\(\int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx\) [1235]

Optimal result

Integrand size = 27, antiderivative size = 256 \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=\frac {(i a+b)^3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(i a-b)^3 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{35 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f} \] Output:

(I*a+b)^3*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f-(I 
*a-b)^3*(c+I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f+2*(a 
^3*d+3*a^2*b*c-3*a*b^2*d-b^3*c)*(c+d*tan(f*x+e))^(1/2)/f+2/3*b*(3*a^2-b^2) 
*(c+d*tan(f*x+e))^(3/2)/f-4/35*b^2*(-8*a*d+b*c)*(c+d*tan(f*x+e))^(5/2)/d^2 
/f+2/7*b^2*(a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(5/2)/d/f
 

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.96 \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=\frac {\frac {4 b^2 (-b c+8 a d) (c+d \tan (e+f x))^{5/2}}{d}+10 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}-\frac {35}{3} (i a+b)^3 d \left (-3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )+\frac {35}{3} (i a-b)^3 d \left (-3 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )}{35 d f} \] Input:

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

Output:

((4*b^2*(-(b*c) + 8*a*d)*(c + d*Tan[e + f*x])^(5/2))/d + 10*b^2*(a + b*Tan 
[e + f*x])*(c + d*Tan[e + f*x])^(5/2) - (35*(I*a + b)^3*d*(-3*(c - I*d)^(3 
/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + Sqrt[c + d*Tan[e + f 
*x]]*(4*c - (3*I)*d + d*Tan[e + f*x])))/3 + (35*(I*a - b)^3*d*(-3*(c + I*d 
)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c + d*Tan[e 
 + f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])))/3)/(35*d*f)
 

Rubi [A] (warning: unable to verify)

Time = 1.66 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.593, Rules used = {3042, 4049, 27, 3042, 4113, 3042, 4011, 3042, 4011, 3042, 4022, 3042, 4020, 25, 73, 221}

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.

\(\displaystyle \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2 \int -\frac {1}{2} (c+d \tan (e+f x))^{3/2} \left (-7 d a^3+5 b^2 d a+2 b^2 (b c-8 a d) \tan ^2(e+f x)+2 b^3 c-7 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )dx}{7 d}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} \left (-7 d a^3+5 b^2 d a+2 b^2 (b c-8 a d) \tan ^2(e+f x)+2 b^3 c-7 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )dx}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} \left (-7 d a^3+5 b^2 d a+2 b^2 (b c-8 a d) \tan (e+f x)^2+2 b^3 c-7 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )dx}{7 d}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} \left (-7 a \left (a^2-3 b^2\right ) d-7 b \left (3 a^2-b^2\right ) \tan (e+f x) d\right )dx+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} \left (-7 a \left (a^2-3 b^2\right ) d-7 b \left (3 a^2-b^2\right ) \tan (e+f x) d\right )dx+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (-7 d \left (c a^3-3 b d a^2-3 b^2 c a+b^3 d\right )-7 d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )dx-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (-7 d \left (c a^3-3 b d a^2-3 b^2 c a+b^3 d\right )-7 d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )dx-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int \frac {7 d \left (-\left (\left (c^2-d^2\right ) a^3\right )+6 b c d a^2+3 b^2 \left (c^2-d^2\right ) a-2 b^3 c d\right )-7 d \left (2 c d a^3+3 b \left (c^2-d^2\right ) a^2-6 b^2 c d a-b^3 \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\int \frac {7 d \left (-\left (\left (c^2-d^2\right ) a^3\right )+6 b c d a^2+3 b^2 \left (c^2-d^2\right ) a-2 b^3 c d\right )-7 d \left (2 c d a^3+3 b \left (c^2-d^2\right ) a^2-6 b^2 c d a-b^3 \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {7}{2} d (a-i b)^3 (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {7}{2} d (a+i b)^3 (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {7}{2} d (a-i b)^3 (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {7}{2} d (a+i b)^3 (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {7 i d (a-i b)^3 (c-i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {7 i d (a+i b)^3 (c+i d)^2 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {\frac {7 i d (a-i b)^3 (c-i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {7 i d (a+i b)^3 (c+i d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {7 (a-i b)^3 (c-i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {7 (a+i b)^3 (c+i d)^2 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}{7 d f}-\frac {-\frac {14 b d \left (3 a^2-b^2\right ) (c+d \tan (e+f x))^{3/2}}{3 f}-\frac {14 d \left (a^3 d+3 a^2 b c-3 a b^2 d-b^3 c\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {7 d (a-i b)^3 (c-i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}-\frac {7 d (a+i b)^3 (c+i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {4 b^2 (b c-8 a d) (c+d \tan (e+f x))^{5/2}}{5 d f}}{7 d}\)

Input:

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

Output:

(2*b^2*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2))/(7*d*f) - ((-7*(a 
- I*b)^3*(c - I*d)^(3/2)*d*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f - (7*(a + 
 I*b)^3*(c + I*d)^(3/2)*d*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f - (14*d*(3 
*a^2*b*c - b^3*c + a^3*d - 3*a*b^2*d)*Sqrt[c + d*Tan[e + f*x]])/f - (14*b* 
(3*a^2 - b^2)*d*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (4*b^2*(b*c - 8*a*d)*( 
c + d*Tan[e + f*x])^(5/2))/(5*d*f))/(7*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 221

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

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 4020

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

rule 4022

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

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3375\) vs. \(2(224)=448\).

Time = 0.48 (sec) , antiderivative size = 3376, normalized size of antiderivative = 13.19

method	result	size
parts	\(\text {Expression too large to display}\)	\(3376\)
derivativedivides	\(\text {Expression too large to display}\)	\(3498\)
default	\(\text {Expression too large to display}\)	\(3498\)

Input:

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

Output:

a^3*(2/f*d*(c+d*tan(f*x+e))^(1/2)-1/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e 
))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2) 
+2*c)^(1/2)*(c^2+d^2)^(1/2)*c+1/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^( 
1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c 
)^(1/2)*c^2-1/4/f*d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^ 
(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+2/f*d/(2*( 
c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1 
/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-1/f*d/(2*(c^2+d^2)^(1/2)- 
2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2)) 
/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)+1/4/f/d*ln((c+d*tan(f*x+e) 
)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*( 
c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c-1/4/f/d*ln((c+d*tan(f*x+e))^(1 
/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+ 
d^2)^(1/2)+2*c)^(1/2)*c^2+1/4/f*d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^( 
1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1 
/2)-2/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1 
/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+1/f*d/(2*(c 
^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan( 
f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2))+b^3*(2/7/f/ 
d^2*(c+d*tan(f*x+e))^(7/2)-2/5/f/d^2*c*(c+d*tan(f*x+e))^(5/2)-2/3/f*(c+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7872 vs. \(2 (218) = 436\).

Time = 2.57 (sec) , antiderivative size = 7872, normalized size of antiderivative = 30.75 \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

Timed out. \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,24,9]%%%}+%%%{10,[0,22,9]%%%}+%%%{45,[0,20,9]%%%} 
+%%%{120,
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 78.41 (sec) , antiderivative size = 22352, normalized size of antiderivative = 87.31 \[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

(c + d*tan(e + f*x))^(1/2)*(((6*b^3*c - 6*a*b^2*d)/(d^2*f) - (4*b^3*c)/(d^ 
2*f))*(c^2 + d^2) - 2*c*(2*c*((6*b^3*c - 6*a*b^2*d)/(d^2*f) - (4*b^3*c)/(d 
^2*f)) - (6*b*(a*d - b*c)^2)/(d^2*f) + (2*b^3*(c^2 + d^2))/(d^2*f)) + (2*( 
a*d - b*c)^3)/(d^2*f)) - atan(((((8*(4*a^3*d^5*f^2 - 12*a*b^2*d^5*f^2 - 4* 
b^3*c*d^4*f^2 + 4*a^3*c^2*d^3*f^2 - 4*b^3*c^3*d^2*f^2 + 12*a^2*b*c*d^4*f^2 
 - 12*a*b^2*c^2*d^3*f^2 + 12*a^2*b*c^3*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan 
(e + f*x))^(1/2)*(-(((8*a^6*c^3*f^2 - 8*b^6*c^3*f^2 + 48*a*b^5*d^3*f^2 + 4 
8*a^5*b*d^3*f^2 - 24*a^6*c*d^2*f^2 + 24*b^6*c*d^2*f^2 + 120*a^2*b^4*c^3*f^ 
2 - 120*a^4*b^2*c^3*f^2 - 160*a^3*b^3*d^3*f^2 - 144*a*b^5*c^2*d*f^2 - 144* 
a^5*b*c^2*d*f^2 - 360*a^2*b^4*c*d^2*f^2 + 480*a^3*b^3*c^2*d*f^2 + 360*a^4* 
b^2*c*d^2*f^2)^2/64 - f^4*(a^12*c^6 + a^12*d^6 + b^12*c^6 + b^12*d^6 + 6*a 
^2*b^10*c^6 + 15*a^4*b^8*c^6 + 20*a^6*b^6*c^6 + 15*a^8*b^4*c^6 + 6*a^10*b^ 
2*c^6 + 6*a^2*b^10*d^6 + 15*a^4*b^8*d^6 + 20*a^6*b^6*d^6 + 15*a^8*b^4*d^6 
+ 6*a^10*b^2*d^6 + 3*a^12*c^2*d^4 + 3*a^12*c^4*d^2 + 3*b^12*c^2*d^4 + 3*b^ 
12*c^4*d^2 + 18*a^2*b^10*c^2*d^4 + 18*a^2*b^10*c^4*d^2 + 45*a^4*b^8*c^2*d^ 
4 + 45*a^4*b^8*c^4*d^2 + 60*a^6*b^6*c^2*d^4 + 60*a^6*b^6*c^4*d^2 + 45*a^8* 
b^4*c^2*d^4 + 45*a^8*b^4*c^4*d^2 + 18*a^10*b^2*c^2*d^4 + 18*a^10*b^2*c^4*d 
^2))^(1/2) + a^6*c^3*f^2 - b^6*c^3*f^2 + 6*a*b^5*d^3*f^2 + 6*a^5*b*d^3*f^2 
 - 3*a^6*c*d^2*f^2 + 3*b^6*c*d^2*f^2 + 15*a^2*b^4*c^3*f^2 - 15*a^4*b^2*c^3 
*f^2 - 20*a^3*b^3*d^3*f^2 - 18*a*b^5*c^2*d*f^2 - 18*a^5*b*c^2*d*f^2 - 4...
 

Reduce [F]

\[ \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x))^{3/2} \, dx=\left (\int \sqrt {d \tan \left (f x +e \right )+c}d x \right ) a^{3} c +\left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{4}d x \right ) b^{3} d +3 \left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{3}d x \right ) a \,b^{2} d +\left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{3}d x \right ) b^{3} c +3 \left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}d x \right ) a^{2} b d +3 \left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}d x \right ) a \,b^{2} c +\left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )d x \right ) a^{3} d +3 \left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )d x \right ) a^{2} b c \] Input:

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

Output:

int(sqrt(tan(e + f*x)*d + c),x)*a**3*c + int(sqrt(tan(e + f*x)*d + c)*tan( 
e + f*x)**4,x)*b**3*d + 3*int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**3,x)* 
a*b**2*d + int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**3,x)*b**3*c + 3*int( 
sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2,x)*a**2*b*d + 3*int(sqrt(tan(e + 
f*x)*d + c)*tan(e + f*x)**2,x)*a*b**2*c + int(sqrt(tan(e + f*x)*d + c)*tan 
(e + f*x),x)*a**3*d + 3*int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x),x)*a**2* 
b*c