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

3.1.51.1 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*b*polylog(5,-I*exp(I*(c+d*x^(1/2))))/d^5-4 
8*b*polylog(5,I*exp(I*(c+d*x^(1/2))))/d^5-48*I*b*polylog(4,-I*exp(I*(c+d*x 
^(1/2))))*x^(1/2)/d^4+48*I*b*polylog(4,I*exp(I*(c+d*x^(1/2))))*x^(1/2)/d^4
 

3.1.51.2 Mathematica [A] (verified)

Time = 0.23 (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)
 

3.1.51.3 Rubi [A] (verified)

Time = 0.47 (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
 

3.1.51.3.1 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]]
 

3.1.51.4 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)
 

3.1.51.5 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)
 

3.1.51.6 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)
 

3.1.51.7 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 = 0.44 (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=\frac {2 \, {\left (d \sqrt {x} + c\right )}^{5} a - 10 \, {\left (d \sqrt {x} + c\right )}^{4} a c + 20 \, {\left (d \sqrt {x} + c\right )}^{3} a c^{2} - 20 \, {\left (d \sqrt {x} + c\right )}^{2} a c^{3} + 10 \, {\left (d \sqrt {x} + c\right )} a c^{4} + 10 \, b c^{4} \log \left (\sec \left (d \sqrt {x} + c\right ) + \tan \left (d \sqrt {x} + c\right )\right ) - 10 \, {\left (i \, {\left (d \sqrt {x} + c\right )}^{4} b - 4 i \, {\left (d \sqrt {x} + c\right )}^{3} b c + 6 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 4 i \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \arctan \left (\cos \left (d \sqrt {x} + c\right ), \sin \left (d \sqrt {x} + c\right ) + 1\right ) - 10 \, {\left (i \, {\left (d \sqrt {x} + c\right )}^{4} b - 4 i \, {\left (d \sqrt {x} + c\right )}^{3} b c + 6 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 4 i \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \arctan \left (\cos \left (d \sqrt {x} + c\right ), -\sin \left (d \sqrt {x} + c\right ) + 1\right ) - 40 \, {\left (i \, {\left (d \sqrt {x} + c\right )}^{3} b - 3 i \, {\left (d \sqrt {x} + c\right )}^{2} b c + 3 i \, {\left (d \sqrt {x} + c\right )} b c^{2} - i \, b c^{3}\right )} {\rm Li}_2\left (i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) - 40 \, {\left (-i \, {\left (d \sqrt {x} + c\right )}^{3} b + 3 i \, {\left (d \sqrt {x} + c\right )}^{2} b c - 3 i \, {\left (d \sqrt {x} + c\right )} b c^{2} + i \, b c^{3}\right )} {\rm Li}_2\left (-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}\right ) + 5 \, {\left ({\left (d \sqrt {x} + c\right )}^{4} b - 4 \, {\left (d \sqrt {x} + c\right )}^{3} b c + 6 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 4 \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} + 2 \, \sin \left (d \sqrt {x} + c\right ) + 1\right ) - 5 \, {\left ({\left (d \sqrt {x} + c\right )}^{4} b - 4 \, {\left (d \sqrt {x} + c\right )}^{3} b c + 6 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 4 \, {\left (d \sqrt {x} + c\right )} b c^{3}\right )} \log \left (\cos \left (d \sqrt {x} + c\right )^{2} + \sin \left (d \sqrt {x} + c\right )^{2} - 2 \, \sin \left (d \sqrt {x} + c\right ) + 1\right ) - 240 \, b {\rm Li}_{5}(i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 240 \, b {\rm Li}_{5}(-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) - 240 \, {\left (-i \, {\left (d \sqrt {x} + c\right )} b + i \, b c\right )} {\rm Li}_{4}(i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) - 240 \, {\left (i \, {\left (d \sqrt {x} + c\right )} b - i \, b c\right )} {\rm Li}_{4}(-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) + 120 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c + b c^{2}\right )} {\rm Li}_{3}(i \, e^{\left (i \, d \sqrt {x} + i \, c\right )}) - 120 \, {\left ({\left (d \sqrt {x} + c\right )}^{2} b - 2 \, {\left (d \sqrt {x} + c\right )} b c + b c^{2}\right )} {\rm Li}_{3}(-i \, e^{\left (i \, d \sqrt {x} + i \, c\right )})}{5 \, d^{5}} \]

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)...
 

3.1.51.8 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)
 

3.1.51.9 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)