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

Optimal. Leaf size=67 \[ -\frac {2 (e x)^{3/2} (2 A b-a B)}{3 a^2 e^4 \sqrt {a+b x^3}}-\frac {2 A}{3 a e (e x)^{3/2} \sqrt {a+b x^3}} \]

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {453, 264} \begin {gather*} -\frac {2 (e x)^{3/2} (2 A b-a B)}{3 a^2 e^4 \sqrt {a+b x^3}}-\frac {2 A}{3 a e (e x)^{3/2} \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 264

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 45, normalized size = 0.67 \begin {gather*} \frac {x \left (-2 a A+2 a B x^3-4 A b x^3\right )}{3 a^2 (e x)^{5/2} \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.04, size = 71, normalized size = 1.06 \begin {gather*} \frac {2 \sqrt {a+b x^3} \left (-a A e^3+a B e^3 x^3-2 A b e^3 x^3\right )}{3 a^2 e (e x)^{3/2} \left (a e^3+b e^3 x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.65, size = 57, normalized size = 0.85 \begin {gather*} \frac {2 \, {\left ({\left (B a - 2 \, A b\right )} x^{3} - A a\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{3 \, {\left (a^{2} b e^{3} x^{5} + a^{3} e^{3} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 39, normalized size = 0.58 \begin {gather*} -\frac {2 \left (2 A \,x^{3} b -B a \,x^{3}+A a \right ) x}{3 \sqrt {b \,x^{3}+a}\, \left (e x \right )^{\frac {5}{2}} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.75, size = 70, normalized size = 1.04 \begin {gather*} -\frac {\left (\frac {2\,A}{3\,a\,b\,e^2}+\frac {x^3\,\left (4\,A\,b-2\,B\,a\right )}{3\,a^2\,b\,e^2}\right )\,\sqrt {b\,x^3+a}}{x^4\,\sqrt {e\,x}+\frac {a\,x\,\sqrt {e\,x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((2*A)/(3*a*b*e^2) + (x^3*(4*A*b - 2*B*a))/(3*a^2*b*e^2))*(a + b*x^3)^(1/2))/(x^4*(e*x)^(1/2) + (a*x*(e*x)^(
1/2))/b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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