\(\int \tan ^4(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\) [511]

Optimal result

Integrand size = 23, antiderivative size = 209 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=-\frac {i (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {i (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{315 b^3 d}-\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{63 b^2 d}+\frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d} \] Output:

-I*(a-I*b)^(3/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+I*(a+I*b) 
^(3/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d+2*b*(a+b*tan(d*x+c) 
)^(1/2)/d+2/315*(8*a^2-63*b^2)*(a+b*tan(d*x+c))^(5/2)/b^3/d-8/63*a*tan(d*x 
+c)*(a+b*tan(d*x+c))^(5/2)/b^2/d+2/9*tan(d*x+c)^2*(a+b*tan(d*x+c))^(5/2)/b 
/d
 

Mathematica [A] (verified)

Time = 4.02 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.62 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\frac {\cos ^2(c+d x) \left (-315 i \left (a^2-b^2\right ) \left (\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}\right ) (a+b \tan (c+d x))^2-630 a b \left (\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}\right ) (a+b \tan (c+d x))^2+\frac {2 (a+b \tan (c+d x))^{5/2} \left (8 a^4-66 a^2 b^2+413 b^4+35 b^4 \sec ^4(c+d x)-4 a b \left (a^2+44 b^2\right ) \tan (c+d x)+b^2 \sec ^2(c+d x) \left (3 a^2-133 b^2+50 a b \tan (c+d x)\right )\right )}{b^3}\right )}{315 d (a \cos (c+d x)+b \sin (c+d x))^2} \] Input:

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

Output:

(Cos[c + d*x]^2*((-315*I)*(a^2 - b^2)*(ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sq 
rt[a - I*b]]/Sqrt[a - I*b] - ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b 
]]/Sqrt[a + I*b])*(a + b*Tan[c + d*x])^2 - 630*a*b*(ArcTanh[Sqrt[a + b*Tan 
[c + d*x]]/Sqrt[a - I*b]]/Sqrt[a - I*b] + ArcTanh[Sqrt[a + b*Tan[c + d*x]] 
/Sqrt[a + I*b]]/Sqrt[a + I*b])*(a + b*Tan[c + d*x])^2 + (2*(a + b*Tan[c + 
d*x])^(5/2)*(8*a^4 - 66*a^2*b^2 + 413*b^4 + 35*b^4*Sec[c + d*x]^4 - 4*a*b* 
(a^2 + 44*b^2)*Tan[c + d*x] + b^2*Sec[c + d*x]^2*(3*a^2 - 133*b^2 + 50*a*b 
*Tan[c + d*x])))/b^3))/(315*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)
 

Rubi [A] (warning: unable to verify)

Time = 1.30 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.739, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4114, 3042, 3963, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2 \int -\frac {1}{2} \tan (c+d x) (a+b \tan (c+d x))^{3/2} \left (4 a \tan ^2(c+d x)+9 b \tan (c+d x)+4 a\right )dx}{9 b}+\frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\int \tan (c+d x) (a+b \tan (c+d x))^{3/2} \left (4 a \tan ^2(c+d x)+9 b \tan (c+d x)+4 a\right )dx}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\int \tan (c+d x) (a+b \tan (c+d x))^{3/2} \left (4 a \tan (c+d x)^2+9 b \tan (c+d x)+4 a\right )dx}{9 b}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {2 \int -\frac {1}{2} (a+b \tan (c+d x))^{3/2} \left (8 a^2+\left (8 a^2-63 b^2\right ) \tan ^2(c+d x)\right )dx}{7 b}+\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {\int (a+b \tan (c+d x))^{3/2} \left (8 a^2+\left (8 a^2-63 b^2\right ) \tan ^2(c+d x)\right )dx}{7 b}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {\int (a+b \tan (c+d x))^{3/2} \left (8 a^2+\left (8 a^2-63 b^2\right ) \tan (c+d x)^2\right )dx}{7 b}}{9 b}\)

\(\Big \downarrow \) 4114

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {63 b^2 \int (a+b \tan (c+d x))^{3/2}dx+\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}}{7 b}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {63 b^2 \int (a+b \tan (c+d x))^{3/2}dx+\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}}{7 b}}{9 b}\)

\(\Big \downarrow \) 3963

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {63 b^2 \left (\int \frac {a^2+2 b \tan (c+d x) a-b^2}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}}{7 b}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {63 b^2 \left (\int \frac {a^2+2 b \tan (c+d x) a-b^2}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}}{7 b}}{9 b}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}+63 b^2 \left (\frac {1}{2} (a-i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )}{7 b}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}+63 b^2 \left (\frac {1}{2} (a-i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )}{7 b}}{9 b}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}+63 b^2 \left (\frac {i (a-i b)^2 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^2 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )}{7 b}}{9 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}+63 b^2 \left (-\frac {i (a-i b)^2 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b)^2 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )}{7 b}}{9 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}+63 b^2 \left (\frac {(a-i b)^2 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^2 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )}{7 b}}{9 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2}}{9 b d}-\frac {\frac {8 a \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {\frac {2 \left (8 a^2-63 b^2\right ) (a+b \tan (c+d x))^{5/2}}{5 b d}+63 b^2 \left (\frac {(a-i b)^{3/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 b \sqrt {a+b \tan (c+d x)}}{d}\right )}{7 b}}{9 b}\)

Input:

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

Output:

(2*Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(5/2))/(9*b*d) - ((8*a*Tan[c + d*x] 
*(a + b*Tan[c + d*x])^(5/2))/(7*b*d) - ((2*(8*a^2 - 63*b^2)*(a + b*Tan[c + 
 d*x])^(5/2))/(5*b*d) + 63*b^2*(((a - I*b)^(3/2)*ArcTan[Tan[c + d*x]/Sqrt[ 
a - I*b]])/d + ((a + I*b)^(3/2)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d + (2 
*b*Sqrt[a + b*Tan[c + d*x]])/d))/(7*b))/(9*b)
 

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 3963

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + 
b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d 
*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 
+ b^2, 0] && GtQ[n, 1]
 

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 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 4114

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*tan[(e_.) 
+ (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m 
 + 1))), x] + Simp[(A - C)   Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a 
, b, e, f, A, C, m}, x] && NeQ[A*b^2 + 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. \(898\) vs. \(2(177)=354\).

Time = 0.33 (sec) , antiderivative size = 899, normalized size of antiderivative = 4.30

method	result	size
derivativedivides	\(\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9 d \,b^{3}}-\frac {4 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d \,b^{3}}+\frac {2 a^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d \,b^{3}}-\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 b d}+\frac {2 b \sqrt {a +b \tan \left (d x +c \right )}}{d}-\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {a^{2}+b^{2}}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}+\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2}}{4 d b}-\frac {b \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}-\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {2 b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {a^{2}+b^{2}}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}-\frac {\ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2}}{4 d b}+\frac {b \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}-\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {2 b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\)	\(899\)
default	\(\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9 d \,b^{3}}-\frac {4 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d \,b^{3}}+\frac {2 a^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d \,b^{3}}-\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 b d}+\frac {2 b \sqrt {a +b \tan \left (d x +c \right )}}{d}-\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {a^{2}+b^{2}}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}+\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2}}{4 d b}-\frac {b \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}-\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {2 b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {a^{2}+b^{2}}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}-\frac {\ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2}}{4 d b}+\frac {b \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}-\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {2 b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\)	\(899\)

Input:

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

Output:

2/9/d/b^3*(a+b*tan(d*x+c))^(9/2)-4/7/d/b^3*a*(a+b*tan(d*x+c))^(7/2)+2/5/d/ 
b^3*a^2*(a+b*tan(d*x+c))^(5/2)-2/5*(a+b*tan(d*x+c))^(5/2)/b/d+2*b*(a+b*tan 
(d*x+c))^(1/2)/d-1/4/d/b*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))*(a^2+b^2)^(1/2)*(2*(a^2+b^2)^(1/2)+ 
2*a)^(1/2)*a+1/4/d/b*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*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/d*b/(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+b^2)^(1/2)+2/d*b/(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*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))*(a^2+b^2)^(1/2)*(2*(a 
^2+b^2)^(1/2)+2*a)^(1/2)*a-1/4/d/b*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*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/d*b/(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+b^2)^(1/2)+2/d*b/(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/...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 873 vs. \(2 (171) = 342\).

Time = 0.10 (sec) , antiderivative size = 873, normalized size of antiderivative = 4.18 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/630*(315*b^3*d*sqrt(-(a^3 - 3*a*b^2 + d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + 
 b^6)/d^4))/d^2)*log(-(3*a^4*b + 2*a^2*b^3 - b^5)*sqrt(b*tan(d*x + c) + a) 
 + (a*d^3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) + (3*a^2*b^2 - b^4)*d)* 
sqrt(-(a^3 - 3*a*b^2 + d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))/d^2)) 
 - 315*b^3*d*sqrt(-(a^3 - 3*a*b^2 + d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6 
)/d^4))/d^2)*log(-(3*a^4*b + 2*a^2*b^3 - b^5)*sqrt(b*tan(d*x + c) + a) - ( 
a*d^3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) + (3*a^2*b^2 - b^4)*d)*sqrt 
(-(a^3 - 3*a*b^2 + d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))/d^2)) - 3 
15*b^3*d*sqrt(-(a^3 - 3*a*b^2 - d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^ 
4))/d^2)*log(-(3*a^4*b + 2*a^2*b^3 - b^5)*sqrt(b*tan(d*x + c) + a) + (a*d^ 
3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) - (3*a^2*b^2 - b^4)*d)*sqrt(-(a 
^3 - 3*a*b^2 - d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))/d^2)) + 315*b 
^3*d*sqrt(-(a^3 - 3*a*b^2 - d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))/ 
d^2)*log(-(3*a^4*b + 2*a^2*b^3 - b^5)*sqrt(b*tan(d*x + c) + a) - (a*d^3*sq 
rt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) - (3*a^2*b^2 - b^4)*d)*sqrt(-(a^3 - 
 3*a*b^2 - d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))/d^2)) + 4*(35*b^4 
*tan(d*x + c)^4 + 50*a*b^3*tan(d*x + c)^3 + 8*a^4 - 63*a^2*b^2 + 315*b^4 + 
 3*(a^2*b^2 - 21*b^4)*tan(d*x + c)^2 - 2*(2*a^3*b + 63*a*b^3)*tan(d*x + c) 
)*sqrt(b*tan(d*x + c) + a))/(b^3*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(tan(d*x+c)^4*(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)^4, x)
 

Giac [F(-2)]

Exception generated. \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^4*(a+b*tan(d*x+c))^(3/2),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,[0,29,11]%%%}+%%%{12,[0,27,11]%%%}+%%%{66,[0,25,11]% 
%%}+%%%{2
 

Mupad [B] (verification not implemented)

Time = 42.04 (sec) , antiderivative size = 1355, normalized size of antiderivative = 6.48 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

(a + b*tan(c + d*x))^(1/2)*(2*a*(2*a*((4*a^2)/(b^3*d) - (2*(a^2 + b^2))/(b 
^3*d)) - (8*a^3)/(b^3*d) + (4*a*(a^2 + b^2))/(b^3*d)) - ((4*a^2)/(b^3*d) - 
 (2*(a^2 + b^2))/(b^3*d))*(a^2 + b^2) + (2*a^4)/(b^3*d)) - atan((b^6*(a + 
b*tan(c + d*x))^(1/2)*((b^3*1i)/(4*d^2) - a^3/(4*d^2) + (3*a*b^2)/(4*d^2) 
- (a^2*b*3i)/(4*d^2))^(1/2)*32i)/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^8*16 
i)/d + (32*a^3*b^5)/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d) - (32*a*b^5*(a + 
 b*tan(c + d*x))^(1/2)*((b^3*1i)/(4*d^2) - a^3/(4*d^2) + (3*a*b^2)/(4*d^2) 
 - (a^2*b*3i)/(4*d^2))^(1/2))/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^8*16i)/ 
d + (32*a^3*b^5)/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d) - (a^2*b^4*(a + b*t 
an(c + d*x))^(1/2)*((b^3*1i)/(4*d^2) - a^3/(4*d^2) + (3*a*b^2)/(4*d^2) - ( 
a^2*b*3i)/(4*d^2))^(1/2)*96i)/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^8*16i)/ 
d + (32*a^3*b^5)/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d) + (96*a^3*b^3*(a + 
b*tan(c + d*x))^(1/2)*((b^3*1i)/(4*d^2) - a^3/(4*d^2) + (3*a*b^2)/(4*d^2) 
- (a^2*b*3i)/(4*d^2))^(1/2))/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^8*16i)/d 
 + (32*a^3*b^5)/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d))*((3*a*b^2 - a^2*b*3 
i - a^3 + b^3*1i)/(4*d^2))^(1/2)*2i - atan((b^6*(a + b*tan(c + d*x))^(1/2) 
*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4*d^2)) 
^(1/2)*32i)/((b^8*16i)/d - (16*a*b^7)/d - (a^2*b^6*32i)/d + (32*a^3*b^5)/d 
 - (a^4*b^4*48i)/d + (48*a^5*b^3)/d) + (32*a*b^5*(a + b*tan(c + d*x))^(1/2 
)*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4*d...
 

Reduce [F]

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