\(\int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [574]

Optimal result

Integrand size = 23, antiderivative size = 193 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(a-b) \left (a^2+4 a b+b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {16 a^2 b}{3 d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)} \] 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-16/3*a^2*b/d/tan(d*x+c)^(1/2)-2/3*a^2*(a+b*tan(d*x+c))/d/tan(d*x+c) 
^(3/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.47 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-(a+i b)^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-i \tan (c+d x)\right )-(a-i b)^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (c+d x)\right )-6 b^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)} \] Input:

Integrate[(a + b*Tan[c + d*x])^3/Tan[c + d*x]^(5/2),x]
 

Output:

(-((a + I*b)^3*Hypergeometric2F1[-3/2, 1, -1/2, (-I)*Tan[c + d*x]]) - (a - 
 I*b)^3*Hypergeometric2F1[-3/2, 1, -1/2, I*Tan[c + d*x]] - 6*b^2*(a + b*Ta 
n[c + d*x]))/(3*d*Tan[c + d*x]^(3/2))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.15, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {3042, 4048, 27, 3042, 4111, 27, 3042, 4017, 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.

\(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\tan (c+d x)^{5/2}}dx\)

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4111

\(\displaystyle \frac {1}{3} \left (\int -\frac {3 \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-3 \int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-3 \int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {1}{3} \left (-\frac {6 \int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {1}{3} \left (-\frac {6 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {1}{3} \left (-\frac {6 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{3} \left (-\frac {6 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{3} \left (-\frac {6 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {1}{3} \left (-\frac {6 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (-\frac {6 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-\frac {6 \left (\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{3} \left (-\frac {6 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}-\frac {16 a^2 b}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)}\)

Input:

Int[(a + b*Tan[c + d*x])^3/Tan[c + d*x]^(5/2),x]
 

Output:

((-6*(((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]))/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 - (16*a^2*b)/(d*Sqrt[Tan[c + d*x]]))/3 - (2*a^2*(a + b*Tan[c + 
 d*x]))/(3*d*Tan[c + d*x]^(3/2))
 

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 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 4048

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

rule 4111

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {\frac {\left (-a^{3}+3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2 a^{3}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {6 a^{2} b}{\sqrt {\tan \left (d x +c \right )}}}{d}\)	\(225\)
default	\(\frac {\frac {\left (-a^{3}+3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2 a^{3}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {6 a^{2} b}{\sqrt {\tan \left (d x +c \right )}}}{d}\)	\(225\)
parts	\(\frac {a^{3} \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {b^{3} \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}+\frac {3 a \,b^{2} \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}+\frac {3 a^{2} b \left (-\frac {2}{\sqrt {\tan \left (d x +c \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(393\)

Input:

int((a+b*tan(d*x+c))^3/tan(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(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*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(-3*a^2*b+b^3)*2^(1/2)*(l 
n((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)))-2/3*a^3/tan(d*x+c)^(3/2)-6*a^2*b/tan(d*x+c)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 941 vs. \(2 (167) = 334\).

Time = 0.10 (sec) , antiderivative size = 941, normalized size of antiderivative = 4.88 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^3/tan(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

1/6*(6*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))*tan(d*x + c)^2 + 6*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))*tan(d*x + c)^2 - 3*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))*tan(d*x + c)^2 + 3*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...
 

Sympy [F]

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

integrate((a+b*tan(d*x+c))**3/tan(d*x+c)**(5/2),x)
 

Output:

Integral((a + b*tan(c + d*x))**3/tan(c + d*x)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.11 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {6 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, {\left (9 \, a^{2} b \tan \left (d x + c\right ) + a^{3}\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \] Input:

integrate((a+b*tan(d*x+c))^3/tan(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

-1/12*(6*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)))) + 6*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*ar 
ctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + 3*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) - 
 3*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) + 8*(9*a^2*b*tan(d*x + c) + a^3)/tan(d*x + c)^(3/2))/ 
d
 

Giac [F]

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

integrate((a+b*tan(d*x+c))^3/tan(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 2.24 (sec) , antiderivative size = 1752, normalized size of antiderivative = 9.08 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

int((a + b*tan(c + d*x))^3/tan(c + d*x)^(5/2),x)
 

Output:

2*atanh((32*a^6*d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2 
) - (3*a*b^5)/(2*d^2) - (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) + (5*a^3 
*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2))/(a^9*d^2*16i + 16*b^9*d^2 + a*b^ 
8*d^2*48i + 48*a^8*b*d^2 - 288*a^2*b^7*d^2 - a^3*b^6*d^2*736i + 960*a^4*b^ 
5*d^2 + a^5*b^4*d^2*960i - 736*a^6*b^3*d^2 - a^7*b^2*d^2*288i) - (32*b^6*d 
^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) - (3*a*b^5)/(2* 
d^2) - (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) + (5*a^3*b^3)/d^2 + (a^4* 
b^2*15i)/(4*d^2))^(1/2))/(a^9*d^2*16i + 16*b^9*d^2 + a*b^8*d^2*48i + 48*a^ 
8*b*d^2 - 288*a^2*b^7*d^2 - a^3*b^6*d^2*736i + 960*a^4*b^5*d^2 + a^5*b^4*d 
^2*960i - 736*a^6*b^3*d^2 - a^7*b^2*d^2*288i) + (480*a^2*b^4*d^3*tan(c + d 
*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) - (3*a*b^5)/(2*d^2) - (3*a^ 
5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) + (5*a^3*b^3)/d^2 + (a^4*b^2*15i)/(4* 
d^2))^(1/2))/(a^9*d^2*16i + 16*b^9*d^2 + a*b^8*d^2*48i + 48*a^8*b*d^2 - 28 
8*a^2*b^7*d^2 - a^3*b^6*d^2*736i + 960*a^4*b^5*d^2 + a^5*b^4*d^2*960i - 73 
6*a^6*b^3*d^2 - a^7*b^2*d^2*288i) - (480*a^4*b^2*d^3*tan(c + d*x)^(1/2)*(( 
b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) - (3*a*b^5)/(2*d^2) - (3*a^5*b)/(2*d^2) 
 - (a^2*b^4*15i)/(4*d^2) + (5*a^3*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2)) 
/(a^9*d^2*16i + 16*b^9*d^2 + a*b^8*d^2*48i + 48*a^8*b*d^2 - 288*a^2*b^7*d^ 
2 - a^3*b^6*d^2*736i + 960*a^4*b^5*d^2 + a^5*b^4*d^2*960i - 736*a^6*b^3*d^ 
2 - a^7*b^2*d^2*288i))*(-(6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4...
 

Reduce [F]

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

int((a+b*tan(d*x+c))^3/tan(d*x+c)^(5/2),x)
 

Output:

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