\(\int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) [538]

Optimal result

Integrand size = 23, antiderivative size = 226 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}-\frac {2 a^2 \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {2 a \left (8 a^2+5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}+\frac {2 \left (4 a^2+b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b^2 \left (a^2+b^2\right ) d} \] Output:

-I*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d+I*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)^2/ 
b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)-2/3*a*(8*a^2+5*b^2)*(a+b*tan(d*x+c))^ 
(1/2)/b^3/(a^2+b^2)/d+2/3*(4*a^2+b^2)*tan(d*x+c)*(a+b*tan(d*x+c))^(1/2)/b^ 
2/(a^2+b^2)/d
 

Mathematica [A] (verified)

Time = 2.15 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.12 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\frac {3 b^2 \left (1+\frac {a}{\sqrt {-b^2}}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\left (a^2+b^2\right ) \sqrt {a-\sqrt {-b^2}}}+\frac {3 \left (b^2+a \sqrt {-b^2}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\left (a^2+b^2\right ) \sqrt {a+\sqrt {-b^2}}}-\frac {2 a^2 \left (8 a^2+5 b^2\right )}{b^2 \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {8 a \tan (c+d x)}{b \sqrt {a+b \tan (c+d x)}}+\frac {2 \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)}}}{3 b d} \] Input:

Integrate[Tan[c + d*x]^4/(a + b*Tan[c + d*x])^(3/2),x]
 

Output:

((3*b^2*(1 + a/Sqrt[-b^2])*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - Sqrt[ 
-b^2]]])/((a^2 + b^2)*Sqrt[a - Sqrt[-b^2]]) + (3*(b^2 + a*Sqrt[-b^2])*ArcT 
anh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/((a^2 + b^2)*Sqrt[a + 
Sqrt[-b^2]]) - (2*a^2*(8*a^2 + 5*b^2))/(b^2*(a^2 + b^2)*Sqrt[a + b*Tan[c + 
 d*x]]) - (8*a*Tan[c + d*x])/(b*Sqrt[a + b*Tan[c + d*x]]) + (2*Tan[c + d*x 
]^2)/Sqrt[a + b*Tan[c + d*x]])/(3*b*d)
 

Rubi [A] (warning: unable to verify)

Time = 1.22 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.01, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 4048, 27, 3042, 4130, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Tan[c + d*x]^4/(a + b*Tan[c + d*x])^(3/2),x]
 

Output:

(-2*a^2*Tan[c + d*x]^2)/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]) + ((2*( 
4*a^2 + b^2)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(3*b*d) - ((-3*(a + I* 
b)*b^2*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) - (3*(a - I*b 
)*b^2*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + (2*a*(8*a^2 
+ 5*b^2)*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(3*b))/(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 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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1978\) vs. \(2(198)=396\).

Time = 0.26 (sec) , antiderivative size = 1979, normalized size of antiderivative = 8.76

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1979\)
default	\(\text {Expression too large to display}\)	\(1979\)

Input:

int(tan(d*x+c)^4/(a+b*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2034 vs. \(2 (192) = 384\).

Time = 0.12 (sec) , antiderivative size = 2034, normalized size of antiderivative = 9.00 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-1/6*(3*((a^2*b^4 + b^6)*d*tan(d*x + c) + (a^3*b^3 + a*b^5)*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 - b^3)*sqrt(b*tan(d*x + c) + a) + ((a^8 + 2*a^6*b^2 - 2*a^2*b^6 
- b^8)*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)) + 2*(3*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))) - 3*((a^2*b^4 + b^6)*d*tan(d*x + c) + (a^3*b^3 + a*b^5 
)*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 - b^3)*sqrt(b*tan(d*x + c) + a) - ((a^8 + 2*a^6* 
b^2 - 2*a^2*b^6 - b^8)*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)) 
 + 2*(3*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...
 

Sympy [F]

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

integrate(tan(d*x+c)**4/(a+b*tan(d*x+c))**(3/2),x)
 

Output:

Integral(tan(c + d*x)**4/(a + b*tan(c + d*x))**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

integrate(tan(d*x + c)^4/(b*tan(d*x + c) + a)^(3/2), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 5.83 (sec) , antiderivative size = 2282, normalized size of antiderivative = 10.10 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^4/(a + b*tan(c + d*x))^(3/2),x)
 

Output:

(log(8*b^9*d^2 - ((((-1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i 
))^(1/2)*(64*a*b^11*d^4 + ((-1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b 
*d^2*3i))^(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 
))/2 + 256*a^3*b^9*d^4 + 384*a^5*b^7*d^4 + 256*a^7*b^5*d^4 + 64*a^9*b^3*d^ 
4))/2 - (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))*(-1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b 
*d^2*3i))^(1/2))/2 + 24*a^2*b^7*d^2 + 24*a^4*b^5*d^2 + 8*a^6*b^3*d^2)*(-1/ 
(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1/2))/2 - log(8*b^9* 
d^2 - ((-1/(4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)))^(1/2)* 
(64*a*b^11*d^4 - (-1/(4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i 
)))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 6 
40*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) + 25 
6*a^3*b^9*d^4 + 384*a^5*b^7*d^4 + 256*a^7*b^5*d^4 + 64*a^9*b^3*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))*(-1/(4*(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)) 
)^(1/2) + 24*a^2*b^7*d^2 + 24*a^4*b^5*d^2 + 8*a^6*b^3*d^2)*(-1/(4*(a^3*d^2 
 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i)))^(1/2) + 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)*(64*a*b^11*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 + ...
 

Reduce [F]

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

int(tan(d*x+c)^4/(a+b*tan(d*x+c))^(3/2),x)
 

Output:

int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**4)/(tan(c + d*x)**2*b**2 + 2*t 
an(c + d*x)*a*b + a**2),x)