\(\int x^{3/2} (a+b \sec (c+d \sqrt {x}))^2 \, dx\) [56]

Optimal result

Integrand size = 22, antiderivative size = 451 \[ \int x^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {6 i b^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 a b \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 a b \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d} \] Output:

6*I*b^2*polylog(4,-exp(2*I*(c+d*x^(1/2))))/d^5+2/5*a^2*x^(5/2)-8*I*a*b*x^2 
*arctan(exp(I*(c+d*x^(1/2))))/d+8*b^2*x^(3/2)*ln(1+exp(2*I*(c+d*x^(1/2)))) 
/d^2+96*I*a*b*x^(1/2)*polylog(4,I*exp(I*(c+d*x^(1/2))))/d^4+16*I*a*b*x^(3/ 
2)*polylog(2,-I*exp(I*(c+d*x^(1/2))))/d^2-2*I*b^2*x^2/d-48*a*b*x*polylog(3 
,-I*exp(I*(c+d*x^(1/2))))/d^3+48*a*b*x*polylog(3,I*exp(I*(c+d*x^(1/2))))/d 
^3+12*b^2*x^(1/2)*polylog(3,-exp(2*I*(c+d*x^(1/2))))/d^4-12*I*b^2*x*polylo 
g(2,-exp(2*I*(c+d*x^(1/2))))/d^3-96*I*a*b*x^(1/2)*polylog(4,-I*exp(I*(c+d* 
x^(1/2))))/d^4-16*I*a*b*x^(3/2)*polylog(2,I*exp(I*(c+d*x^(1/2))))/d^2+96*a 
*b*polylog(5,-I*exp(I*(c+d*x^(1/2))))/d^5-96*a*b*polylog(5,I*exp(I*(c+d*x^ 
(1/2))))/d^5+2*b^2*x^2*tan(c+d*x^(1/2))/d
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 443, normalized size of antiderivative = 0.98 \[ \int x^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 \left (-5 i b^2 d^4 x^2+a^2 d^5 x^{5/2}-20 i a b d^4 x^2 \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )+20 b^2 d^3 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )+40 i a b d^3 x^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )-40 i a b d^3 x^{3/2} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )-30 i b^2 d^2 x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-120 a b d^2 x \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )+120 a b d^2 x \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )+30 b^2 d \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-240 i a b d \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )+240 i a b d \sqrt {x} \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )+15 i b^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )+240 a b \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )-240 a b \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )+5 b^2 d^4 x^2 \tan \left (c+d \sqrt {x}\right )\right )}{5 d^5} \] Input:

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

Output:

(2*((-5*I)*b^2*d^4*x^2 + a^2*d^5*x^(5/2) - (20*I)*a*b*d^4*x^2*ArcTan[E^(I* 
(c + d*Sqrt[x]))] + 20*b^2*d^3*x^(3/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))] 
+ (40*I)*a*b*d^3*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (40*I)*a 
*b*d^3*x^(3/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - (30*I)*b^2*d^2*x*Poly 
Log[2, -E^((2*I)*(c + d*Sqrt[x]))] - 120*a*b*d^2*x*PolyLog[3, (-I)*E^(I*(c 
 + d*Sqrt[x]))] + 120*a*b*d^2*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))] + 30*b 
^2*d*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))] - (240*I)*a*b*d*Sqrt[x 
]*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))] + (240*I)*a*b*d*Sqrt[x]*PolyLog[4 
, I*E^(I*(c + d*Sqrt[x]))] + (15*I)*b^2*PolyLog[4, -E^((2*I)*(c + d*Sqrt[x 
]))] + 240*a*b*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))] - 240*a*b*PolyLog[5, 
 I*E^(I*(c + d*Sqrt[x]))] + 5*b^2*d^4*x^2*Tan[c + d*Sqrt[x]]))/(5*d^5)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, 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.

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

\(\Big \downarrow \) 4692

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4678

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {1}{5} a^2 x^{5/2}-\frac {4 i a b x^2 \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {48 a b \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 a b \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {48 i a b \sqrt {x} \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {48 i a b \sqrt {x} \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 a b x \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b x \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 i a b x^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b x^{3/2} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {3 i b^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {6 b^2 \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 i b^2 x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d}-\frac {i b^2 x^2}{d}\right )\)

Input:

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

Output:

2*(((-I)*b^2*x^2)/d + (a^2*x^(5/2))/5 - ((4*I)*a*b*x^2*ArcTan[E^(I*(c + d* 
Sqrt[x]))])/d + (4*b^2*x^(3/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ( 
(8*I)*a*b*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((8*I)*a*b 
*x^(3/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - ((6*I)*b^2*x*PolyLog[2 
, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - (24*a*b*x*PolyLog[3, (-I)*E^(I*(c + d 
*Sqrt[x]))])/d^3 + (24*a*b*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 + (6 
*b^2*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^4 - ((48*I)*a*b*Sqr 
t[x]*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))])/d^4 + ((48*I)*a*b*Sqrt[x]*Pol 
yLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + ((3*I)*b^2*PolyLog[4, -E^((2*I)*(c 
 + d*Sqrt[x]))])/d^5 + (48*a*b*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 
 - (48*a*b*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 + (b^2*x^2*Tan[c + d*S 
qrt[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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x**(3/2)*(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. 2869 vs. \(2 (346) = 692\).

Time = 0.29 (sec) , antiderivative size = 2869, normalized size of antiderivative = 6.36 \[ \int x^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

2/5*((d*sqrt(x) + c)^5*a^2 - 5*(d*sqrt(x) + c)^4*a^2*c + 10*(d*sqrt(x) + c 
)^3*a^2*c^2 - 10*(d*sqrt(x) + c)^2*a^2*c^3 + 5*(d*sqrt(x) + c)*a^2*c^4 + 1 
0*a*b*c^4*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) + 5*(6*b^2*c^4 - 6* 
((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a 
*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^3 + ((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) 
 + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^3)*c 
os(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) + c)^4*a*b - 4*I*(d*sqrt(x) + c)^3*a 
*b*c + 6*I*(d*sqrt(x) + c)^2*a*b*c^2 - 4*I*(d*sqrt(x) + c)*a*b*c^3)*sin(2* 
d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), sin(d*sqrt(x) + c) + 1) - 6* 
((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a 
*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^3 + ((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) 
 + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^3)*c 
os(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) + c)^4*a*b - 4*I*(d*sqrt(x) + c)^3*a 
*b*c + 6*I*(d*sqrt(x) + c)^2*a*b*c^2 - 4*I*(d*sqrt(x) + c)*a*b*c^3)*sin(2* 
d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), -sin(d*sqrt(x) + c) + 1) + 4 
*(4*(d*sqrt(x) + c)^3*b^2 - 9*(d*sqrt(x) + c)^2*b^2*c + 9*(d*sqrt(x) + c)* 
b^2*c^2 - 3*b^2*c^3 + (4*(d*sqrt(x) + c)^3*b^2 - 9*(d*sqrt(x) + c)^2*b^2*c 
 + 9*(d*sqrt(x) + c)*b^2*c^2 - 3*b^2*c^3)*cos(2*d*sqrt(x) + 2*c) - (-4*I*( 
d*sqrt(x) + c)^3*b^2 + 9*I*(d*sqrt(x) + c)^2*b^2*c - 9*I*(d*sqrt(x) + c)*b 
^2*c^2 + 3*I*b^2*c^3)*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(2*d*sqrt(x) +...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*(sqrt(x)*cos(sqrt(x)*d + c)*tan((sqrt(x)*d + c)/2)**2*a**2*d**5*x**2 + 
120*sqrt(x)*cos(sqrt(x)*d + c)*tan((sqrt(x)*d + c)/2)**2*a*b*d - sqrt(x)*c 
os(sqrt(x)*d + c)*tan((sqrt(x)*d + c)/2)**2*b**2*d**5*x**2 - 120*sqrt(x)*c 
os(sqrt(x)*d + c)*tan((sqrt(x)*d + c)/2)**2*b**2*d - sqrt(x)*cos(sqrt(x)*d 
 + c)*a**2*d**5*x**2 + 40*sqrt(x)*cos(sqrt(x)*d + c)*a*b*d**3*x + 120*sqrt 
(x)*cos(sqrt(x)*d + c)*a*b*d + sqrt(x)*cos(sqrt(x)*d + c)*b**2*d**5*x**2 - 
 40*sqrt(x)*cos(sqrt(x)*d + c)*b**2*d**3*x - 120*sqrt(x)*cos(sqrt(x)*d + c 
)*b**2*d + 120*cos(sqrt(x)*d + c)*int(sqrt(x)/(tan((sqrt(x)*d + c)/2)**4 - 
 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*tan((sqrt(x)*d + c)/2)**2*a*b*d**3 - 
120*cos(sqrt(x)*d + c)*int(sqrt(x)/(tan((sqrt(x)*d + c)/2)**4 - 2*tan((sqr 
t(x)*d + c)/2)**2 + 1),x)*tan((sqrt(x)*d + c)/2)**2*b**2*d**3 - 120*cos(sq 
rt(x)*d + c)*int(sqrt(x)/(tan((sqrt(x)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c 
)/2)**2 + 1),x)*a*b*d**3 + 120*cos(sqrt(x)*d + c)*int(sqrt(x)/(tan((sqrt(x 
)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*b**2*d**3 + 240*cos(s 
qrt(x)*d + c)*int(tan((sqrt(x)*d + c)/2)/(tan((sqrt(x)*d + c)/2)**4 - 2*ta 
n((sqrt(x)*d + c)/2)**2 + 1),x)*tan((sqrt(x)*d + c)/2)**2*a*b*d**2 - 240*c 
os(sqrt(x)*d + c)*int(tan((sqrt(x)*d + c)/2)/(tan((sqrt(x)*d + c)/2)**4 - 
2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*tan((sqrt(x)*d + c)/2)**2*b**2*d**2 - 
240*cos(sqrt(x)*d + c)*int(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**2 + 240*cos(sqrt(x)*d +...