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

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Rubi [A]  time = 0.07, antiderivative size = 131, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{\sqrt {b x^{3} + a}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{\sqrt {b x^{3} + a}}\,{d x} \]

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)

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{3}+A \right ) \left (e x \right )^{m}}{\sqrt {b \,x^{3}+a}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{\sqrt {b x^{3} + a}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^3+A\right )\,{\left (e\,x\right )}^m}{\sqrt {b\,x^3+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 4.87, size = 119, normalized size = 0.91 \[ \frac {A e^{m} x x^{m} \Gamma \left (\frac {m}{3} + \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{3} + \frac {1}{3} \\ \frac {m}{3} + \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} + \frac {B e^{m} x^{4} x^{m} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{3} + \frac {4}{3} \\ \frac {m}{3} + \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {m}{3} + \frac {7}{3}\right )} \]

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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