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

Optimal. Leaf size=121 \[ \frac{e^2 (e x)^{3/2} \sqrt{a+b x^3} (4 A b-3 a B)}{12 b^2}-\frac{a e^{7/2} (4 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{12 b^{5/2}}+\frac{B (e x)^{9/2} \sqrt{a+b x^3}}{6 b e} \]

[Out]

((4*A*b - 3*a*B)*e^2*(e*x)^(3/2)*Sqrt[a + b*x^3])/(12*b^2) + (B*(e*x)^(9/2)*Sqrt[a + b*x^3])/(6*b*e) - (a*(4*A
*b - 3*a*B)*e^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(12*b^(5/2))

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Rubi [A]  time = 0.0860604, antiderivative size = 121, 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 = {459, 321, 329, 275, 217, 206} \[ \frac{e^2 (e x)^{3/2} \sqrt{a+b x^3} (4 A b-3 a B)}{12 b^2}-\frac{a e^{7/2} (4 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{12 b^{5/2}}+\frac{B (e x)^{9/2} \sqrt{a+b x^3}}{6 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A*b - 3*a*B)*e^2*(e*x)^(3/2)*Sqrt[a + b*x^3])/(12*b^2) + (B*(e*x)^(9/2)*Sqrt[a + b*x^3])/(6*b*e) - (a*(4*A
*b - 3*a*B)*e^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + b*x^3])])/(12*b^(5/2))

Rule 459

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && 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{(e x)^{7/2} \left (A+B x^3\right )}{\sqrt{a+b x^3}} \, dx &=\frac{B (e x)^{9/2} \sqrt{a+b x^3}}{6 b e}-\frac{\left (-6 A b+\frac{9 a B}{2}\right ) \int \frac{(e x)^{7/2}}{\sqrt{a+b x^3}} \, dx}{6 b}\\ &=\frac{(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt{a+b x^3}}{12 b^2}+\frac{B (e x)^{9/2} \sqrt{a+b x^3}}{6 b e}-\frac{\left (a (4 A b-3 a B) e^3\right ) \int \frac{\sqrt{e x}}{\sqrt{a+b x^3}} \, dx}{8 b^2}\\ &=\frac{(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt{a+b x^3}}{12 b^2}+\frac{B (e x)^{9/2} \sqrt{a+b x^3}}{6 b e}-\frac{\left (a (4 A b-3 a B) e^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^6}{e^3}}} \, dx,x,\sqrt{e x}\right )}{4 b^2}\\ &=\frac{(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt{a+b x^3}}{12 b^2}+\frac{B (e x)^{9/2} \sqrt{a+b x^3}}{6 b e}-\frac{\left (a (4 A b-3 a B) e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^2}{e^3}}} \, dx,x,(e x)^{3/2}\right )}{12 b^2}\\ &=\frac{(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt{a+b x^3}}{12 b^2}+\frac{B (e x)^{9/2} \sqrt{a+b x^3}}{6 b e}-\frac{\left (a (4 A b-3 a B) e^2\right ) \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 )}{12 b^2}\\ &=\frac{(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt{a+b x^3}}{12 b^2}+\frac{B (e x)^{9/2} \sqrt{a+b x^3}}{6 b e}-\frac{a (4 A b-3 a B) e^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} (e x)^{3/2}}{e^{3/2} \sqrt{a+b x^3}}\right )}{12 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.121261, size = 97, normalized size = 0.8 \[ \frac{e^3 \sqrt{e x} \left (\sqrt{b} x^{3/2} \sqrt{a+b x^3} \left (-3 a B+4 A b+2 b B x^3\right )+a (3 a B-4 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a+b x^3}}\right )\right )}{12 b^{5/2} \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.059, size = 6861, normalized size = 56.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(1/2),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 (e x\right )^{\frac{7}{2}}}{\sqrt{b x^{3} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 4.16013, size = 551, normalized size = 4.55 \begin{align*} \left [-\frac{{\left (3 \, B a^{2} - 4 \, A a b\right )} e^{3} \sqrt{\frac{e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e + 4 \,{\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt{b x^{3} + a} \sqrt{e x} \sqrt{\frac{e}{b}}\right ) - 4 \,{\left (2 \, B b e^{3} x^{4} -{\left (3 \, B a - 4 \, A b\right )} e^{3} x\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{48 \, b^{2}}, -\frac{{\left (3 \, B a^{2} - 4 \, A a b\right )} e^{3} \sqrt{-\frac{e}{b}} \arctan \left (\frac{2 \, \sqrt{b x^{3} + a} \sqrt{e x} b x \sqrt{-\frac{e}{b}}}{2 \, b e x^{3} + a e}\right ) - 2 \,{\left (2 \, B b e^{3} x^{4} -{\left (3 \, B a - 4 \, A b\right )} e^{3} x\right )} \sqrt{b x^{3} + a} \sqrt{e x}}{24 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

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

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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((e*x)**(7/2)*(B*x**3+A)/(b*x**3+a)**(1/2),x)

[Out]

Timed out

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Giac [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((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

Timed out