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

Optimal. Leaf size=118 \[ \frac{(e x)^{3/2} \sqrt{a+b x^3} (a B+2 A b)}{3 a e^4}+\frac{(a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{3 \sqrt{b} e^{5/2}}-\frac{2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0844829, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{(e x)^{3/2} \sqrt{a+b x^3} (a B+2 A b)}{3 a e^4}+\frac{(a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{3 \sqrt{b} e^{5/2}}-\frac{2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^3])/(3*a*e^4) - (2*A*(a + b*x^3)^(3/2))/(3*a*e*(e*x)^(3/2)) + ((2*A*b
+ a*B)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(3*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{\sqrt{a+b x^3} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx &=-\frac{2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac{(2 A b+a B) \int \sqrt{e x} \sqrt{a+b x^3} \, dx}{a e^3}\\ &=\frac{(2 A b+a B) (e x)^{3/2} \sqrt{a+b x^3}}{3 a e^4}-\frac{2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac{(2 A b+a B) \int \frac{\sqrt{e x}}{\sqrt{a+b x^3}} \, dx}{2 e^3}\\ &=\frac{(2 A b+a B) (e x)^{3/2} \sqrt{a+b x^3}}{3 a e^4}-\frac{2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac{(2 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 )}{e^4}\\ &=\frac{(2 A b+a B) (e x)^{3/2} \sqrt{a+b x^3}}{3 a e^4}-\frac{2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac{(2 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 )}{3 e^4}\\ &=\frac{(2 A b+a B) (e x)^{3/2} \sqrt{a+b x^3}}{3 a e^4}-\frac{2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac{(2 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 )}{3 e^4}\\ &=\frac{(2 A b+a B) (e x)^{3/2} \sqrt{a+b x^3}}{3 a e^4}-\frac{2 A \left (a+b x^3\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac{(2 A b+a B) \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{3 \sqrt{b} e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.187416, size = 87, normalized size = 0.74 \[ \frac{x \sqrt{a+b x^3} \left (\frac{x^{3/2} (a B+2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \sqrt{\frac{b x^3}{a}+1}}-2 A+B x^3\right )}{3 (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.059, size = 6668, normalized size = 56.5 \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)*(b*x^3+a)^(1/2)/(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 )} \sqrt{b x^{3} + a}}{\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)*(b*x^3+a)^(1/2)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 4.32924, size = 486, normalized size = 4.12 \begin{align*} \left [\frac{{\left (B a + 2 \, 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 (B b x^{3} - 2 \, A b\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{12 \, b e^{3} x^{2}}, -\frac{{\left (B a + 2 \, 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 (B b x^{3} - 2 \, A b\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{6 \, b e^{3} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/(e*x)^(5/2),x, algorithm="fricas")

[Out]

[1/12*((B*a + 2*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*(B*b*x^3 - 2*A*b)*sqrt(b*x^3 + a)*sqrt(e*x))/(b*e^3*x^2), -1/6*((B*a + 2*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*(B*b*x^3 - 2*A*b)*sqrt(b*x^3 +
a)*sqrt(e*x))/(b*e^3*x^2)]

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Sympy [A]  time = 22.1409, size = 160, normalized size = 1.36 \begin{align*} - \frac{2 A \sqrt{a}}{3 e^{\frac{5}{2}} x^{\frac{3}{2}} \sqrt{1 + \frac{b x^{3}}{a}}} + \frac{2 A \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} x^{\frac{3}{2}}}{\sqrt{a}} \right )}}{3 e^{\frac{5}{2}}} - \frac{2 A b x^{\frac{3}{2}}}{3 \sqrt{a} e^{\frac{5}{2}} \sqrt{1 + \frac{b x^{3}}{a}}} + \frac{B \sqrt{a} x^{\frac{3}{2}} \sqrt{1 + \frac{b x^{3}}{a}}}{3 e^{\frac{5}{2}}} + \frac{B a \operatorname{asinh}{\left (\frac{\sqrt{b} x^{\frac{3}{2}}}{\sqrt{a}} \right )}}{3 \sqrt{b} e^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)*(b*x**3+a)**(1/2)/(e*x)**(5/2),x)

[Out]

-2*A*sqrt(a)/(3*e**(5/2)*x**(3/2)*sqrt(1 + b*x**3/a)) + 2*A*sqrt(b)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(3*e**(5/2
)) - 2*A*b*x**(3/2)/(3*sqrt(a)*e**(5/2)*sqrt(1 + b*x**3/a)) + B*sqrt(a)*x**(3/2)*sqrt(1 + b*x**3/a)/(3*e**(5/2
)) + B*a*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(3*sqrt(b)*e**(5/2))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{3} + A\right )} \sqrt{b x^{3} + a}}{\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)*(b*x^3+a)^(1/2)/(e*x)^(5/2),x, algorithm="giac")

[Out]

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