\(\int x (a+b \sec (c+d \sqrt {x}))^2 \, dx\) [38]

Optimal result

Integrand size = 18, antiderivative size = 355 \[ \int x \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=-\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 i a b x^{3/2} \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {12 i a b x \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {24 a b \sqrt {x} \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 i a b \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {24 i a b \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d} \] Output:

-2*I*b^2*x^(3/2)/d+1/2*a^2*x^2-8*I*a*b*x^(3/2)*arctan(exp(I*(c+d*x^(1/2))) 
)/d+6*b^2*x*ln(1+exp(2*I*(c+d*x^(1/2))))/d^2+12*I*a*b*x*polylog(2,-I*exp(I 
*(c+d*x^(1/2))))/d^2-12*I*a*b*x*polylog(2,I*exp(I*(c+d*x^(1/2))))/d^2-6*I* 
b^2*x^(1/2)*polylog(2,-exp(2*I*(c+d*x^(1/2))))/d^3-24*a*b*x^(1/2)*polylog( 
3,-I*exp(I*(c+d*x^(1/2))))/d^3+24*a*b*x^(1/2)*polylog(3,I*exp(I*(c+d*x^(1/ 
2))))/d^3+3*b^2*polylog(3,-exp(2*I*(c+d*x^(1/2))))/d^4-24*I*a*b*polylog(4, 
-I*exp(I*(c+d*x^(1/2))))/d^4+24*I*a*b*polylog(4,I*exp(I*(c+d*x^(1/2))))/d^ 
4+2*b^2*x^(3/2)*tan(c+d*x^(1/2))/d
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.98 \[ \int x \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {-4 i b^2 d^3 x^{3/2}+a^2 d^4 x^2-16 i a b d^3 x^{3/2} \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )+12 b^2 d^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )+24 i a b d^2 x \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )-24 i a b d^2 x \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )-12 i b^2 d \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-48 a b d \sqrt {x} \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )+48 a b d \sqrt {x} \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )+6 b^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-48 i a b \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )+48 i a b \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )+4 b^2 d^3 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{2 d^4} \] Input:

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

Output:

((-4*I)*b^2*d^3*x^(3/2) + a^2*d^4*x^2 - (16*I)*a*b*d^3*x^(3/2)*ArcTan[E^(I 
*(c + d*Sqrt[x]))] + 12*b^2*d^2*x*Log[1 + E^((2*I)*(c + d*Sqrt[x]))] + (24 
*I)*a*b*d^2*x*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (24*I)*a*b*d^2*x*Po 
lyLog[2, I*E^(I*(c + d*Sqrt[x]))] - (12*I)*b^2*d*Sqrt[x]*PolyLog[2, -E^((2 
*I)*(c + d*Sqrt[x]))] - 48*a*b*d*Sqrt[x]*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[ 
x]))] + 48*a*b*d*Sqrt[x]*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))] + 6*b^2*PolyL 
og[3, -E^((2*I)*(c + d*Sqrt[x]))] - (48*I)*a*b*PolyLog[4, (-I)*E^(I*(c + d 
*Sqrt[x]))] + (48*I)*a*b*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))] + 4*b^2*d^3*x 
^(3/2)*Tan[c + d*Sqrt[x]])/(2*d^4)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4692, 3042, 4678, 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.

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

\(\Big \downarrow \) 4692

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4678

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {a^2 x^2}{4}-\frac {4 i a b x^{3/2} \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {12 i a b \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {12 i a b \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {12 a b \sqrt {x} \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 a b \sqrt {x} \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i a b x \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}-\frac {3 i b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {3 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {i b^2 x^{3/2}}{d}\right )\)

Input:

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

Output:

2*(((-I)*b^2*x^(3/2))/d + (a^2*x^2)/4 - ((4*I)*a*b*x^(3/2)*ArcTan[E^(I*(c 
+ d*Sqrt[x]))])/d + (3*b^2*x*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((6 
*I)*a*b*x*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((6*I)*a*b*x*PolyL 
og[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - ((3*I)*b^2*Sqrt[x]*PolyLog[2, -E^((2 
*I)*(c + d*Sqrt[x]))])/d^3 - (12*a*b*Sqrt[x]*PolyLog[3, (-I)*E^(I*(c + d*S 
qrt[x]))])/d^3 + (12*a*b*Sqrt[x]*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 
+ (3*b^2*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/(2*d^4) - ((12*I)*a*b*Pol 
yLog[4, (-I)*E^(I*(c + d*Sqrt[x]))])/d^4 + ((12*I)*a*b*PolyLog[4, I*E^(I*( 
c + d*Sqrt[x]))])/d^4 + (b^2*x^(3/2)*Tan[c + d*Sqrt[x]])/d)
 

Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4678

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
 

rule 4692

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x*(a + b*sec(c + d*sqrt(x)))**2, 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. 1991 vs. \(2 (270) = 540\).

Time = 0.22 (sec) , antiderivative size = 1991, normalized size of antiderivative = 5.61 \[ \int x \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*((d*sqrt(x) + c)^4*a^2 - 4*(d*sqrt(x) + c)^3*a^2*c + 6*(d*sqrt(x) + c) 
^2*a^2*c^2 - 4*(d*sqrt(x) + c)*a^2*c^3 - 8*a*b*c^3*log(sec(d*sqrt(x) + c) 
+ tan(d*sqrt(x) + c)) - 4*(4*b^2*c^3 + 4*((d*sqrt(x) + c)^3*a*b - 3*(d*sqr 
t(x) + c)^2*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2 + ((d*sqrt(x) + c)^3*a*b - 3 
*(d*sqrt(x) + c)^2*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2)*cos(2*d*sqrt(x) + 2* 
c) + (I*(d*sqrt(x) + c)^3*a*b - 3*I*(d*sqrt(x) + c)^2*a*b*c + 3*I*(d*sqrt( 
x) + c)*a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), sin(d 
*sqrt(x) + c) + 1) + 4*((d*sqrt(x) + c)^3*a*b - 3*(d*sqrt(x) + c)^2*a*b*c 
+ 3*(d*sqrt(x) + c)*a*b*c^2 + ((d*sqrt(x) + c)^3*a*b - 3*(d*sqrt(x) + c)^2 
*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2)*cos(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) 
 + c)^3*a*b - 3*I*(d*sqrt(x) + c)^2*a*b*c + 3*I*(d*sqrt(x) + c)*a*b*c^2)*s 
in(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), -sin(d*sqrt(x) + c) + 1 
) - 6*((d*sqrt(x) + c)^2*b^2 - 2*(d*sqrt(x) + c)*b^2*c + b^2*c^2 + ((d*sqr 
t(x) + c)^2*b^2 - 2*(d*sqrt(x) + c)*b^2*c + b^2*c^2)*cos(2*d*sqrt(x) + 2*c 
) - (-I*(d*sqrt(x) + c)^2*b^2 + 2*I*(d*sqrt(x) + c)*b^2*c - I*b^2*c^2)*sin 
(2*d*sqrt(x) + 2*c))*arctan2(sin(2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*c 
) + 1) + 4*((d*sqrt(x) + c)^3*b^2 - 3*(d*sqrt(x) + c)^2*b^2*c + 3*(d*sqrt( 
x) + c)*b^2*c^2)*cos(2*d*sqrt(x) + 2*c) + 6*((d*sqrt(x) + c)*b^2 - b^2*c + 
 ((d*sqrt(x) + c)*b^2 - b^2*c)*cos(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) + c) 
*b^2 - I*b^2*c)*sin(2*d*sqrt(x) + 2*c))*dilog(-e^(2*I*d*sqrt(x) + 2*I*c...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx =\text {Too large to display} \] Input:

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

Output:

(24*cos(sqrt(x)*d + c)*int((sqrt(x)*tan((sqrt(x)*d + c)/2))/(tan((sqrt(x)* 
d + c)/2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*a*b*d**3 - 24*cos(sqrt( 
x)*d + c)*int((sqrt(x)*tan((sqrt(x)*d + c)/2))/(tan((sqrt(x)*d + c)/2)**4 
- 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*b**2*d**3 + 8*cos(sqrt(x)*d + c)*int 
(x/(tan((sqrt(x)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*a*b*d* 
*4 + 24*cos(sqrt(x)*d + c)*int(1/(tan((sqrt(x)*d + c)/2)**4 - 2*tan((sqrt( 
x)*d + c)/2)**2 + 1),x)*a*b*d**2 - 24*cos(sqrt(x)*d + c)*int(1/(tan((sqrt( 
x)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*b**2*d**2 + 12*cos(s 
qrt(x)*d + c)*log(tan((sqrt(x)*d + c)/2)**2 + 1)*a*b - 12*cos(sqrt(x)*d + 
c)*log(tan((sqrt(x)*d + c)/2)**2 + 1)*b**2 - 12*cos(sqrt(x)*d + c)*log(tan 
((sqrt(x)*d + c)/2) - 1)*a*b + 12*cos(sqrt(x)*d + c)*log(tan((sqrt(x)*d + 
c)/2) - 1)*b**2 - 12*cos(sqrt(x)*d + c)*log(tan((sqrt(x)*d + c)/2) + 1)*a* 
b + 12*cos(sqrt(x)*d + c)*log(tan((sqrt(x)*d + c)/2) + 1)*b**2 + cos(sqrt( 
x)*d + c)*a**2*d**4*x**2 - cos(sqrt(x)*d + c)*a*b*d**4*x**2 - 6*cos(sqrt(x 
)*d + c)*a*b*d**2*x + 6*cos(sqrt(x)*d + c)*b**2*d**2*x - 4*sqrt(x)*sin(sqr 
t(x)*d + c)*a*b*d**3*x - 12*sqrt(x)*sin(sqrt(x)*d + c)*a*b*d + 4*sqrt(x)*s 
in(sqrt(x)*d + c)*b**2*d**3*x + 12*sqrt(x)*sin(sqrt(x)*d + c)*b**2*d - 12* 
a*b*d**2*x + 12*b**2*d**2*x)/(2*cos(sqrt(x)*d + c)*d**4)