\(\int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) [547]

Optimal result

Integrand size = 23, antiderivative size = 226 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}-\frac {2 a^2 \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {4 a^3 \left (2 a^2+5 b^2\right )}{3 b^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {2 \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d} \] Output:

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

Mathematica [A] (verified)

Time = 2.11 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.36 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\frac {3 \left (-b^2\right )^{3/2} \left (a^2-b^2+2 a \sqrt {-b^2}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\left (a^2+b^2\right )^2 \sqrt {a-\sqrt {-b^2}}}+\frac {3 \left (-b^2\right )^{3/2} \left (-a^2+b^2+2 a \sqrt {-b^2}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\left (a^2+b^2\right )^2 \sqrt {a+\sqrt {-b^2}}}+\frac {2 a^2 \left (8 a^2+9 b^2\right )}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {24 a b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}}+\frac {6 b^2 \tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}}-\frac {12 a b^4}{\left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}}{3 b^3 d} \] Input:

Integrate[Tan[c + d*x]^4/(a + b*Tan[c + d*x])^(5/2),x]
 

Output:

((3*(-b^2)^(3/2)*(a^2 - b^2 + 2*a*Sqrt[-b^2])*ArcTanh[Sqrt[a + b*Tan[c + d 
*x]]/Sqrt[a - Sqrt[-b^2]]])/((a^2 + b^2)^2*Sqrt[a - Sqrt[-b^2]]) + (3*(-b^ 
2)^(3/2)*(-a^2 + b^2 + 2*a*Sqrt[-b^2])*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sq 
rt[a + Sqrt[-b^2]]])/((a^2 + b^2)^2*Sqrt[a + Sqrt[-b^2]]) + (2*a^2*(8*a^2 
+ 9*b^2))/((a^2 + b^2)*(a + b*Tan[c + d*x])^(3/2)) + (24*a*b*Tan[c + d*x]) 
/(a + b*Tan[c + d*x])^(3/2) + (6*b^2*Tan[c + d*x]^2)/(a + b*Tan[c + d*x])^ 
(3/2) - (12*a*b^4)/((a^2 + b^2)^2*Sqrt[a + b*Tan[c + d*x]]))/(3*b^3*d)
 

Rubi [A] (warning: unable to verify)

Time = 1.31 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.13, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 4048, 27, 3042, 4118, 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.

\(\displaystyle \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^4}{(a+b \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4118

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Tan[c + d*x]^4/(a + b*Tan[c + d*x])^(5/2),x]
 

Output:

(-2*a^2*Tan[c + d*x]^2)/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ( 
(4*a^3*(2*a^2 + 5*b^2))/(b^2*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]) + ((3 
*(a + I*b)^2*b^2*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + ( 
3*(a - I*b)^2*b^2*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + 
(2*(a^2 + b^2)*(4*a^2 + 3*b^2)*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(b*(a^2 + 
b^2)))/(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 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]
 

rule 4118

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_. 
)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(c^2*C - B*c*d + A*d^2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d^2*f*(n + 1)*(c^2 + d^2))), x] + Simp[1/(d*(c^2 
 + d^2))   Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b* 
(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d + a*C*d) 
*Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, 
c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n 
, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2372\) vs. \(2(196)=392\).

Time = 0.26 (sec) , antiderivative size = 2373, normalized size of antiderivative = 10.50

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2373\)
default	\(\text {Expression too large to display}\)	\(2373\)

Input:

int(tan(d*x+c)^4/(a+b*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

4/d*b/(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 
^4+1/d*b^3/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*ta 
n(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/b/(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^4+4/d*b/(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^4-2/d*b^3/(a^2+b^2)^3/(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-1/4/d/b/(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^5-2/d*b/(a^2+b^2)^3/(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/(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^6+1/4/d/b/(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^5-1/d/b/(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)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3383 vs. \(2 (190) = 380\).

Time = 0.15 (sec) , antiderivative size = 3383, normalized size of antiderivative = 14.97 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(tan(d*x+c)**4/(a+b*tan(d*x+c))**(5/2),x)
 

Output:

Integral(tan(c + d*x)**4/(a + b*tan(c + d*x))**(5/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate(tan(d*x + c)^4/(b*tan(d*x + c) + a)^(5/2), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 9.10 (sec) , antiderivative size = 3739, normalized size of antiderivative = 16.54 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^4/(a + b*tan(c + d*x))^(5/2),x)
 

Output:

(log(16*a*b^15*d^2 - ((-1/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2* 
5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(((-1/(a^5*d^2 - b^5*d^2*1i 
+ 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(8 
96*a^6*b^15*d^4 - 160*a^2*b^19*d^4 - 128*a^4*b^17*d^4 - 32*b^21*d^4 + 3136 
*a^8*b^13*d^4 + 4928*a^10*b^11*d^4 + 4480*a^12*b^9*d^4 + 2432*a^14*b^7*d^4 
 + 736*a^16*b^5*d^4 + 96*a^18*b^3*d^4 + ((-1/(a^5*d^2 - b^5*d^2*1i + 5*a*b 
^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(a + b*ta 
n(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))/2))/2 - (a + b*tan(c + d*x))^(1/2)*(320*a^4*b^14*d^3 - 16*b^ 
18*d^3 + 1024*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)))/2 + 96*a^3*b^13*d^2 + 240*a^5*b^11*d^2 + 
320*a^7*b^9*d^2 + 240*a^9*b^7*d^2 + 96*a^11*b^5*d^2 + 16*a^13*b^3*d^2)*(-1 
/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10 
*a^3*b^2*d^2))^(1/2))/2 - log(16*a*b^15*d^2 - ((-1/(4*(a^5*d^2 - b^5*d^2*1 
i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2)))^(1/2) 
*(896*a^6*b^15*d^4 - 160*a^2*b^19*d^4 - 128*a^4*b^17*d^4 - 32*b^21*d^4 + 3 
136*a^8*b^13*d^4 + 4928*a^10*b^11*d^4 + 4480*a^12*b^9*d^4 + 2432*a^14*b^7* 
d^4 + 736*a^16*b^5*d^4 + 96*a^18*b^3*d^4 - (-1/(4*(a^5*d^2 - b^5*d^2*1i...
 

Reduce [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \] Input:

int(tan(d*x+c)^4/(a+b*tan(d*x+c))^(5/2),x)
 

Output:

int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**4)/(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)