\(\int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\) [615]

Optimal result

Integrand size = 25, antiderivative size = 280 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\frac {i (i a-b)^{3/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {a \left (a^2+24 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 b^{3/2} d}-\frac {i (i a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\left (a^2+8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 b d}-\frac {a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{12 b d}+\frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d} \] Output:

I*(I*a-b)^(3/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/ 
2))/d-1/8*a*(a^2+24*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c)) 
^(1/2))/b^(3/2)/d-I*(I*a+b)^(3/2)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/( 
a+b*tan(d*x+c))^(1/2))/d-1/8*(a^2+8*b^2)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c)) 
^(1/2)/b/d-1/12*a*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2)/b/d+1/3*tan(d*x+ 
c)^(1/2)*(a+b*tan(d*x+c))^(5/2)/b/d
 

Mathematica [A] (verified)

Time = 2.57 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.12 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\frac {-3 a^{3/2} \left (a^2+24 b^2\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}+\sqrt {b} \left (24 \sqrt [4]{-1} (-a+i b)^{3/2} b \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}-24 \sqrt [4]{-1} (a+i b)^{3/2} b \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {\tan (c+d x)} \left (3 \left (a^3-8 a b^2\right )+b \left (17 a^2-24 b^2\right ) \tan (c+d x)+22 a b^2 \tan ^2(c+d x)+8 b^3 \tan ^3(c+d x)\right )\right )}{24 b^{3/2} d \sqrt {a+b \tan (c+d x)}} \] Input:

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

Output:

(-3*a^(3/2)*(a^2 + 24*b^2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*S 
qrt[1 + (b*Tan[c + d*x])/a] + Sqrt[b]*(24*(-1)^(1/4)*(-a + I*b)^(3/2)*b*Ar 
cTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x 
]]]*Sqrt[a + b*Tan[c + d*x]] - 24*(-1)^(1/4)*(a + I*b)^(3/2)*b*ArcTan[((-1 
)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a 
 + b*Tan[c + d*x]] + Sqrt[Tan[c + d*x]]*(3*(a^3 - 8*a*b^2) + b*(17*a^2 - 2 
4*b^2)*Tan[c + d*x] + 22*a*b^2*Tan[c + d*x]^2 + 8*b^3*Tan[c + d*x]^3)))/(2 
4*b^(3/2)*d*Sqrt[a + b*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.56 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\int \frac {a \left (a^2+24 b^2\right ) \tan ^2(c+d x)+16 b \left (a^2-b^2\right ) \tan (c+d x)+a \left (a^2-8 b^2\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {\left (a^2+8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {1}{2} \int \frac {a \left (a^2+24 b^2\right ) \tan ^2(c+d x)+16 b \left (a^2-b^2\right ) \tan (c+d x)+a \left (a^2-8 b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {\left (a^2+8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {1}{2} \int \frac {a \left (a^2+24 b^2\right ) \tan (c+d x)^2+16 b \left (a^2-b^2\right ) \tan (c+d x)+a \left (a^2-8 b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {\left (a^2+8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\)

\(\Big \downarrow \) 4138

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {\int \frac {a \left (a^2+24 b^2\right ) \tan ^2(c+d x)+16 b \left (a^2-b^2\right ) \tan (c+d x)+a \left (a^2-8 b^2\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}+\frac {\left (a^2+8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {\int \frac {a \left (a^2+24 b^2\right ) \tan ^2(c+d x)+16 b \left (a^2-b^2\right ) \tan (c+d x)+a \left (a^2-8 b^2\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}+\frac {\left (a^2+8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\)

\(\Big \downarrow \) 2257

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {\int \left (\frac {a \left (a^2+24 b^2\right )}{\sqrt {a+b \tan (c+d x)}}-\frac {16 \left (2 a b^2-b \left (a^2-b^2\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}+\frac {\left (a^2+8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {3}{4} \left (\frac {\left (a^2+8 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\frac {a \left (a^2+24 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-8 i b (-b+i a)^{3/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+8 i b (b+i a)^{3/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )}{6 b}\)

Input:

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

Output:

(Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(5/2))/(3*b*d) - ((a*Sqrt[Tan[c + 
 d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*d) + (3*(((-8*I)*(I*a - b)^(3/2)*b*A 
rcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + (a*(a 
^2 + 24*b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]] 
])/Sqrt[b] + (8*I)*b*(I*a + b)^(3/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d 
*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + ((a^2 + 8*b^2)*Sqrt[Tan[c + d*x]]*Sqr 
t[a + b*Tan[c + d*x]])/d))/4)/(6*b)
 

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 = 0.69 (sec) , antiderivative size = 1346642, normalized size of antiderivative = 4809.44

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

Input:

int(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(3/2),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. 3676 vs. \(2 (224) = 448\).

Time = 1.03 (sec) , antiderivative size = 7354, normalized size of antiderivative = 26.26 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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