\(\int \frac {A+B x^3}{(e x)^{7/2} (a+b x^3)^{5/2}} \, dx\) [288]

Optimal result

Integrand size = 26, antiderivative size = 320 \[ \int \frac {A+B x^3}{(e x)^{7/2} \left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 A}{5 a e (e x)^{5/2} \left (a+b x^3\right )^{3/2}}-\frac {2 (14 A b-5 a B) \sqrt {e x}}{45 a^2 e^4 \left (a+b x^3\right )^{3/2}}-\frac {16 (14 A b-5 a B) \sqrt {e x}}{135 a^3 e^4 \sqrt {a+b x^3}}-\frac {16 (14 A b-5 a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{135 \sqrt [4]{3} a^{10/3} e^4 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \] Output:

-2/5*A/a/e/(e*x)^(5/2)/(b*x^3+a)^(3/2)-2/45*(14*A*b-5*B*a)*(e*x)^(1/2)/a^2 
/e^4/(b*x^3+a)^(3/2)-16/135*(14*A*b-5*B*a)*(e*x)^(1/2)/a^3/e^4/(b*x^3+a)^( 
1/2)-16/405*(14*A*b-5*B*a)*(e*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)*Inverse 
JacobiAM(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)/a^(10/3)/e^4/(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 = 10.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.38 \[ \int \frac {A+B x^3}{(e x)^{7/2} \left (a+b x^3\right )^{5/2}} \, dx=\frac {x \left (-224 A b^2 x^6+a^2 \left (-54 A+110 B x^3\right )+a \left (-308 A b x^3+80 b B x^6\right )+32 (-14 A b+5 a B) x^3 \left (a+b x^3\right ) \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-\frac {b x^3}{a}\right )\right )}{135 a^3 (e x)^{7/2} \left (a+b x^3\right )^{3/2}} \] Input:

Integrate[(A + B*x^3)/((e*x)^(7/2)*(a + b*x^3)^(5/2)),x]
 

Output:

(x*(-224*A*b^2*x^6 + a^2*(-54*A + 110*B*x^3) + a*(-308*A*b*x^3 + 80*b*B*x^ 
6) + 32*(-14*A*b + 5*a*B)*x^3*(a + b*x^3)*Sqrt[1 + (b*x^3)/a]*Hypergeometr 
ic2F1[1/6, 1/2, 7/6, -((b*x^3)/a)]))/(135*a^3*(e*x)^(7/2)*(a + b*x^3)^(3/2 
))
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {955, 819, 819, 851, 766}

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.

\(\displaystyle \int \frac {A+B x^3}{(e x)^{7/2} \left (a+b x^3\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 955

\(\displaystyle -\frac {(14 A b-5 a B) \int \frac {1}{\sqrt {e x} \left (b x^3+a\right )^{5/2}}dx}{5 a e^3}-\frac {2 A}{5 a e (e x)^{5/2} \left (a+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 819

\(\displaystyle -\frac {(14 A b-5 a B) \left (\frac {8 \int \frac {1}{\sqrt {e x} \left (b x^3+a\right )^{3/2}}dx}{9 a}+\frac {2 \sqrt {e x}}{9 a e \left (a+b x^3\right )^{3/2}}\right )}{5 a e^3}-\frac {2 A}{5 a e (e x)^{5/2} \left (a+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 819

\(\displaystyle -\frac {(14 A b-5 a B) \left (\frac {8 \left (\frac {2 \int \frac {1}{\sqrt {e x} \sqrt {b x^3+a}}dx}{3 a}+\frac {2 \sqrt {e x}}{3 a e \sqrt {a+b x^3}}\right )}{9 a}+\frac {2 \sqrt {e x}}{9 a e \left (a+b x^3\right )^{3/2}}\right )}{5 a e^3}-\frac {2 A}{5 a e (e x)^{5/2} \left (a+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 851

\(\displaystyle -\frac {(14 A b-5 a B) \left (\frac {8 \left (\frac {4 \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {e x}}{3 a e}+\frac {2 \sqrt {e x}}{3 a e \sqrt {a+b x^3}}\right )}{9 a}+\frac {2 \sqrt {e x}}{9 a e \left (a+b x^3\right )^{3/2}}\right )}{5 a e^3}-\frac {2 A}{5 a e (e x)^{5/2} \left (a+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 766

\(\displaystyle -\frac {(14 A b-5 a B) \left (\frac {8 \left (\frac {2 \sqrt {e x} \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right ) \sqrt {\frac {a^{2/3} e^2-\sqrt [3]{a} \sqrt [3]{b} e^2 x+b^{2/3} e^2 x^2}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{3 \sqrt [4]{3} a^{4/3} e^2 \sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} e x \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right )}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}}}+\frac {2 \sqrt {e x}}{3 a e \sqrt {a+b x^3}}\right )}{9 a}+\frac {2 \sqrt {e x}}{9 a e \left (a+b x^3\right )^{3/2}}\right )}{5 a e^3}-\frac {2 A}{5 a e (e x)^{5/2} \left (a+b x^3\right )^{3/2}}\)

Input:

Int[(A + B*x^3)/((e*x)^(7/2)*(a + b*x^3)^(5/2)),x]
 

Output:

(-2*A)/(5*a*e*(e*x)^(5/2)*(a + b*x^3)^(3/2)) - ((14*A*b - 5*a*B)*((2*Sqrt[ 
e*x])/(9*a*e*(a + b*x^3)^(3/2)) + (8*((2*Sqrt[e*x])/(3*a*e*Sqrt[a + b*x^3] 
) + (2*Sqrt[e*x]*(a^(1/3)*e + b^(1/3)*e*x)*Sqrt[(a^(2/3)*e^2 - a^(1/3)*b^( 
1/3)*e^2*x + b^(2/3)*e^2*x^2)/(a^(1/3)*e + (1 + Sqrt[3])*b^(1/3)*e*x)^2]*E 
llipticF[ArcCos[(a^(1/3)*e + (1 - Sqrt[3])*b^(1/3)*e*x)/(a^(1/3)*e + (1 + 
Sqrt[3])*b^(1/3)*e*x)], (2 + Sqrt[3])/4])/(3*3^(1/4)*a^(4/3)*e^2*Sqrt[(b^( 
1/3)*e*x*(a^(1/3)*e + b^(1/3)*e*x))/(a^(1/3)*e + (1 + Sqrt[3])*b^(1/3)*e*x 
)^2]*Sqrt[a + b*x^3])))/(9*a)))/(5*a*e^3)
 

Defintions of rubi rules used

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 819

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( 
c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 
1) + 1)/(a*n*(p + 1))   Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a 
, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 955

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), 
 x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1))   Int[(e 
*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* 
c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || 
(LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 12.21 (sec) , antiderivative size = 829, normalized size of antiderivative = 2.59

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(829\)
risch	\(\text {Expression too large to display}\)	\(2182\)
default	\(\text {Expression too large to display}\)	\(7299\)

Input:

int((B*x^3+A)/(e*x)^(7/2)/(b*x^3+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

((b*x^3+a)*e*x)^(1/2)/(e*x)^(1/2)/(b*x^3+a)^(1/2)*(-2/5/e^4/a^3*A*(b*e*x^4 
+a*e*x)^(1/2)/x^3-2/9/e^4/a^2/b^2*(A*b-B*a)*(b*e*x^4+a*e*x)^(1/2)/(x^3+a/b 
)^2-2/27/e^3*x/a^3*(17*A*b-8*B*a)/((x^3+a/b)*b*e*x)^(1/2)+2*(-2/5*b/a^3/e^ 
3*A-2/27/a^3*(17*A*b-8*B*a)/e^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)/(-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)/(b*e*x*(x-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))*(x+1/ 
2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*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)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.56 \[ \int \frac {A+B x^3}{(e x)^{7/2} \left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 \, {\left (16 \, {\left ({\left (5 \, B a b^{2} - 14 \, A b^{3}\right )} x^{9} + 2 \, {\left (5 \, B a^{2} b - 14 \, A a b^{2}\right )} x^{6} + {\left (5 \, B a^{3} - 14 \, A a^{2} b\right )} x^{3}\right )} \sqrt {a e} {\rm weierstrassPInverse}\left (0, -\frac {4 \, b}{a}, \frac {1}{x}\right ) - {\left (8 \, {\left (5 \, B a^{2} b - 14 \, A a b^{2}\right )} x^{6} - 27 \, A a^{3} + 11 \, {\left (5 \, B a^{3} - 14 \, A a^{2} b\right )} x^{3}\right )} \sqrt {b x^{3} + a} \sqrt {e x}\right )}}{135 \, {\left (a^{4} b^{2} e^{4} x^{9} + 2 \, a^{5} b e^{4} x^{6} + a^{6} e^{4} x^{3}\right )}} \] Input:

integrate((B*x^3+A)/(e*x)^(7/2)/(b*x^3+a)^(5/2),x, algorithm="fricas")
 

Output:

-2/135*(16*((5*B*a*b^2 - 14*A*b^3)*x^9 + 2*(5*B*a^2*b - 14*A*a*b^2)*x^6 + 
(5*B*a^3 - 14*A*a^2*b)*x^3)*sqrt(a*e)*weierstrassPInverse(0, -4*b/a, 1/x) 
- (8*(5*B*a^2*b - 14*A*a*b^2)*x^6 - 27*A*a^3 + 11*(5*B*a^3 - 14*A*a^2*b)*x 
^3)*sqrt(b*x^3 + a)*sqrt(e*x))/(a^4*b^2*e^4*x^9 + 2*a^5*b*e^4*x^6 + a^6*e^ 
4*x^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x^3}{(e x)^{7/2} \left (a+b x^3\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x**3+A)/(e*x)**(7/2)/(b*x**3+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B x^3}{(e x)^{7/2} \left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {5}{2}} \left (e x\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((B*x^3+A)/(e*x)^(7/2)/(b*x^3+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)/((b*x^3 + a)^(5/2)*(e*x)^(7/2)), x)
 

Giac [F]

\[ \int \frac {A+B x^3}{(e x)^{7/2} \left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {5}{2}} \left (e x\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((B*x^3+A)/(e*x)^(7/2)/(b*x^3+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)/((b*x^3 + a)^(5/2)*(e*x)^(7/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^3}{(e x)^{7/2} \left (a+b x^3\right )^{5/2}} \, dx=\int \frac {B\,x^3+A}{{\left (e\,x\right )}^{7/2}\,{\left (b\,x^3+a\right )}^{5/2}} \,d x \] Input:

int((A + B*x^3)/((e*x)^(7/2)*(a + b*x^3)^(5/2)),x)
 

Output:

int((A + B*x^3)/((e*x)^(7/2)*(a + b*x^3)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {A+B x^3}{(e x)^{7/2} \left (a+b x^3\right )^{5/2}} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{3}+a}}{b^{2} x^{10}+2 a b \,x^{7}+a^{2} x^{4}}d x \right )}{e^{4}} \] Input:

int((B*x^3+A)/(e*x)^(7/2)/(b*x^3+a)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(e)*int((sqrt(x)*sqrt(a + b*x**3))/(a**2*x**4 + 2*a*b*x**7 + b**2*x** 
10),x))/e**4