3.503 \(\int \frac{(e x)^m (A+B x^3)}{\sqrt{a+b x^3}} \, dx\)

Optimal. Leaf size=131 \[ \frac{2 B \sqrt{a+b x^3} (e x)^{m+1}}{b e (2 m+5)}-\frac{\sqrt{\frac{b x^3}{a}+1} (e x)^{m+1} (2 a B (m+1)-A b (2 m+5)) \, _2F_1\left (\frac{1}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{b e (m+1) (2 m+5) \sqrt{a+b x^3}} \]

[Out]

(2*B*(e*x)^(1 + m)*Sqrt[a + b*x^3])/(b*e*(5 + 2*m)) - ((2*a*B*(1 + m) - A*b*(5 + 2*m))*(e*x)^(1 + m)*Sqrt[1 +
(b*x^3)/a]*Hypergeometric2F1[1/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(b*e*(1 + m)*(5 + 2*m)*Sqrt[a + b*x^3])

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Rubi [A]  time = 0.0708553, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {459, 365, 364} \[ \frac{2 B \sqrt{a+b x^3} (e x)^{m+1}}{b e (2 m+5)}-\frac{\sqrt{\frac{b x^3}{a}+1} (e x)^{m+1} (2 a B (m+1)-A b (2 m+5)) \, _2F_1\left (\frac{1}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{b e (m+1) (2 m+5) \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*(e*x)^(1 + m)*Sqrt[a + b*x^3])/(b*e*(5 + 2*m)) - ((2*a*B*(1 + m) - A*b*(5 + 2*m))*(e*x)^(1 + m)*Sqrt[1 +
(b*x^3)/a]*Hypergeometric2F1[1/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(b*e*(1 + m)*(5 + 2*m)*Sqrt[a + b*x^3])

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 365

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(e x)^m \left (A+B x^3\right )}{\sqrt{a+b x^3}} \, dx &=\frac{2 B (e x)^{1+m} \sqrt{a+b x^3}}{b e (5+2 m)}-\frac{\left (a B (1+m)-A b \left (\frac{5}{2}+m\right )\right ) \int \frac{(e x)^m}{\sqrt{a+b x^3}} \, dx}{b \left (\frac{5}{2}+m\right )}\\ &=\frac{2 B (e x)^{1+m} \sqrt{a+b x^3}}{b e (5+2 m)}-\frac{\left (\left (a B (1+m)-A b \left (\frac{5}{2}+m\right )\right ) \sqrt{1+\frac{b x^3}{a}}\right ) \int \frac{(e x)^m}{\sqrt{1+\frac{b x^3}{a}}} \, dx}{b \left (\frac{5}{2}+m\right ) \sqrt{a+b x^3}}\\ &=\frac{2 B (e x)^{1+m} \sqrt{a+b x^3}}{b e (5+2 m)}-\frac{(2 a B (1+m)-A b (5+2 m)) (e x)^{1+m} \sqrt{1+\frac{b x^3}{a}} \, _2F_1\left (\frac{1}{2},\frac{1+m}{3};\frac{4+m}{3};-\frac{b x^3}{a}\right )}{b e (1+m) (5+2 m) \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0766648, size = 110, normalized size = 0.84 \[ \frac{x \sqrt{\frac{b x^3}{a}+1} (e x)^m \left (A (m+4) \, _2F_1\left (\frac{1}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )+B (m+1) x^3 \, _2F_1\left (\frac{1}{2},\frac{m+4}{3};\frac{m+7}{3};-\frac{b x^3}{a}\right )\right )}{(m+1) (m+4) \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*Sqrt[1 + (b*x^3)/a]*(A*(4 + m)*Hypergeometric2F1[1/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)] + B*(1 +
m)*x^3*Hypergeometric2F1[1/2, (4 + m)/3, (7 + m)/3, -((b*x^3)/a)]))/((1 + m)*(4 + m)*Sqrt[a + b*x^3])

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex \right ) ^{m} \left ( B{x}^{3}+A \right ){\frac{1}{\sqrt{b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(1/2),x)

[Out]

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(1/2),x)

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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 )^{m}}{\sqrt{b x^{3} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^3 + A)*(e*x)^m/sqrt(b*x^3 + a), 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 )} \left (e x\right )^{m}}{\sqrt{b x^{3} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^3 + A)*(e*x)^m/sqrt(b*x^3 + a), x)

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Sympy [C]  time = 5.49894, size = 119, normalized size = 0.91 \begin{align*} \frac{A e^{m} x x^{m} \Gamma \left (\frac{m}{3} + \frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{m}{3} + \frac{1}{3} \\ \frac{m}{3} + \frac{4}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt{a} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right )} + \frac{B e^{m} x^{4} x^{m} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{m}{3} + \frac{4}{3} \\ \frac{m}{3} + \frac{7}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt{a} \Gamma \left (\frac{m}{3} + \frac{7}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*e**m*x*x**m*gamma(m/3 + 1/3)*hyper((1/2, m/3 + 1/3), (m/3 + 4/3,), b*x**3*exp_polar(I*pi)/a)/(3*sqrt(a)*gamm
a(m/3 + 4/3)) + B*e**m*x**4*x**m*gamma(m/3 + 4/3)*hyper((1/2, m/3 + 4/3), (m/3 + 7/3,), b*x**3*exp_polar(I*pi)
/a)/(3*sqrt(a)*gamma(m/3 + 7/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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