (a^2-3 b^2\right )+5 b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\int \sqrt {\tan (c+d x)} \left (5 a \left (a^2-3 b^2\right )+5 b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {1}{5} \left (\int \frac {5 a \left (a^2-3 b^2\right ) \tan (c+d x)-5 b \left (3 a^2-b^2\right )}{\sqrt {\tan (c+d x)}}dx+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\int \frac {5 a \left (a^2-3 b^2\right ) \tan (c+d x)-5 b \left (3 a^2-b^2\right )}{\sqrt {\tan (c+d x)}}dx+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {1}{5} \left (\frac {2 \int -\frac {5 \left (b \left (3 a^2-b^2\right )-a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (-\frac {10 \int \frac {b \left (3 a^2-b^2\right )-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\)

Input:

Int[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^3,x]
 

Output:

(2*b^2*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]))/(5*d) + ((-10*(-1/2*((a + 
b)*(a^2 - 4*a*b + b^2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) 
+ ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])) + ((a - b)*(a^2 + 4*a*b 
 + b^2)*(-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))/d + 
(10*b*(3*a^2 - b^2)*Sqrt[Tan[c + d*x]])/d + (8*a*b^2*Tan[c + d*x]^(3/2))/d 
)/5
 

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 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 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 4011
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

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 4113
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 
Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.15

method result size
derivativedivides \(\frac {\frac {2 b^{3} \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} \tan \left (d x +c \right )^{\frac {3}{2}}+6 a^{2} b \sqrt {\tan \left (d x +c \right )}-2 b^{3} \sqrt {\tan \left (d x +c \right )}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{d}\) \(250\)
default \(\frac {\frac {2 b^{3} \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} \tan \left (d x +c \right )^{\frac {3}{2}}+6 a^{2} b \sqrt {\tan \left (d x +c \right )}-2 b^{3} \sqrt {\tan \left (d x +c \right )}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{d}\) \(250\)
parts \(\frac {a^{3} \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}+\frac {b^{3} \left (\frac {2 \tan \left (d x +c \right )^{\frac {5}{2}}}{5}-2 \sqrt {\tan \left (d x +c \right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {3 a \,b^{2} \left (\frac {2 \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {3 a^{2} b \left (2 \sqrt {\tan \left (d x +c \right )}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\) \(416\)

Input:

int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(2/5*b^3*tan(d*x+c)^(5/2)+2*a*b^2*tan(d*x+c)^(3/2)+6*a^2*b*tan(d*x+c)^ 
(1/2)-2*b^3*tan(d*x+c)^(1/2)+1/4*(-3*a^2*b+b^3)*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*arcta 
n(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*( 
a^3-3*a*b^2)*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*arc 
tan(-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. 933 vs. \(2 (190) = 380\).

Time = 0.09 (sec) , antiderivative size = 933, normalized size of antiderivative = 4.28 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

1/10*(10*sqrt(1/2)*d*sqrt((a^6 - 6*a^5*b + 3*a^4*b^2 + 20*a^3*b^3 + 3*a^2* 
b^4 - 6*a*b^5 + b^6)/d^2)*arctan((2*sqrt(1/2)*(a^3 + 3*a^2*b - 3*a*b^2 - b 
^3)*d*sqrt((a^6 - 6*a^5*b + 3*a^4*b^2 + 20*a^3*b^3 + 3*a^2*b^4 - 6*a*b^5 + 
 b^6)/d^2)*sqrt(tan(d*x + c)) + d^2*sqrt((a^6 + 6*a^5*b + 3*a^4*b^2 - 20*a 
^3*b^3 + 3*a^2*b^4 + 6*a*b^5 + b^6)/d^2)*sqrt((a^6 - 6*a^5*b + 3*a^4*b^2 + 
 20*a^3*b^3 + 3*a^2*b^4 - 6*a*b^5 + b^6)/d^2))/(a^6 - 15*a^4*b^2 + 15*a^2* 
b^4 - b^6)) + 10*sqrt(1/2)*d*sqrt((a^6 - 6*a^5*b + 3*a^4*b^2 + 20*a^3*b^3 
+ 3*a^2*b^4 - 6*a*b^5 + b^6)/d^2)*arctan((2*sqrt(1/2)*(a^3 + 3*a^2*b - 3*a 
*b^2 - b^3)*d*sqrt((a^6 - 6*a^5*b + 3*a^4*b^2 + 20*a^3*b^3 + 3*a^2*b^4 - 6 
*a*b^5 + b^6)/d^2)*sqrt(tan(d*x + c)) - d^2*sqrt((a^6 + 6*a^5*b + 3*a^4*b^ 
2 - 20*a^3*b^3 + 3*a^2*b^4 + 6*a*b^5 + b^6)/d^2)*sqrt((a^6 - 6*a^5*b + 3*a 
^4*b^2 + 20*a^3*b^3 + 3*a^2*b^4 - 6*a*b^5 + b^6)/d^2))/(a^6 - 15*a^4*b^2 + 
 15*a^2*b^4 - b^6)) + 5*sqrt(1/2)*d*sqrt((a^6 + 6*a^5*b + 3*a^4*b^2 - 20*a 
^3*b^3 + 3*a^2*b^4 + 6*a*b^5 + b^6)/d^2)*log(-a^3 - 3*a^2*b + 3*a*b^2 + b^ 
3 + 2*sqrt(1/2)*d*sqrt((a^6 + 6*a^5*b + 3*a^4*b^2 - 20*a^3*b^3 + 3*a^2*b^4 
 + 6*a*b^5 + b^6)/d^2)*sqrt(tan(d*x + c)) - (a^3 + 3*a^2*b - 3*a*b^2 - b^3 
)*tan(d*x + c)) - 5*sqrt(1/2)*d*sqrt((a^6 + 6*a^5*b + 3*a^4*b^2 - 20*a^3*b 
^3 + 3*a^2*b^4 + 6*a*b^5 + b^6)/d^2)*log(-a^3 - 3*a^2*b + 3*a*b^2 + b^3 - 
2*sqrt(1/2)*d*sqrt((a^6 + 6*a^5*b + 3*a^4*b^2 - 20*a^3*b^3 + 3*a^2*b^4 + 6 
*a*b^5 + b^6)/d^2)*sqrt(tan(d*x + c)) - (a^3 + 3*a^2*b - 3*a*b^2 - b^3)...
 

Sympy [F]

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

integrate(tan(d*x+c)**(1/2)*(a+b*tan(d*x+c))**3,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.09 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\frac {8 \, b^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 40 \, a b^{2} \tan \left (d x + c\right )^{\frac {3}{2}} + 10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 40 \, {\left (3 \, a^{2} b - b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{20 \, d} \] Input:

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/20*(8*b^3*tan(d*x + c)^(5/2) + 40*a*b^2*tan(d*x + c)^(3/2) + 10*sqrt(2)* 
(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d 
*x + c)))) + 10*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2 
)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 5*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - 
 b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 5*sqrt(2)*(a^3 
+ 3*a^2*b - 3*a*b^2 - b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) 
+ 1) + 40*(3*a^2*b - b^3)*sqrt(tan(d*x + c)))/d
 

Giac [F(-2)]

Exception generated. \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3,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 = 3.42 (sec) , antiderivative size = 1742, normalized size of antiderivative = 7.99 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^(1/2)*(a + b*tan(c + d*x))^3,x)
 

Output:

(2*b^3*tan(c + d*x)^(5/2))/(5*d) - atan((((8*(4*b^3*d^2 - 12*a^2*b*d^2)*(( 
6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*1 
5i)/(4*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 
 15*a^4*b^2))/d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 2 
0*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)*1i - ((8*(4*b^3*d^2 - 12*a^2*b*d^2 
)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b 
^2*15i)/(4*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b 
^4 - 15*a^4*b^2))/d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i 
 - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)*1i)/(((8*(4*b^3*d^2 - 12*a^2*b 
*d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a 
^4*b^2*15i)/(4*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a 
^2*b^4 - 15*a^4*b^2))/d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4 
*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2) - (16*(3*a*b^8 - a^9 + 8*a 
^3*b^6 + 6*a^5*b^4))/d^3 + ((8*(4*b^3*d^2 - 12*a^2*b*d^2)*((6*a*b^5 + 6*a^ 
5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^( 
1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/ 
d^2)*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20*a^3*b^3 + a^ 
4*b^2*15i)/(4*d^2))^(1/2)))*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^ 
4*15i - 20*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)*2i - atan((((8*(4*b^3*d^2 
 - 12*a^2*b*d^2)*((6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*15i - ...
 

Reduce [F]

\[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx=\frac {2 \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2} b^{3}+30 \sqrt {\tan \left (d x +c \right )}\, a^{2} b -10 \sqrt {\tan \left (d x +c \right )}\, b^{3}-15 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) a^{2} b d +5 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) b^{3} d +5 \left (\int \sqrt {\tan \left (d x +c \right )}d x \right ) a^{3} d +15 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \right ) a \,b^{2} d}{5 d} \] Input:

int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3,x)
 

Output:

(2*sqrt(tan(c + d*x))*tan(c + d*x)**2*b**3 + 30*sqrt(tan(c + d*x))*a**2*b 
- 10*sqrt(tan(c + d*x))*b**3 - 15*int(sqrt(tan(c + d*x))/tan(c + d*x),x)*a 
**2*b*d + 5*int(sqrt(tan(c + d*x))/tan(c + d*x),x)*b**3*d + 5*int(sqrt(tan 
(c + d*x)),x)*a**3*d + 15*int(sqrt(tan(c + d*x))*tan(c + d*x)**2,x)*a*b**2 
*d)/(5*d)