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

Optimal. Leaf size=352 \[ \frac {27\ 3^{3/4} a^{5/3} \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} (5 a B+16 A b) F\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{640 e^4 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {\sqrt {e x} \left (a+b x^3\right )^{5/2} (5 a B+16 A b)}{40 a e^4}+\frac {3 \sqrt {e x} \left (a+b x^3\right )^{3/2} (5 a B+16 A b)}{80 e^4}+\frac {27 a \sqrt {e x} \sqrt {a+b x^3} (5 a B+16 A b)}{320 e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}} \]

[Out]

-2/5*A*(b*x^3+a)^(7/2)/a/e/(e*x)^(5/2)+3/80*(16*A*b+5*B*a)*(b*x^3+a)^(3/2)*(e*x)^(1/2)/e^4+1/40*(16*A*b+5*B*a)
*(b*x^3+a)^(5/2)*(e*x)^(1/2)/a/e^4+27/320*a*(16*A*b+5*B*a)*(e*x)^(1/2)*(b*x^3+a)^(1/2)/e^4+27/640*3^(3/4)*a^(5
/3)*(16*A*b+5*B*a)*(a^(1/3)+b^(1/3)*x)*((a^(1/3)+b^(1/3)*x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(
1/2)/(a^(1/3)+b^(1/3)*x*(1-3^(1/2)))*(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))*EllipticF((1-(a^(1/3)+b^(1/3)*x*(1-3^(1/2
)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(e*x)^(1/2)*((a^(2/3)-a^(1/3)*b^(1/3)*
x+b^(2/3)*x^2)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/e^4/(b*x^3+a)^(1/2)/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^
(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 352, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {453, 279, 329, 225} \[ \frac {27\ 3^{3/4} a^{5/3} \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} (5 a B+16 A b) F\left (\cos ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{640 e^4 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {\sqrt {e x} \left (a+b x^3\right )^{5/2} (5 a B+16 A b)}{40 a e^4}+\frac {3 \sqrt {e x} \left (a+b x^3\right )^{3/2} (5 a B+16 A b)}{80 e^4}+\frac {27 a \sqrt {e x} \sqrt {a+b x^3} (5 a B+16 A b)}{320 e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(27*a*(16*A*b + 5*a*B)*Sqrt[e*x]*Sqrt[a + b*x^3])/(320*e^4) + (3*(16*A*b + 5*a*B)*Sqrt[e*x]*(a + b*x^3)^(3/2))
/(80*e^4) + ((16*A*b + 5*a*B)*Sqrt[e*x]*(a + b*x^3)^(5/2))/(40*a*e^4) - (2*A*(a + b*x^3)^(7/2))/(5*a*e*(e*x)^(
5/2)) + (27*3^(3/4)*a^(5/3)*(16*A*b + 5*a*B)*Sqrt[e*x]*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x
+ b^(2/3)*x^2)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)^2]*EllipticF[ArcCos[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a
^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(640*e^4*Sqrt[(b^(1/3)*x*(a^(1/3) + b^(1/3)*x))/(a^(1/3)
+ (1 + Sqrt[3])*b^(1/3)*x)^2]*Sqrt[a + b*x^3])

Rule 225

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]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(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]), x]] /; FreeQ[{a, b}, x]

Rule 279

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[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 453

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] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right )^{5/2} \left (A+B x^3\right )}{(e x)^{7/2}} \, dx &=-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {(16 A b+5 a B) \int \frac {\left (a+b x^3\right )^{5/2}}{\sqrt {e x}} \, dx}{5 a e^3}\\ &=\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {(3 (16 A b+5 a B)) \int \frac {\left (a+b x^3\right )^{3/2}}{\sqrt {e x}} \, dx}{16 e^3}\\ &=\frac {3 (16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{3/2}}{80 e^4}+\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {(27 a (16 A b+5 a B)) \int \frac {\sqrt {a+b x^3}}{\sqrt {e x}} \, dx}{160 e^3}\\ &=\frac {27 a (16 A b+5 a B) \sqrt {e x} \sqrt {a+b x^3}}{320 e^4}+\frac {3 (16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{3/2}}{80 e^4}+\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {\left (81 a^2 (16 A b+5 a B)\right ) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^3}} \, dx}{640 e^3}\\ &=\frac {27 a (16 A b+5 a B) \sqrt {e x} \sqrt {a+b x^3}}{320 e^4}+\frac {3 (16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{3/2}}{80 e^4}+\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {\left (81 a^2 (16 A b+5 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{320 e^4}\\ &=\frac {27 a (16 A b+5 a B) \sqrt {e x} \sqrt {a+b x^3}}{320 e^4}+\frac {3 (16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{3/2}}{80 e^4}+\frac {(16 A b+5 a B) \sqrt {e x} \left (a+b x^3\right )^{5/2}}{40 a e^4}-\frac {2 A \left (a+b x^3\right )^{7/2}}{5 a e (e x)^{5/2}}+\frac {27\ 3^{3/4} a^{5/3} (16 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}} F\left (\cos ^{-1}\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 )}{640 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}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 88, normalized size = 0.25 \[ \frac {2 x \sqrt {a+b x^3} \left (\frac {a^2 x^3 (5 a B+16 A b) \, _2F_1\left (-\frac {5}{2},\frac {1}{6};\frac {7}{6};-\frac {b x^3}{a}\right )}{\sqrt {\frac {b x^3}{a}+1}}-A \left (a+b x^3\right )^3\right )}{5 a (e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[a + b*x^3]*(-(A*(a + b*x^3)^3) + (a^2*(16*A*b + 5*a*B)*x^3*Hypergeometric2F1[-5/2, 1/6, 7/6, -((b*x^
3)/a)])/Sqrt[1 + (b*x^3)/a]))/(5*a*(e*x)^(7/2))

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B b^{2} x^{9} + {\left (2 \, B a b + A b^{2}\right )} x^{6} + {\left (B a^{2} + 2 \, A a b\right )} x^{3} + A a^{2}\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^2*x^9 + (2*B*a*b + A*b^2)*x^6 + (B*a^2 + 2*A*a*b)*x^3 + A*a^2)*sqrt(b*x^3 + a)*sqrt(e*x)/(e^4*x^
4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {5}{2}}}{\left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.10, size = 4422, normalized size = 12.56 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/320*(b*x^3+a)^(1/2)*(2592*I*A*3^(1/2)*(-a*b^2)^(2/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)
))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/
2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x
*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))
*x^3*a^2*b*e-64*I*A*3^(1/2)*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(
-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^6*b^3-810*B*(-(I*
3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+
I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(
-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*
(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^5*a^3*b^2*e-810*B*(-a*b^2)^(2/3)*(-(I*3^(1/2)-3)*x*b/(I*3^
(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a
*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2
)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3
^(1/2))/(I*3^(1/2)-3))^(1/2))*x^3*a^3*e+810*I*B*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)
))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/
2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x
*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))
*x^5*a^3*b^2*e-2592*A*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3
)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(
1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1
/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^5*a^2*b^3*e+120*B*(e*x*(b*x^3+a
))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^
(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^9*b^3+192*A*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*
x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^
2)^(1/3)))^(1/2)*x^6*b^3-384*A*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2
)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*a^2*b+5184*A*(-a
*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-
a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3
^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2
),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^4*a^2*b^2*e+1620*B*(-a*b^2)^(1/3)*(-(I*3^
(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*
3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a
*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I
*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^4*a^3*b*e-2592*A*(-a*b^2)^(2/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/
2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^
2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*E
llipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1
/2))/(I*3^(1/2)-3))^(1/2))*x^3*a^2*b*e+810*I*B*3^(1/2)*(-a*b^2)^(2/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+
(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(
1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(
I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1
/2)-3))^(1/2))*x^3*a^3*e+128*I*A*3^(1/2)*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))
*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*a^2*b-
1620*I*B*3^(1/2)*(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a
*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-
(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(
-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^4*a^3*b*e+1104*A*(e*x
*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/
3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^3*a*b^2+705*B*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3
)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2
*b*x-(-a*b^2)^(1/3)))^(1/2)*x^3*a^2*b+420*B*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/
3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^6
*a*b^2-40*I*B*3^(1/2)*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2
)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^9*b^3-5184*I*A*3^(1/2)*
(-a*b^2)^(1/3)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x
+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(
I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(
1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^4*a^2*b^2*e-235*I*B*3^(1/2)*(e*x*(b*x^
3+a))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I
*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^3*a^2*b-368*I*A*3^(1/2)*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^
(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/
3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^3*a*b^2+2592*I*A*3^(1/2)*(-(I*3^(1/2)-3)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1
/3)))^(1/2)*((I*3^(1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))/(1+I*3^(1/2))/(-b*x+(-a*b^2)^(1/3)))^(1/2)*((I*3^
(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3
)*x*b/(I*3^(1/2)-1)/(-b*x+(-a*b^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/
2))*x^5*a^2*b^3*e-140*I*B*3^(1/2)*(e*x*(b*x^3+a))^(1/2)*(-a*b^2)^(1/3)*(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(
1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)*x^6*a*b^2)/x^
2/b/(-a*b^2)^(1/3)/e^3/(e*x)^(1/2)/(e*x*(b*x^3+a))^(1/2)/(I*3^(1/2)-3)/(1/b^2*e*x*(-b*x+(-a*b^2)^(1/3))*(I*3^(
1/2)*(-a*b^2)^(1/3)+2*b*x+(-a*b^2)^(1/3))*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3)))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {5}{2}}}{\left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^3+A\right )\,{\left (b\,x^3+a\right )}^{5/2}}{{\left (e\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 85.78, size = 311, normalized size = 0.88 \[ \frac {A a^{\frac {5}{2}} \Gamma \left (- \frac {5}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{6}, - \frac {1}{2} \\ \frac {1}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {1}{6}\right )} + \frac {2 A a^{\frac {3}{2}} b \sqrt {x} \Gamma \left (\frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{6} \\ \frac {7}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right )} + \frac {A \sqrt {a} b^{2} x^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{6} \\ \frac {13}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {13}{6}\right )} + \frac {B a^{\frac {5}{2}} \sqrt {x} \Gamma \left (\frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{6} \\ \frac {7}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right )} + \frac {2 B a^{\frac {3}{2}} b x^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{6} \\ \frac {13}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {13}{6}\right )} + \frac {B \sqrt {a} b^{2} x^{\frac {13}{2}} \Gamma \left (\frac {13}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {13}{6} \\ \frac {19}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {7}{2}} \Gamma \left (\frac {19}{6}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(5/2)*gamma(-5/6)*hyper((-5/6, -1/2), (1/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)*x**(5/2)*gamma(1/6))
+ 2*A*a**(3/2)*b*sqrt(x)*gamma(1/6)*hyper((-1/2, 1/6), (7/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)*gamma(7/6
)) + A*sqrt(a)*b**2*x**(7/2)*gamma(7/6)*hyper((-1/2, 7/6), (13/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)*gamm
a(13/6)) + B*a**(5/2)*sqrt(x)*gamma(1/6)*hyper((-1/2, 1/6), (7/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)*gamm
a(7/6)) + 2*B*a**(3/2)*b*x**(7/2)*gamma(7/6)*hyper((-1/2, 7/6), (13/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(7/2)
*gamma(13/6)) + B*sqrt(a)*b**2*x**(13/2)*gamma(13/6)*hyper((-1/2, 13/6), (19/6,), b*x**3*exp_polar(I*pi)/a)/(3
*e**(7/2)*gamma(19/6))

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