3.6.20 \(\int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) [520]

3.6.20.1 Optimal result

Integrand size = 23, antiderivative size = 158 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {i (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {i (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d} \]

output

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

3.6.20.2 Mathematica [A] (verified)

Time = 2.31 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.30 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {21 i (a-i b)^{5/2} b \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-21 i (a+i b)^{5/2} b \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\frac {1}{2} \sec ^3(c+d x) \left (3 a \left (3 a^2-46 b^2\right ) \cos (c+d x)+\left (3 a^3-58 a b^2\right ) \cos (3 (c+d x))-2 b \left (-9 a^2+4 b^2+\left (-9 a^2+10 b^2\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right ) \sqrt {a+b \tan (c+d x)}}{21 b d} \]

input

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

output

((21*I)*(a - I*b)^(5/2)*b*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] 
- (21*I)*(a + I*b)^(5/2)*b*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] 
 + (Sec[c + d*x]^3*(3*a*(3*a^2 - 46*b^2)*Cos[c + d*x] + (3*a^3 - 58*a*b^2) 
*Cos[3*(c + d*x)] - 2*b*(-9*a^2 + 4*b^2 + (-9*a^2 + 10*b^2)*Cos[2*(c + d*x 
)])*Sin[c + d*x])*Sqrt[a + b*Tan[c + d*x]])/2)/(21*b*d)
 

3.6.20.3 Rubi [A] (warning: unable to verify)

Time = 0.87 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.87, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 4026, 25, 3042, 3963, 3042, 4011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4026

\(\displaystyle \int -(a+b \tan (c+d x))^{5/2}dx+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\int (a+b \tan (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\int (a+b \tan (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 3963

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

\(\displaystyle -\int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 4022

\(\displaystyle -\frac {1}{2} (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (a-i b)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {1}{2} (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (a-i b)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {i (a-i b)^3 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b)^3 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i (a-i b)^3 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^3 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {(a+i b)^3 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}-\frac {(a-i b)^3 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(a-i b)^{5/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 (a+b \tan (c+d x))^{7/2}}{7 b d}-\frac {2 b (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a b \sqrt {a+b \tan (c+d x)}}{d}\)

input

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

output

-(((a - I*b)^(5/2)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d) - ((a + I*b)^(5/ 
2)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d - (4*a*b*Sqrt[a + b*Tan[c + d*x]] 
)/d - (2*b*(a + b*Tan[c + d*x])^(3/2))/(3*d) + (2*(a + b*Tan[c + d*x])^(7/ 
2))/(7*b*d)
 

3.6.20.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4011

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

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 4026

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( 
m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* 
x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ 
[m, -1] &&  !(EqQ[m, 2] && EqQ[a, 0])
 

3.6.20.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1193\) vs. \(2(130)=260\).

Time = 0.11 (sec) , antiderivative size = 1194, normalized size of antiderivative = 7.56

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1194\)
default	\(\text {Expression too large to display}\)	\(1194\)

input

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

output

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

3.6.20.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1175 vs. \(2 (124) = 248\).

Time = 0.26 (sec) , antiderivative size = 1175, normalized size of antiderivative = 7.44 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

input

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

output

1/42*(21*b*d*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 1 
00*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8*b - 14 
*a^4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a) + ((a^2 - b^2)*d^3*sq 
rt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) + 2* 
(5*a^5*b^2 - 10*a^3*b^4 + a*b^6)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + d^ 
2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4)) 
/d^2)) - 21*b*d*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 
- 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8*b - 
 14*a^4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a) - ((a^2 - b^2)*d^3 
*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) + 
 2*(5*a^5*b^2 - 10*a^3*b^4 + a*b^6)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 + 
 d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^ 
4))/d^2)) - 21*b*d*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b 
^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8* 
b - 14*a^4*b^5 - 8*a^2*b^7 + b^9)*sqrt(b*tan(d*x + c) + a) + ((a^2 - b^2)* 
d^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4 
) - 2*(5*a^5*b^2 - 10*a^3*b^4 + a*b^6)*d)*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^ 
4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10) 
/d^4))/d^2)) + 21*b*d*sqrt(-(a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^ 
8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((...
 

3.6.20.6 Sympy [F]

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

input

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

output

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

3.6.20.7 Maxima [F]

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

input

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

output

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

3.6.20.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.6.20.9 Mupad [B] (verification not implemented)

Time = 27.91 (sec) , antiderivative size = 2165, normalized size of antiderivative = 13.70 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

input

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

output

atan(((((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d* 
x))^(1/2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2) 
/(4*d^2))^(1/2))*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a 
^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 
+ 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b 
^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i - (((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2 
))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*(-(5*a*b^4 - a^4*b*5i + a^5 - 
 b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*(-(5*a*b^4 - a^4*b*5i 
+ a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(a + b*tan 
(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*(-(5*a*b^ 
4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i) 
/((((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^ 
(1/2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b^2)/(4* 
d^2))^(1/2))*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10*a^3*b 
^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15 
*a^4*b^4 - a^6*b^2))/d^2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*1 
0i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(6*a^4*b^7 - b^11 + 8*a^6*b^5 + 3*a^ 
8*b^3))/d^3 + (((8*(8*a*b^5*d^2 + 8*a^3*b^3*d^2))/d^3 + 64*a*b^2*(a + b*ta 
n(c + d*x))^(1/2)*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3*10i - 10* 
a^3*b^2)/(4*d^2))^(1/2))*(-(5*a*b^4 - a^4*b*5i + a^5 - b^5*1i + a^2*b^3...