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

Optimal. Leaf size=133 \[ \frac{\sqrt{\frac{b x^3}{a}+1} (e x)^{m+1} (2 a B (m+1)+A b (7-2 m)) \, _2F_1\left (\frac{3}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{9 a^2 b e (m+1) \sqrt{a+b x^3}}+\frac{2 (e x)^{m+1} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}} \]

[Out]

(2*(A*b - a*B)*(e*x)^(1 + m))/(9*a*b*e*(a + b*x^3)^(3/2)) + ((A*b*(7 - 2*m) + 2*a*B*(1 + m))*(e*x)^(1 + m)*Sqr
t[1 + (b*x^3)/a]*Hypergeometric2F1[3/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(9*a^2*b*e*(1 + m)*Sqrt[a + b*x^3
])

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Rubi [A]  time = 0.0789417, antiderivative size = 133, 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 = {457, 365, 364} \[ \frac{\sqrt{\frac{b x^3}{a}+1} (e x)^{m+1} (2 a B (m+1)+A b (7-2 m)) \, _2F_1\left (\frac{3}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{9 a^2 b e (m+1) \sqrt{a+b x^3}}+\frac{2 (e x)^{m+1} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*(e*x)^(1 + m))/(9*a*b*e*(a + b*x^3)^(3/2)) + ((A*b*(7 - 2*m) + 2*a*B*(1 + m))*(e*x)^(1 + m)*Sqr
t[1 + (b*x^3)/a]*Hypergeometric2F1[3/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(9*a^2*b*e*(1 + m)*Sqrt[a + b*x^3
])

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

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 )}{\left (a+b x^3\right )^{5/2}} \, dx &=\frac{2 (A b-a B) (e x)^{1+m}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac{\left (2 \left (-A b \left (-\frac{7}{2}+m\right )+a B (1+m)\right )\right ) \int \frac{(e x)^m}{\left (a+b x^3\right )^{3/2}} \, dx}{9 a b}\\ &=\frac{2 (A b-a B) (e x)^{1+m}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac{\left (2 \left (-A b \left (-\frac{7}{2}+m\right )+a B (1+m)\right ) \sqrt{1+\frac{b x^3}{a}}\right ) \int \frac{(e x)^m}{\left (1+\frac{b x^3}{a}\right )^{3/2}} \, dx}{9 a^2 b \sqrt{a+b x^3}}\\ &=\frac{2 (A b-a B) (e x)^{1+m}}{9 a b e \left (a+b x^3\right )^{3/2}}+\frac{(A b (7-2 m)+2 a B (1+m)) (e x)^{1+m} \sqrt{1+\frac{b x^3}{a}} \, _2F_1\left (\frac{3}{2},\frac{1+m}{3};\frac{4+m}{3};-\frac{b x^3}{a}\right )}{9 a^2 b e (1+m) \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0890926, size = 113, normalized size = 0.85 \[ \frac{x \sqrt{\frac{b x^3}{a}+1} (e x)^m \left (A (m+4) \, _2F_1\left (\frac{5}{2},\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )+B (m+1) x^3 \, _2F_1\left (\frac{5}{2},\frac{m+4}{3};\frac{m+7}{3};-\frac{b x^3}{a}\right )\right )}{a^2 (m+1) (m+4) \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*Sqrt[1 + (b*x^3)/a]*(A*(4 + m)*Hypergeometric2F1[5/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)] + B*(1 +
m)*x^3*Hypergeometric2F1[5/2, (4 + m)/3, (7 + m)/3, -((b*x^3)/a)]))/(a^2*(1 + m)*(4 + m)*Sqrt[a + b*x^3])

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex \right ) ^{m} \left ( B{x}^{3}+A \right ) \left ( b{x}^{3}+a \right ) ^{-{\frac{5}{2}}}}\, 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)^(5/2),x)

[Out]

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(5/2),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}}{{\left (b x^{3} + a\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)*(e*x)^m/(b*x^3 + a)^(5/2), 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} \left (e x\right )^{m}}{b^{3} x^{9} + 3 \, a b^{2} x^{6} + 3 \, a^{2} b x^{3} + a^{3}}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral((B*x^3 + A)*sqrt(b*x^3 + a)*(e*x)^m/(b^3*x^9 + 3*a*b^2*x^6 + 3*a^2*b*x^3 + a^3), 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((e*x)**m*(B*x**3+A)/(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{{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="giac")

[Out]

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