\(\int x^{3/2} (a+b \sec (c+d \sqrt {x})) \, dx\) [51]

Optimal result

Integrand size = 20, antiderivative size = 284 \[ \int x^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2}{5} a x^{5/2}-\frac {4 i b x^2 \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i b x^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 b x \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 i b \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 b \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 b \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5} \] Output:

2/5*a*x^(5/2)-4*I*b*x^2*arctan(exp(I*(c+d*x^(1/2))))/d+8*I*b*x^(3/2)*polyl 
og(2,-I*exp(I*(c+d*x^(1/2))))/d^2-8*I*b*x^(3/2)*polylog(2,I*exp(I*(c+d*x^( 
1/2))))/d^2-24*b*x*polylog(3,-I*exp(I*(c+d*x^(1/2))))/d^3+24*b*x*polylog(3 
,I*exp(I*(c+d*x^(1/2))))/d^3-48*I*b*x^(1/2)*polylog(4,-I*exp(I*(c+d*x^(1/2 
))))/d^4+48*I*b*x^(1/2)*polylog(4,I*exp(I*(c+d*x^(1/2))))/d^4+48*b*polylog 
(5,-I*exp(I*(c+d*x^(1/2))))/d^5-48*b*polylog(5,I*exp(I*(c+d*x^(1/2))))/d^5
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.99 \[ \int x^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2 \left (a d^5 x^{5/2}-10 i b d^4 x^2 \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )+20 i b d^3 x^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )-20 i b d^3 x^{3/2} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )-60 b d^2 x \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )+60 b d^2 x \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )-120 i b d \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )+120 i b d \sqrt {x} \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )+120 b \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )-120 b \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )\right )}{5 d^5} \] Input:

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

Output:

(2*(a*d^5*x^(5/2) - (10*I)*b*d^4*x^2*ArcTan[E^(I*(c + d*Sqrt[x]))] + (20*I 
)*b*d^3*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (20*I)*b*d^3*x^(3 
/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - 60*b*d^2*x*PolyLog[3, (-I)*E^(I* 
(c + d*Sqrt[x]))] + 60*b*d^2*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))] - (120* 
I)*b*d*Sqrt[x]*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))] + (120*I)*b*d*Sqrt[x 
]*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))] + 120*b*PolyLog[5, (-I)*E^(I*(c + d* 
Sqrt[x]))] - 120*b*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))]))/(5*d^5)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2010, 2009}

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.

\(\displaystyle \int x^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx\)

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2}{5} a x^{5/2}-\frac {4 i b x^2 \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {48 b \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 b \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 i b \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i b \sqrt {x} \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 b x \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 b x \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 i b x^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i b x^{3/2} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}\)

Input:

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

Output:

(2*a*x^(5/2))/5 - ((4*I)*b*x^2*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + ((8*I)*b 
*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((8*I)*b*x^(3/2)*Po 
lyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (24*b*x*PolyLog[3, (-I)*E^(I*(c + 
 d*Sqrt[x]))])/d^3 + (24*b*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 - (( 
48*I)*b*Sqrt[x]*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))])/d^4 + ((48*I)*b*Sq 
rt[x]*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (48*b*PolyLog[5, (-I)*E^( 
I*(c + d*Sqrt[x]))])/d^5 - (48*b*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

Maple [F]

\[\int x^{\frac {3}{2}} \left (a +b \sec \left (c +d \sqrt {x}\right )\right )d x\]

Input:

int(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x)
 

Output:

int(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x)
 

Fricas [F]

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

integrate(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x, algorithm="fricas")
 

Output:

integral(b*x^(3/2)*sec(d*sqrt(x) + c) + a*x^(3/2), x)
 

Sympy [F]

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

integrate(x**(3/2)*(a+b*sec(c+d*x**(1/2))),x)
 

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 738 vs. \(2 (208) = 416\).

Time = 0.21 (sec) , antiderivative size = 738, normalized size of antiderivative = 2.60 \[ \int x^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx =\text {Too large to display} \] Input:

integrate(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x, algorithm="maxima")
 

Output:

1/5*(2*(d*sqrt(x) + c)^5*a - 10*(d*sqrt(x) + c)^4*a*c + 20*(d*sqrt(x) + c) 
^3*a*c^2 - 20*(d*sqrt(x) + c)^2*a*c^3 + 10*(d*sqrt(x) + c)*a*c^4 + 10*b*c^ 
4*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) - 10*(I*(d*sqrt(x) + c)^4*b 
 - 4*I*(d*sqrt(x) + c)^3*b*c + 6*I*(d*sqrt(x) + c)^2*b*c^2 - 4*I*(d*sqrt(x 
) + c)*b*c^3)*arctan2(cos(d*sqrt(x) + c), sin(d*sqrt(x) + c) + 1) - 10*(I* 
(d*sqrt(x) + c)^4*b - 4*I*(d*sqrt(x) + c)^3*b*c + 6*I*(d*sqrt(x) + c)^2*b* 
c^2 - 4*I*(d*sqrt(x) + c)*b*c^3)*arctan2(cos(d*sqrt(x) + c), -sin(d*sqrt(x 
) + c) + 1) - 40*(I*(d*sqrt(x) + c)^3*b - 3*I*(d*sqrt(x) + c)^2*b*c + 3*I* 
(d*sqrt(x) + c)*b*c^2 - I*b*c^3)*dilog(I*e^(I*d*sqrt(x) + I*c)) - 40*(-I*( 
d*sqrt(x) + c)^3*b + 3*I*(d*sqrt(x) + c)^2*b*c - 3*I*(d*sqrt(x) + c)*b*c^2 
 + I*b*c^3)*dilog(-I*e^(I*d*sqrt(x) + I*c)) + 5*((d*sqrt(x) + c)^4*b - 4*( 
d*sqrt(x) + c)^3*b*c + 6*(d*sqrt(x) + c)^2*b*c^2 - 4*(d*sqrt(x) + c)*b*c^3 
)*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*sin(d*sqrt(x) + c) + 
 1) - 5*((d*sqrt(x) + c)^4*b - 4*(d*sqrt(x) + c)^3*b*c + 6*(d*sqrt(x) + c) 
^2*b*c^2 - 4*(d*sqrt(x) + c)*b*c^3)*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt( 
x) + c)^2 - 2*sin(d*sqrt(x) + c) + 1) - 240*b*polylog(5, I*e^(I*d*sqrt(x) 
+ I*c)) + 240*b*polylog(5, -I*e^(I*d*sqrt(x) + I*c)) - 240*(-I*(d*sqrt(x) 
+ c)*b + I*b*c)*polylog(4, I*e^(I*d*sqrt(x) + I*c)) - 240*(I*(d*sqrt(x) + 
c)*b - I*b*c)*polylog(4, -I*e^(I*d*sqrt(x) + I*c)) + 120*((d*sqrt(x) + c)^ 
2*b - 2*(d*sqrt(x) + c)*b*c + b*c^2)*polylog(3, I*e^(I*d*sqrt(x) + I*c)...
 

Giac [F]

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

integrate(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x, algorithm="giac")
 

Output:

integrate((b*sec(d*sqrt(x) + c) + a)*x^(3/2), x)
 

Mupad [F(-1)]

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

int(x^(3/2)*(a + b/cos(c + d*x^(1/2))),x)
 

Output:

int(x^(3/2)*(a + b/cos(c + d*x^(1/2))), x)
 

Reduce [F]

\[ \int x^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {2 \sqrt {x}\, a \,x^{2}}{5}+\frac {2 \sqrt {x}\, b \,x^{2}}{5}-2 \left (\int \frac {\sqrt {x}\, \tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2} x}{\tan \left (\frac {\sqrt {x}\, d}{2}+\frac {c}{2}\right )^{2}-1}d x \right ) b \] Input:

int(x^(3/2)*(a+b*sec(c+d*x^(1/2))),x)
 

Output:

(2*(sqrt(x)*a*x**2 + sqrt(x)*b*x**2 - 5*int((sqrt(x)*tan((sqrt(x)*d + c)/2 
)**2*x)/(tan((sqrt(x)*d + c)/2)**2 - 1),x)*b))/5