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

Optimal. Leaf size=152 \[ \frac{(e x)^{3/2} \left (a+b x^3\right )^{3/2} (a B+4 A b)}{6 a e^4}+\frac{(e x)^{3/2} \sqrt{a+b x^3} (a B+4 A b)}{4 e^4}+\frac{a (a B+4 A b) \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{4 \sqrt{b} e^{5/2}}-\frac{2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}} \]

[Out]

((4*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^3])/(4*e^4) + ((4*A*b + a*B)*(e*x)^(3/2)*(a + b*x^3)^(3/2))/(6*a*e^4)
- (2*A*(a + b*x^3)^(5/2))/(3*a*e*(e*x)^(3/2)) + (a*(4*A*b + a*B)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a
+ b*x^3])])/(4*Sqrt[b]*e^(5/2))

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Rubi [A]  time = 0.104042, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {453, 279, 329, 275, 217, 206} \[ \frac{(e x)^{3/2} \left (a+b x^3\right )^{3/2} (a B+4 A b)}{6 a e^4}+\frac{(e x)^{3/2} \sqrt{a+b x^3} (a B+4 A b)}{4 e^4}+\frac{a (a B+4 A b) \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{4 \sqrt{b} e^{5/2}}-\frac{2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^3])/(4*e^4) + ((4*A*b + a*B)*(e*x)^(3/2)*(a + b*x^3)^(3/2))/(6*a*e^4)
- (2*A*(a + b*x^3)^(5/2))/(3*a*e*(e*x)^(3/2)) + (a*(4*A*b + a*B)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a
+ b*x^3])])/(4*Sqrt[b]*e^(5/2))

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 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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.145976, size = 126, normalized size = 0.83 \[ \frac{x \sqrt{a+b x^3} \left (\sqrt{b} \sqrt{\frac{b x^3}{a}+1} \left (-8 a A+5 a B x^3+4 A b x^3+2 b B x^6\right )+3 \sqrt{a} x^{3/2} (a B+4 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )\right )}{12 \sqrt{b} (e x)^{5/2} \sqrt{\frac{b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x^3]*(Sqrt[b]*Sqrt[1 + (b*x^3)/a]*(-8*a*A + 4*A*b*x^3 + 5*a*B*x^3 + 2*b*B*x^6) + 3*Sqrt[a]*(4*A*
b + a*B)*x^(3/2)*ArcSinh[(Sqrt[b]*x^(3/2))/Sqrt[a]]))/(12*Sqrt[b]*(e*x)^(5/2)*Sqrt[1 + (b*x^3)/a])

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Maple [C]  time = 0.049, size = 7108, normalized size = 46.8 \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)^(3/2)*(B*x^3+A)/(e*x)^(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{{\left (B x^{3} + A\right )}{\left (b x^{3} + a\right )}^{\frac{3}{2}}}{\left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 4.1967, size = 590, normalized size = 3.88 \begin{align*} \left [\frac{3 \,{\left (B a^{2} + 4 \, A a b\right )} \sqrt{b e} x^{2} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \,{\left (2 \, b x^{4} + a x\right )} \sqrt{b x^{3} + a} \sqrt{b e} \sqrt{e x}\right ) + 4 \,{\left (2 \, B b^{2} x^{6} +{\left (5 \, B a b + 4 \, A b^{2}\right )} x^{3} - 8 \, A a b\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{48 \, b e^{3} x^{2}}, -\frac{3 \,{\left (B a^{2} + 4 \, A a b\right )} \sqrt{-b e} x^{2} \arctan \left (\frac{2 \, \sqrt{b x^{3} + a} \sqrt{-b e} \sqrt{e x} x}{2 \, b e x^{3} + a e}\right ) - 2 \,{\left (2 \, B b^{2} x^{6} +{\left (5 \, B a b + 4 \, A b^{2}\right )} x^{3} - 8 \, A a b\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{24 \, b e^{3} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*(B*a^2 + 4*A*a*b)*sqrt(b*e)*x^2*log(-8*b^2*e*x^6 - 8*a*b*e*x^3 - a^2*e - 4*(2*b*x^4 + a*x)*sqrt(b*x^3
+ a)*sqrt(b*e)*sqrt(e*x)) + 4*(2*B*b^2*x^6 + (5*B*a*b + 4*A*b^2)*x^3 - 8*A*a*b)*sqrt(b*x^3 + a)*sqrt(e*x))/(b
*e^3*x^2), -1/24*(3*(B*a^2 + 4*A*a*b)*sqrt(-b*e)*x^2*arctan(2*sqrt(b*x^3 + a)*sqrt(-b*e)*sqrt(e*x)*x/(2*b*e*x^
3 + a*e)) - 2*(2*B*b^2*x^6 + (5*B*a*b + 4*A*b^2)*x^3 - 8*A*a*b)*sqrt(b*x^3 + a)*sqrt(e*x))/(b*e^3*x^2)]

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Sympy [B]  time = 62.0412, size = 289, normalized size = 1.9 \begin{align*} - \frac{2 A a^{\frac{3}{2}}}{3 e^{\frac{5}{2}} x^{\frac{3}{2}} \sqrt{1 + \frac{b x^{3}}{a}}} + \frac{A \sqrt{a} b x^{\frac{3}{2}} \sqrt{1 + \frac{b x^{3}}{a}}}{3 e^{\frac{5}{2}}} - \frac{2 A \sqrt{a} b x^{\frac{3}{2}}}{3 e^{\frac{5}{2}} \sqrt{1 + \frac{b x^{3}}{a}}} + \frac{A a \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} x^{\frac{3}{2}}}{\sqrt{a}} \right )}}{e^{\frac{5}{2}}} + \frac{B a^{\frac{3}{2}} x^{\frac{3}{2}} \sqrt{1 + \frac{b x^{3}}{a}}}{3 e^{\frac{5}{2}}} + \frac{B a^{\frac{3}{2}} x^{\frac{3}{2}}}{12 e^{\frac{5}{2}} \sqrt{1 + \frac{b x^{3}}{a}}} + \frac{B \sqrt{a} b x^{\frac{9}{2}}}{4 e^{\frac{5}{2}} \sqrt{1 + \frac{b x^{3}}{a}}} + \frac{B a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x^{\frac{3}{2}}}{\sqrt{a}} \right )}}{4 \sqrt{b} e^{\frac{5}{2}}} + \frac{B b^{2} x^{\frac{15}{2}}}{6 \sqrt{a} e^{\frac{5}{2}} \sqrt{1 + \frac{b x^{3}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A*a**(3/2)/(3*e**(5/2)*x**(3/2)*sqrt(1 + b*x**3/a)) + A*sqrt(a)*b*x**(3/2)*sqrt(1 + b*x**3/a)/(3*e**(5/2))
- 2*A*sqrt(a)*b*x**(3/2)/(3*e**(5/2)*sqrt(1 + b*x**3/a)) + A*a*sqrt(b)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/e**(5/2
) + B*a**(3/2)*x**(3/2)*sqrt(1 + b*x**3/a)/(3*e**(5/2)) + B*a**(3/2)*x**(3/2)/(12*e**(5/2)*sqrt(1 + b*x**3/a))
+ B*sqrt(a)*b*x**(9/2)/(4*e**(5/2)*sqrt(1 + b*x**3/a)) + B*a**2*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(4*sqrt(b)*e*
*(5/2)) + B*b**2*x**(15/2)/(6*sqrt(a)*e**(5/2)*sqrt(1 + b*x**3/a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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