\(\int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) [624]

Optimal result

Integrand size = 25, antiderivative size = 277 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=-\frac {i (i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {5 a \left (a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 \sqrt {b} d}-\frac {i (i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{12 d}+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d} \] Output:

-I*(I*a-b)^(5/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1 
/2))/d+5/8*a*(a^2-8*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c)) 
^(1/2))/b^(1/2)/d-I*(I*a+b)^(5/2)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/( 
a+b*tan(d*x+c))^(1/2))/d+1/8*(11*a^2-8*b^2)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+ 
c))^(1/2)/d+13/12*a*b*tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1/2)/d+1/3*b^2*ta 
n(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 2.10 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.08 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {-24 (-1)^{3/4} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+24 (-1)^{3/4} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 \left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+26 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+8 b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+\frac {15 a^{3/2} \left (a^2-8 b^2\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{24 d} \] Input:

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

Output:

(-24*(-1)^(3/4)*(-a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Ta 
n[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 24*(-1)^(3/4)*(a + I*b)^(5/2)*Arc 
Tan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]] 
] + 3*(11*a^2 - 8*b^2)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + 26*a* 
b*Tan[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]] + 8*b^2*Tan[c + d*x]^(5/2)*S 
qrt[a + b*Tan[c + d*x]] + (15*a^(3/2)*(a^2 - 8*b^2)*ArcSinh[(Sqrt[b]*Sqrt[ 
Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b* 
Tan[c + d*x]]))/(24*d)
 

Rubi [A] (verified)

Time = 1.85 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4130, 27, 3042, 4138, 2035, 2257, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

\(\displaystyle \frac {1}{3} \int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (13 a b^2 \tan ^2(c+d x)+6 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (6 a^2-5 b^2\right )\right )}{2 \sqrt {a+b \tan (c+d x)}}dx+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (13 a b^2 \tan ^2(c+d x)+6 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (6 a^2-5 b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \int \frac {\tan (c+d x)^{3/2} \left (13 a b^2 \tan (c+d x)^2+6 b \left (3 a^2-b^2\right ) \tan (c+d x)+a \left (6 a^2-5 b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {1}{6} \left (\frac {\int -\frac {3 \sqrt {\tan (c+d x)} \left (13 a^2 b^2-\left (11 a^2-8 b^2\right ) \tan ^2(c+d x) b^2-8 a \left (a^2-3 b^2\right ) \tan (c+d x) b\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{2 b}+\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {3 \int \frac {\sqrt {\tan (c+d x)} \left (13 a^2 b^2-\left (11 a^2-8 b^2\right ) \tan ^2(c+d x) b^2-8 a \left (a^2-3 b^2\right ) \tan (c+d x) b\right )}{\sqrt {a+b \tan (c+d x)}}dx}{4 b}\right )+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {3 \int \frac {\sqrt {\tan (c+d x)} \left (13 a^2 b^2-\left (11 a^2-8 b^2\right ) \tan (c+d x)^2 b^2-8 a \left (a^2-3 b^2\right ) \tan (c+d x) b\right )}{\sqrt {a+b \tan (c+d x)}}dx}{4 b}\right )+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {1}{6} \left (\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\int \frac {16 \left (3 a^2-b^2\right ) \tan (c+d x) b^3-5 a \left (a^2-8 b^2\right ) \tan ^2(c+d x) b^2+a \left (11 a^2-8 b^2\right ) b^2}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b}-\frac {b \left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )}{4 b}\right )+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\int \frac {16 \left (3 a^2-b^2\right ) \tan (c+d x) b^3-5 a \left (a^2-8 b^2\right ) \tan ^2(c+d x) b^2+a \left (11 a^2-8 b^2\right ) b^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}-\frac {b \left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )}{4 b}\right )+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\int \frac {16 \left (3 a^2-b^2\right ) \tan (c+d x) b^3-5 a \left (a^2-8 b^2\right ) \tan (c+d x)^2 b^2+a \left (11 a^2-8 b^2\right ) b^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}-\frac {b \left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )}{4 b}\right )+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4138

\(\displaystyle \frac {1}{6} \left (\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\int \frac {16 \left (3 a^2-b^2\right ) \tan (c+d x) b^3-5 a \left (a^2-8 b^2\right ) \tan ^2(c+d x) b^2+a \left (11 a^2-8 b^2\right ) b^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 b d}-\frac {b \left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )}{4 b}\right )+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {1}{6} \left (\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\int \frac {16 \left (3 a^2-b^2\right ) \tan (c+d x) b^3-5 a \left (a^2-8 b^2\right ) \tan ^2(c+d x) b^2+a \left (11 a^2-8 b^2\right ) b^2}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d}-\frac {b \left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )}{4 b}\right )+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 2257

\(\displaystyle \frac {1}{6} \left (\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\int \left (\frac {16 \left (\left (3 a^2-b^2\right ) \tan (c+d x) b^3+a \left (a^2-3 b^2\right ) b^2\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}-\frac {5 a b^2 \left (a^2-8 b^2\right )}{\sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}}{b d}-\frac {b \left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )}{4 b}\right )+\frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{3 d}+\frac {1}{6} \left (\frac {13 a b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {3 \left (-\frac {b \left (11 a^2-8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {-5 a b^{3/2} \left (a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+8 i b^2 (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+8 i b^2 (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b d}\right )}{4 b}\right )\)

Input:

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

Output:

(b^2*Tan[c + d*x]^(5/2)*Sqrt[a + b*Tan[c + d*x]])/(3*d) + ((13*a*b*Tan[c + 
 d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(2*d) - (3*(((8*I)*(I*a - b)^(5/2)*b 
^2*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - 5 
*a*b^(3/2)*(a^2 - 8*b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*T 
an[c + d*x]]] + (8*I)*b^2*(I*a + b)^(5/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[ 
c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(b*d) - (b*(11*a^2 - 8*b^2)*Sqrt[Tan 
[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d))/(4*b))/6
 

Defintions of rubi rules used

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

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

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 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 4138

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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 0.59 (sec) , antiderivative size = 1311074, normalized size of antiderivative = 4733.12

\[\text {output too large to display}\]

Input:

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

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4740 vs. \(2 (221) = 442\).

Time = 1.18 (sec) , antiderivative size = 9482, normalized size of antiderivative = 34.23 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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