\(\int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) [539]

Optimal result

Integrand size = 23, antiderivative size = 165 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}-\frac {2 a^2 \tan (c+d x)}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \left (2 a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.81 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.47 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}-\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}+\frac {4 a}{b \sqrt {a+b \tan (c+d x)}}-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )}{(i a+b) \sqrt {a+b \tan (c+d x)}}+\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )}{(i a-b) \sqrt {a+b \tan (c+d x)}}+\frac {2 \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}}{b d} \] Input:

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

Output:

((I*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/Sqrt[a - I*b] - (I*Ar 
cTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/Sqrt[a + I*b] + (4*a)/(b*Sq 
rt[a + b*Tan[c + d*x]]) - (a*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c 
+ d*x])/(a - I*b)])/((I*a + b)*Sqrt[a + b*Tan[c + d*x]]) + (a*Hypergeometr 
ic2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)])/((I*a - b)*Sqrt[a + b 
*Tan[c + d*x]]) + (2*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]])/(b*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.90 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 4048, 27, 3042, 4113, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-2*a^2*Tan[c + d*x])/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]) + (((I*a 
- b)*b*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) - (b*(I*a + b 
)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + (2*(2*a^2 + b^2) 
*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(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 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. \(1767\) vs. \(2(145)=290\).

Time = 0.26 (sec) , antiderivative size = 1768, normalized size of antiderivative = 10.72

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1768\)
default	\(\text {Expression too large to display}\)	\(1768\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2011 vs. \(2 (141) = 282\).

Time = 0.12 (sec) , antiderivative size = 2011, normalized size of antiderivative = 12.19 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(((a^2*b^3 + b^5)*d*tan(d*x + c) + (a^3*b^2 + a*b^4)*d)*sqrt(((a^6 + 3 
*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 
 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)* 
d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log(-(3* 
a^2 - b^2)*sqrt(b*tan(d*x + c) + a) + (2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a* 
b^6)*d^3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8* 
b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) - (3*a^4 - 4*a^2* 
b^2 + b^4)*d)*sqrt(((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b 
^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a 
^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 3*a 
^2*b^4 + b^6)*d^2))) - ((a^2*b^3 + b^5)*d*tan(d*x + c) + (a^3*b^2 + a*b^4) 
*d)*sqrt(((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b^2 - 6*a^2 
*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6 
*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b 
^6)*d^2))*log(-(3*a^2 - b^2)*sqrt(b*tan(d*x + c) + a) - (2*(a^7 + 3*a^5*b^ 
2 + 3*a^3*b^4 + a*b^6)*d^3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6* 
a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) 
 - (3*a^4 - 4*a^2*b^2 + b^4)*d)*sqrt(((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)* 
d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 
 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b^2)/((...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 3.51 (sec) , antiderivative size = 2869, normalized size of antiderivative = 17.39 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

atan((((1i/(4*(a^3*d^2*1i + b^3*d^2 - a*b^2*d^2*3i - 3*a^2*b*d^2)))^(1/2)* 
((1i/(4*(a^3*d^2*1i + b^3*d^2 - a*b^2*d^2*3i - 3*a^2*b*d^2)))^(1/2)*(a + b 
*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 640*a^5*b^8*d^5 + 
 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5) - 32*b^12*d^4 - 96*a 
^2*b^10*d^4 - 64*a^4*b^8*d^4 + 64*a^6*b^6*d^4 + 96*a^8*b^4*d^4 + 32*a^10*b 
^2*d^4) + (a + b*tan(c + d*x))^(1/2)*(16*b^10*d^3 + 32*a^2*b^8*d^3 - 32*a^ 
6*b^4*d^3 - 16*a^8*b^2*d^3))*(1i/(4*(a^3*d^2*1i + b^3*d^2 - a*b^2*d^2*3i - 
 3*a^2*b*d^2)))^(1/2)*1i + ((1i/(4*(a^3*d^2*1i + b^3*d^2 - a*b^2*d^2*3i - 
3*a^2*b*d^2)))^(1/2)*(32*b^12*d^4 + (1i/(4*(a^3*d^2*1i + b^3*d^2 - a*b^2*d 
^2*3i - 3*a^2*b*d^2)))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 3 
20*a^3*b^10*d^5 + 640*a^5*b^8*d^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64 
*a^11*b^2*d^5) + 96*a^2*b^10*d^4 + 64*a^4*b^8*d^4 - 64*a^6*b^6*d^4 - 96*a^ 
8*b^4*d^4 - 32*a^10*b^2*d^4) + (a + b*tan(c + d*x))^(1/2)*(16*b^10*d^3 + 3 
2*a^2*b^8*d^3 - 32*a^6*b^4*d^3 - 16*a^8*b^2*d^3))*(1i/(4*(a^3*d^2*1i + b^3 
*d^2 - a*b^2*d^2*3i - 3*a^2*b*d^2)))^(1/2)*1i)/(((1i/(4*(a^3*d^2*1i + b^3* 
d^2 - a*b^2*d^2*3i - 3*a^2*b*d^2)))^(1/2)*(32*b^12*d^4 + (1i/(4*(a^3*d^2*1 
i + b^3*d^2 - a*b^2*d^2*3i - 3*a^2*b*d^2)))^(1/2)*(a + b*tan(c + d*x))^(1/ 
2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 640*a^5*b^8*d^5 + 640*a^7*b^6*d^5 + 
 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5) + 96*a^2*b^10*d^4 + 64*a^4*b^8*d^4 - 6 
4*a^6*b^6*d^4 - 96*a^8*b^4*d^4 - 32*a^10*b^2*d^4) + (a + b*tan(c + d*x)...
 

Reduce [F]

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

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

Output:

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