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

Optimal. Leaf size=320 \[ -\frac{16 \sqrt{e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} (14 A b-5 a B) \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{a}+\left (1-\sqrt{3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{135 \sqrt [4]{3} a^{10/3} e^4 \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{16 \sqrt{e x} (14 A b-5 a B)}{135 a^3 e^4 \sqrt{a+b x^3}}-\frac{2 \sqrt{e x} (14 A b-5 a B)}{45 a^2 e^4 \left (a+b x^3\right )^{3/2}}-\frac{2 A}{5 a e (e x)^{5/2} \left (a+b x^3\right )^{3/2}} \]

[Out]

(-2*A)/(5*a*e*(e*x)^(5/2)*(a + b*x^3)^(3/2)) - (2*(14*A*b - 5*a*B)*Sqrt[e*x])/(45*a^2*e^4*(a + b*x^3)^(3/2)) -
(16*(14*A*b - 5*a*B)*Sqrt[e*x])/(135*a^3*e^4*Sqrt[a + b*x^3]) - (16*(14*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[ArcC
os[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(135*3^(1/4)*a^
(10/3)*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])

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Rubi [A]  time = 0.257692, antiderivative size = 320, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {453, 290, 329, 225} \[ -\frac{16 \sqrt{e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} (14 A b-5 a B) F\left (\cos ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}{\left (1+\sqrt{3}\right ) \sqrt [3]{b} x+\sqrt [3]{a}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{135 \sqrt [4]{3} a^{10/3} e^4 \sqrt{\frac{\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt{3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{16 \sqrt{e x} (14 A b-5 a B)}{135 a^3 e^4 \sqrt{a+b x^3}}-\frac{2 \sqrt{e x} (14 A b-5 a B)}{45 a^2 e^4 \left (a+b x^3\right )^{3/2}}-\frac{2 A}{5 a e (e x)^{5/2} \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(5*a*e*(e*x)^(5/2)*(a + b*x^3)^(3/2)) - (2*(14*A*b - 5*a*B)*Sqrt[e*x])/(45*a^2*e^4*(a + b*x^3)^(3/2)) -
(16*(14*A*b - 5*a*B)*Sqrt[e*x])/(135*a^3*e^4*Sqrt[a + b*x^3]) - (16*(14*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[ArcC
os[(a^(1/3) + (1 - Sqrt[3])*b^(1/3)*x)/(a^(1/3) + (1 + Sqrt[3])*b^(1/3)*x)], (2 + Sqrt[3])/4])/(135*3^(1/4)*a^
(10/3)*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 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]

Rule 290

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] + Dist[(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 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 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]

Rubi steps

\begin{align*} \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{(14 A b-5 a B) \int \frac{1}{\sqrt{e x} \left (a+b x^3\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}}-\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{(8 (14 A b-5 a B)) \int \frac{1}{\sqrt{e x} \left (a+b x^3\right )^{3/2}} \, dx}{45 a^2 e^3}\\ &=-\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)) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^3}} \, dx}{135 a^3 e^3}\\ &=-\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{(32 (14 A b-5 a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^6}{e^3}}} \, dx,x,\sqrt{e x}\right )}{135 a^3 e^4}\\ &=-\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}} 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 )}{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}}\\ \end{align*}

Mathematica [C]  time = 0.0750694, size = 121, normalized size = 0.38 \[ \frac{x \left (a^2 \left (110 B x^3-54 A\right )+32 x^3 \left (a+b x^3\right ) \sqrt{\frac{b x^3}{a}+1} (5 a B-14 A b) \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{b x^3}{a}\right )+a \left (80 b B x^6-308 A b x^3\right )-224 A b^2 x^6\right )}{135 a^3 (e x)^{7/2} \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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]*Hypergeometric2F1[1/6, 1/2, 7/6, -((b*x^3)/a)]))/(135*a^3*(e*x)^(7/2)*(a + b*x^3)^(3
/2))

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Maple [C]  time = 0.066, size = 7299, normalized size = 22.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{5}{2}} \left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{3} + A\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{b^{3} e^{4} x^{13} + 3 \, a b^{2} e^{4} x^{10} + 3 \, a^{2} b e^{4} x^{7} + a^{3} e^{4} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^3 + A)*sqrt(b*x^3 + a)*sqrt(e*x)/(b^3*e^4*x^13 + 3*a*b^2*e^4*x^10 + 3*a^2*b*e^4*x^7 + a^3*e^4*x^
4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{5}{2}} \left (e x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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