\(\int \frac {(c x)^{15/2}}{(a+b x^3)^{3/2}} \, dx\) [350]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 19, antiderivative size = 550 \[ \int \frac {(c x)^{15/2}}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 c^2 (c x)^{11/2}}{3 b \sqrt {a+b x^3}}+\frac {11 c^5 (c x)^{5/2} \sqrt {a+b x^3}}{12 b^2}-\frac {55 \left (1+\sqrt {3}\right ) a c^7 \sqrt {c x} \sqrt {a+b x^3}}{24 b^{8/3} \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {55 a^{4/3} c^7 \sqrt {c x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} E\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{8\ 3^{3/4} b^{8/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {55 \left (1-\sqrt {3}\right ) a^{4/3} c^7 \sqrt {c x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{48 \sqrt [4]{3} b^{8/3} \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \] Output:

-2/3*c^2*(c*x)^(11/2)/b/(b*x^3+a)^(1/2)+11/12*c^5*(c*x)^(5/2)*(b*x^3+a)^(1 
/2)/b^2-55/24*(1+3^(1/2))*a*c^7*(c*x)^(1/2)*(b*x^3+a)^(1/2)/b^(8/3)/(a^(1/ 
3)+(1+3^(1/2))*b^(1/3)*x)+55/24*a^(4/3)*c^7*(c*x)^(1/2)*(a^(1/3)+b^(1/3)*x 
)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x) 
^2)^(1/2)*EllipticE((1-(a^(1/3)+(1-3^(1/2))*b^(1/3)*x)^2/(a^(1/3)+(1+3^(1/ 
2))*b^(1/3)*x)^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*3^(1/4)/b^(8/3)/(b^(1/3)* 
x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2)/(b*x^3+a)^( 
1/2)+55/144*(1-3^(1/2))*a^(4/3)*c^7*(c*x)^(1/2)*(a^(1/3)+b^(1/3)*x)*((a^(2 
/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2 
)*InverseJacobiAM(arccos((a^(1/3)+(1-3^(1/2))*b^(1/3)*x)/(a^(1/3)+(1+3^(1/ 
2))*b^(1/3)*x)),1/4*6^(1/2)+1/4*2^(1/2))*3^(3/4)/b^(8/3)/(b^(1/3)*x*(a^(1/ 
3)+b^(1/3)*x)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2)/(b*x^3+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.13 \[ \int \frac {(c x)^{15/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {c^5 (c x)^{5/2} \left (-11 a+2 b x^3+11 a \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{6},\frac {3}{2},\frac {11}{6},-\frac {b x^3}{a}\right )\right )}{8 b^2 \sqrt {a+b x^3}} \] Input:

Integrate[(c*x)^(15/2)/(a + b*x^3)^(3/2),x]
 

Output:

(c^5*(c*x)^(5/2)*(-11*a + 2*b*x^3 + 11*a*Sqrt[1 + (b*x^3)/a]*Hypergeometri 
c2F1[5/6, 3/2, 11/6, -((b*x^3)/a)]))/(8*b^2*Sqrt[a + b*x^3])
 

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 624, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {817, 843, 851, 837, 25, 766, 2420}

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.

\(\displaystyle \int \frac {(c x)^{15/2}}{\left (a+b x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {11 c^3 \int \frac {(c x)^{9/2}}{\sqrt {b x^3+a}}dx}{3 b}-\frac {2 c^2 (c x)^{11/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {11 c^3 \left (\frac {c^2 (c x)^{5/2} \sqrt {a+b x^3}}{4 b}-\frac {5 a c^3 \int \frac {(c x)^{3/2}}{\sqrt {b x^3+a}}dx}{8 b}\right )}{3 b}-\frac {2 c^2 (c x)^{11/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {11 c^3 \left (\frac {c^2 (c x)^{5/2} \sqrt {a+b x^3}}{4 b}-\frac {5 a c^2 \int \frac {c^2 x^2}{\sqrt {b x^3+a}}d\sqrt {c x}}{4 b}\right )}{3 b}-\frac {2 c^2 (c x)^{11/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 837

\(\displaystyle \frac {11 c^3 \left (\frac {c^2 (c x)^{5/2} \sqrt {a+b x^3}}{4 b}-\frac {5 a c^2 \left (-\frac {\left (1-\sqrt {3}\right ) a^{2/3} c^2 \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {c x}}{2 b^{2/3}}-\frac {\int -\frac {2 b^{2/3} x^2 c^2+\left (1-\sqrt {3}\right ) a^{2/3} c^2}{\sqrt {b x^3+a}}d\sqrt {c x}}{2 b^{2/3}}\right )}{4 b}\right )}{3 b}-\frac {2 c^2 (c x)^{11/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {11 c^3 \left (\frac {c^2 (c x)^{5/2} \sqrt {a+b x^3}}{4 b}-\frac {5 a c^2 \left (\frac {\int \frac {2 b^{2/3} x^2 c^2+\left (1-\sqrt {3}\right ) a^{2/3} c^2}{\sqrt {b x^3+a}}d\sqrt {c x}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) a^{2/3} c^2 \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {c x}}{2 b^{2/3}}\right )}{4 b}\right )}{3 b}-\frac {2 c^2 (c x)^{11/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {11 c^3 \left (\frac {c^2 (c x)^{5/2} \sqrt {a+b x^3}}{4 b}-\frac {5 a c^2 \left (\frac {\int \frac {2 b^{2/3} x^2 c^2+\left (1-\sqrt {3}\right ) a^{2/3} c^2}{\sqrt {b x^3+a}}d\sqrt {c x}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a} c \sqrt {c x} \left (\sqrt [3]{a} c+\sqrt [3]{b} c x\right ) \sqrt {\frac {a^{2/3} c^2-\sqrt [3]{a} \sqrt [3]{b} c^2 x+b^{2/3} c^2 x^2}{\left (\sqrt [3]{a} c+\left (1+\sqrt {3}\right ) \sqrt [3]{b} c x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x c+\sqrt [3]{a} c}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x c+\sqrt [3]{a} c}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} b^{2/3} \sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} c x \left (\sqrt [3]{a} c+\sqrt [3]{b} c x\right )}{\left (\sqrt [3]{a} c+\left (1+\sqrt {3}\right ) \sqrt [3]{b} c x\right )^2}}}\right )}{4 b}\right )}{3 b}-\frac {2 c^2 (c x)^{11/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 2420

\(\displaystyle \frac {11 c^3 \left (\frac {c^2 (c x)^{5/2} \sqrt {a+b x^3}}{4 b}-\frac {5 a c^2 \left (\frac {\frac {\left (1+\sqrt {3}\right ) c^3 \sqrt {c x} \sqrt {a+b x^3}}{\sqrt [3]{a} c+\left (1+\sqrt {3}\right ) \sqrt [3]{b} c x}-\frac {\sqrt [4]{3} \sqrt [3]{a} c \sqrt {c x} \left (\sqrt [3]{a} c+\sqrt [3]{b} c x\right ) \sqrt {\frac {a^{2/3} c^2-\sqrt [3]{a} \sqrt [3]{b} c^2 x+b^{2/3} c^2 x^2}{\left (\sqrt [3]{a} c+\left (1+\sqrt {3}\right ) \sqrt [3]{b} c x\right )^2}} E\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x c+\sqrt [3]{a} c}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x c+\sqrt [3]{a} c}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} c x \left (\sqrt [3]{a} c+\sqrt [3]{b} c x\right )}{\left (\sqrt [3]{a} c+\left (1+\sqrt {3}\right ) \sqrt [3]{b} c x\right )^2}}}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a} c \sqrt {c x} \left (\sqrt [3]{a} c+\sqrt [3]{b} c x\right ) \sqrt {\frac {a^{2/3} c^2-\sqrt [3]{a} \sqrt [3]{b} c^2 x+b^{2/3} c^2 x^2}{\left (\sqrt [3]{a} c+\left (1+\sqrt {3}\right ) \sqrt [3]{b} c x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x c+\sqrt [3]{a} c}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x c+\sqrt [3]{a} c}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} b^{2/3} \sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} c x \left (\sqrt [3]{a} c+\sqrt [3]{b} c x\right )}{\left (\sqrt [3]{a} c+\left (1+\sqrt {3}\right ) \sqrt [3]{b} c x\right )^2}}}\right )}{4 b}\right )}{3 b}-\frac {2 c^2 (c x)^{11/2}}{3 b \sqrt {a+b x^3}}\)

Input:

Int[(c*x)^(15/2)/(a + b*x^3)^(3/2),x]
 

Output:

(-2*c^2*(c*x)^(11/2))/(3*b*Sqrt[a + b*x^3]) + (11*c^3*((c^2*(c*x)^(5/2)*Sq 
rt[a + b*x^3])/(4*b) - (5*a*c^2*((((1 + Sqrt[3])*c^3*Sqrt[c*x]*Sqrt[a + b* 
x^3])/(a^(1/3)*c + (1 + Sqrt[3])*b^(1/3)*c*x) - (3^(1/4)*a^(1/3)*c*Sqrt[c* 
x]*(a^(1/3)*c + b^(1/3)*c*x)*Sqrt[(a^(2/3)*c^2 - a^(1/3)*b^(1/3)*c^2*x + b 
^(2/3)*c^2*x^2)/(a^(1/3)*c + (1 + Sqrt[3])*b^(1/3)*c*x)^2]*EllipticE[ArcCo 
s[(a^(1/3)*c + (1 - Sqrt[3])*b^(1/3)*c*x)/(a^(1/3)*c + (1 + Sqrt[3])*b^(1/ 
3)*c*x)], (2 + Sqrt[3])/4])/(Sqrt[(b^(1/3)*c*x*(a^(1/3)*c + b^(1/3)*c*x))/ 
(a^(1/3)*c + (1 + Sqrt[3])*b^(1/3)*c*x)^2]*Sqrt[a + b*x^3]))/(2*b^(2/3)) - 
 ((1 - Sqrt[3])*a^(1/3)*c*Sqrt[c*x]*(a^(1/3)*c + b^(1/3)*c*x)*Sqrt[(a^(2/3 
)*c^2 - a^(1/3)*b^(1/3)*c^2*x + b^(2/3)*c^2*x^2)/(a^(1/3)*c + (1 + Sqrt[3] 
)*b^(1/3)*c*x)^2]*EllipticF[ArcCos[(a^(1/3)*c + (1 - Sqrt[3])*b^(1/3)*c*x) 
/(a^(1/3)*c + (1 + Sqrt[3])*b^(1/3)*c*x)], (2 + Sqrt[3])/4])/(4*3^(1/4)*b^ 
(2/3)*Sqrt[(b^(1/3)*c*x*(a^(1/3)*c + b^(1/3)*c*x))/(a^(1/3)*c + (1 + Sqrt[ 
3])*b^(1/3)*c*x)^2]*Sqrt[a + b*x^3])))/(4*b)))/(3*b)
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 766
Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 817
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( 
n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 837
Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 
3]], s = Denom[Rt[b/a, 3]]}, Simp[(Sqrt[3] - 1)*(s^2/(2*r^2))   Int[1/Sqrt[ 
a + b*x^6], x], x] - Simp[1/(2*r^2)   Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4)/S 
qrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]
 

rule 843
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 851
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = 
 Denominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ 
n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && 
FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 2420
Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = 
 Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqr 
t[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d*s*x* 
(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2 
*r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]) 
)*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 
 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 
- Sqrt[3])*d, 0]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.52 (sec) , antiderivative size = 1148, normalized size of antiderivative = 2.09

method result size
elliptic \(\text {Expression too large to display}\) \(1148\)
risch \(\text {Expression too large to display}\) \(2203\)
default \(\text {Expression too large to display}\) \(2909\)

Input:

int((c*x)^(15/2)/(b*x^3+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/c/x*(c*x)^(1/2)/(b*x^3+a)^(1/2)*(c*x*(b*x^3+a))^(1/2)*(2/3/b^2*c^8*x^3*a 
/((x^3+a/b)*b*c*x)^(1/2)+1/4/b^2*c^7*x^2*(b*c*x^4+a*c*x)^(1/2)-55/24*a*c^8 
/b^2*(x*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*(x+1/2/b*( 
-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))+(1/2/b*(-a*b^2)^(1/3)-1/2*I* 
3^(1/2)/b*(-a*b^2)^(1/3))*((-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2) 
^(1/3))*x/(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(- 
a*b^2)^(1/3)))^(1/2)*(x-1/b*(-a*b^2)^(1/3))^2*(1/b*(-a*b^2)^(1/3)*(x+1/2/b 
*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(-1/2/b*(-a*b^2)^(1/3)-1/2 
*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/3)))^(1/2)*(1/b*(-a*b^2)^( 
1/3)*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(-1/2/b*(-a*b 
^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/3)))^(1/2)*(( 
(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/b*(-a*b^2)^(1/3)+1/ 
b^2*(-a*b^2)^(2/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)) 
*b/(-a*b^2)^(1/3)*EllipticF(((-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^ 
2)^(1/3))*x/(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b* 
(-a*b^2)^(1/3)))^(1/2),((3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/ 
3))*(1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(1/2/b*(-a*b^2)^ 
(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(3/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/ 
b*(-a*b^2)^(1/3)))^(1/2))+(1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^( 
1/3))*EllipticE(((-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))...
 

Fricas [F]

\[ \int \frac {(c x)^{15/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {15}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c*x)^(15/2)/(b*x^3+a)^(3/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x^3 + a)*sqrt(c*x)*c^7*x^7/(b^2*x^6 + 2*a*b*x^3 + a^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c x)^{15/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((c*x)**(15/2)/(b*x**3+a)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c x)^{15/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {15}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c*x)^(15/2)/(b*x^3+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^(15/2)/(b*x^3 + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {(c x)^{15/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {15}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c*x)^(15/2)/(b*x^3+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((c*x)^(15/2)/(b*x^3 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{15/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\int \frac {{\left (c\,x\right )}^{15/2}}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \] Input:

int((c*x)^(15/2)/(a + b*x^3)^(3/2),x)
 

Output:

int((c*x)^(15/2)/(a + b*x^3)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(c x)^{15/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {c}\, c^{7} \left (-22 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a \,x^{2}+4 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b \,x^{5}+55 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{3}+a}\, x}{b^{2} x^{6}+2 a b \,x^{3}+a^{2}}d x \right ) a^{3}+55 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{3}+a}\, x}{b^{2} x^{6}+2 a b \,x^{3}+a^{2}}d x \right ) a^{2} b \,x^{3}\right )}{16 b^{2} \left (b \,x^{3}+a \right )} \] Input:

int((c*x)^(15/2)/(b*x^3+a)^(3/2),x)
 

Output:

(sqrt(c)*c**7*( - 22*sqrt(x)*sqrt(a + b*x**3)*a*x**2 + 4*sqrt(x)*sqrt(a + 
b*x**3)*b*x**5 + 55*int((sqrt(x)*sqrt(a + b*x**3)*x)/(a**2 + 2*a*b*x**3 + 
b**2*x**6),x)*a**3 + 55*int((sqrt(x)*sqrt(a + b*x**3)*x)/(a**2 + 2*a*b*x** 
3 + b**2*x**6),x)*a**2*b*x**3))/(16*b**2*(a + b*x**3))