\(\int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) [548]

Optimal result

Integrand size = 23, antiderivative size = 172 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}-\frac {2 a^2 \tan (c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {4 a^2 \left (a^2+4 b^2\right )}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 1.01 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.28 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {-\frac {4 a}{b}-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )}{i a+b}+\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )}{i a-b}-6 \tan (c+d x)+\frac {3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))}{i a+b}+\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))}{a+i b}}{3 b d (a+b \tan (c+d x))^{3/2}} \] Input:

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

Output:

((-4*a)/b - (a*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a - 
I*b)])/(I*a + b) + (a*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x] 
)/(a + I*b)])/(I*a - b) - 6*Tan[c + d*x] + (3*Hypergeometric2F1[-1/2, 1, 1 
/2, (a + b*Tan[c + d*x])/(a - I*b)]*(a + b*Tan[c + d*x]))/(I*a + b) + ((3* 
I)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)]*(a + b* 
Tan[c + d*x]))/(a + I*b))/(3*b*d*(a + b*Tan[c + d*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.99 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.22, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 4048, 27, 3042, 4111, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4111

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-2*a^2*Tan[c + d*x])/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ((- 
3*(((-I)*(a + I*b)^2*b*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]* 
d) + (I*(a - I*b)^2*b*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d 
)))/(a^2 + b^2) - (4*a^2*(a^2 + 4*b^2))/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c 
+ d*x]]))/(3*b*(a^2 + b^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 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 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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2159\) vs. \(2(148)=296\).

Time = 0.26 (sec) , antiderivative size = 2160, normalized size of antiderivative = 12.56

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2160\)
default	\(\text {Expression too large to display}\)	\(2160\)

Input:

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

Output:

-6/d*a^2/(a^2+b^2)^2/(a+b*tan(d*x+c))^(1/2)+1/2/d*b^2/(a^2+b^2)^3*ln(b*tan 
(d*x+c)+a-(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^( 
1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+1/2/d*b^2/(a^2+b^2)^(7/2)*ln(b*tan(d 
*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/ 
2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-1/2/d*b^2/(a^2+b^2)^(7/2)*ln(b*tan(d 
*x+c)+a-(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/ 
2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-6/d*b^2/(a^2+b^2)^(7/2)/(2*(a^2+b^2) 
^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)-(2*(a^2+b^2)^(1/2)+2*a) 
^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3-5/d*b^4/(a^2+b^2)^(7/2)/(2*(a^2 
+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)-(2*(a^2+b^2)^(1/2) 
+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+2/d*b^2/(a^2+b^2)^(5/2)/(2*( 
a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)-(2*(a^2+b^2)^(1 
/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-6/d*b^2/(a^2+b^2)^(7/2)/( 
2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2) 
^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3-5/d*b^4/(a^2+b^2)^(7 
/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2 
+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+2/d*b^2/(a^2+b^2) 
^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*( 
a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-1/2/d*b^2/(a^2 
+b^2)^3*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3324 vs. \(2 (144) = 288\).

Time = 0.17 (sec) , antiderivative size = 3324, normalized size of antiderivative = 19.33 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^3/(a+b*tan(d*x+c))^(5/2),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,[6,14,5]%%%}+%%%{6,[6,12,5]%%%}+%%%{15,[6,10,5]%%%}+ 
%%%{20,[6
 

Mupad [B] (verification not implemented)

Time = 7.13 (sec) , antiderivative size = 4638, normalized size of antiderivative = 26.97 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^2*5i - 5*a^4*b*d^2 + 10*a^2*b 
^3*d^2 - a^3*b^2*d^2*10i)))^(1/2)*((1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^ 
2*5i - 5*a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i)))^(1/2)*(96*a*b^20* 
d^4 + 736*a^3*b^18*d^4 + 2432*a^5*b^16*d^4 + 4480*a^7*b^14*d^4 + 4928*a^9* 
b^12*d^4 + 3136*a^11*b^10*d^4 + 896*a^13*b^8*d^4 - 128*a^15*b^6*d^4 - 160* 
a^17*b^4*d^4 - 32*a^19*b^2*d^4 + (1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^2* 
5i - 5*a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i)))^(1/2)*(a + b*tan(c 
+ d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18*d^5 + 7680 
*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 13440*a^13*b^10 
*d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 + 64*a^21* 
b^2*d^5)) + (a + b*tan(c + d*x))^(1/2)*(320*a^4*b^14*d^3 - 16*b^18*d^3 + 1 
024*a^6*b^12*d^3 + 1440*a^8*b^10*d^3 + 1024*a^10*b^8*d^3 + 320*a^12*b^6*d^ 
3 - 16*a^16*b^2*d^3))*1i - (1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^2*5i - 5 
*a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i)))^(1/2)*((1i/(4*(a^5*d^2*1i 
 - b^5*d^2 + a*b^4*d^2*5i - 5*a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i 
)))^(1/2)*(96*a*b^20*d^4 + 736*a^3*b^18*d^4 + 2432*a^5*b^16*d^4 + 4480*a^7 
*b^14*d^4 + 4928*a^9*b^12*d^4 + 3136*a^11*b^10*d^4 + 896*a^13*b^8*d^4 - 12 
8*a^15*b^6*d^4 - 160*a^17*b^4*d^4 - 32*a^19*b^2*d^4 - (1i/(4*(a^5*d^2*1i - 
 b^5*d^2 + a*b^4*d^2*5i - 5*a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i)) 
)^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 + ...
 

Reduce [F]

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

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

Output:

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