3.6.65 \(\int \frac {A+B x^3}{(e x)^{5/2} (a+b x^3)^{5/2}} \, dx\) [565]

Optimal. Leaf size=104 \[ -\frac {2 A}{3 a e (e x)^{3/2} \left (a+b x^3\right )^{3/2}}-\frac {2 (4 A b-a B) (e x)^{3/2}}{9 a^2 e^4 \left (a+b x^3\right )^{3/2}}-\frac {4 (4 A b-a B) (e x)^{3/2}}{9 a^3 e^4 \sqrt {a+b x^3}} \]

[Out]

-2/3*A/a/e/(e*x)^(3/2)/(b*x^3+a)^(3/2)-2/9*(4*A*b-B*a)*(e*x)^(3/2)/a^2/e^4/(b*x^3+a)^(3/2)-4/9*(4*A*b-B*a)*(e*
x)^(3/2)/a^3/e^4/(b*x^3+a)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {464, 279, 270} \begin {gather*} -\frac {4 (e x)^{3/2} (4 A b-a B)}{9 a^3 e^4 \sqrt {a+b x^3}}-\frac {2 (e x)^{3/2} (4 A b-a B)}{9 a^2 e^4 \left (a+b x^3\right )^{3/2}}-\frac {2 A}{3 a e (e x)^{3/2} \left (a+b x^3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 270

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 279

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

Rule 464

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

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Mathematica [A]
time = 0.35, size = 67, normalized size = 0.64 \begin {gather*} \frac {2 x \left (-3 a^2 A-12 a A b x^3+3 a^2 B x^3-8 A b^2 x^6+2 a b B x^6\right )}{9 a^3 (e x)^{5/2} \left (a+b x^3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(-3*a^2*A - 12*a*A*b*x^3 + 3*a^2*B*x^3 - 8*A*b^2*x^6 + 2*a*b*B*x^6))/(9*a^3*(e*x)^(5/2)*(a + b*x^3)^(3/2)
)

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Maple [A]
time = 0.35, size = 67, normalized size = 0.64

method	result	size
gosper	\(-\frac {2 x \left (8 A \,b^{2} x^{6}-2 B a b \,x^{6}+12 a A b \,x^{3}-3 a^{2} B \,x^{3}+3 a^{2} A \right )}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} a^{3} \left (e x \right )^{\frac {5}{2}}}\)	\(62\)
default	\(-\frac {2 \left (8 A \,b^{2} x^{6}-2 B a b \,x^{6}+12 a A b \,x^{3}-3 a^{2} B \,x^{3}+3 a^{2} A \right )}{9 \sqrt {e x}\, e^{2} a^{3} \left (b \,x^{3}+a \right )^{\frac {3}{2}} x}\)	\(67\)
risch	\(-\frac {2 A \sqrt {b \,x^{3}+a}}{3 a^{3} x \,e^{2} \sqrt {e x}}-\frac {2 \left (5 A \,b^{2} x^{3}-2 B a b \,x^{3}+6 a b A -3 a^{2} B \right ) x^{2}}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} a^{3} e^{2} \sqrt {e x}}\)	\(82\)
elliptic	\(\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}\, \left (-\frac {2 x \left (A b -B a \right ) \sqrt {b e \,x^{4}+a e x}}{9 e^{3} a^{2} b^{2} \left (x^{3}+\frac {a}{b}\right )^{2}}-\frac {2 x^{2} \left (5 A b -2 B a \right )}{9 e^{2} a^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b e x}}-\frac {2 A \sqrt {b e \,x^{4}+a e x}}{3 e^{3} a^{3} x^{2}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{3}+a}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*(8*A*b^2*x^6-2*B*a*b*x^6+12*A*a*b*x^3-3*B*a^2*x^3+3*A*a^2)/(e*x)^(1/2)/e^2/a^3/(b*x^3+a)^(3/2)/x

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Maxima [A]
time = 0.29, size = 90, normalized size = 0.87 \begin {gather*} -\frac {2}{9} \, {\left (\frac {B {\left (b - \frac {3 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} x^{\frac {9}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}} - A {\left (\frac {{\left (b^{2} - \frac {6 \, {\left (b x^{3} + a\right )} b}{x^{3}}\right )} x^{\frac {9}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {3 \, \sqrt {b x^{3} + a}}{a^{3} x^{\frac {3}{2}}}\right )}\right )} e^{\left (-\frac {5}{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)^(5/2),x, algorithm="maxima")

[Out]

-2/9*(B*(b - 3*(b*x^3 + a)/x^3)*x^(9/2)/((b*x^3 + a)^(3/2)*a^2) - A*((b^2 - 6*(b*x^3 + a)*b/x^3)*x^(9/2)/((b*x
^3 + a)^(3/2)*a^3) - 3*sqrt(b*x^3 + a)/(a^3*x^(3/2))))*e^(-5/2)

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Fricas [A]
time = 2.79, size = 84, normalized size = 0.81 \begin {gather*} \frac {2 \, {\left (2 \, {\left (B a b - 4 \, A b^{2}\right )} x^{6} + 3 \, {\left (B a^{2} - 4 \, A a b\right )} x^{3} - 3 \, A a^{2}\right )} \sqrt {b x^{3} + a} \sqrt {x} e^{\left (-\frac {5}{2}\right )}}{9 \, {\left (a^{3} b^{2} x^{8} + 2 \, a^{4} b x^{5} + a^{5} 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)^(5/2),x, algorithm="fricas")

[Out]

2/9*(2*(B*a*b - 4*A*b^2)*x^6 + 3*(B*a^2 - 4*A*a*b)*x^3 - 3*A*a^2)*sqrt(b*x^3 + a)*sqrt(x)*e^(-5/2)/(a^3*b^2*x^
8 + 2*a^4*b*x^5 + a^5*x^2)

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Sympy [F(-1)] Timed out
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)**(5/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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)^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.80, size = 115, normalized size = 1.11 \begin {gather*} -\frac {\sqrt {b\,x^3+a}\,\left (\frac {2\,A}{3\,a\,b^2\,e^2}-\frac {x^3\,\left (6\,B\,a^2-24\,A\,a\,b\right )}{9\,a^3\,b^2\,e^2}+\frac {x^6\,\left (16\,A\,b^2-4\,B\,a\,b\right )}{9\,a^3\,b^2\,e^2}\right )}{x^7\,\sqrt {e\,x}+\frac {a^2\,x\,\sqrt {e\,x}}{b^2}+\frac {2\,a\,x^4\,\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)^(5/2)),x)

[Out]

-((a + b*x^3)^(1/2)*((2*A)/(3*a*b^2*e^2) - (x^3*(6*B*a^2 - 24*A*a*b))/(9*a^3*b^2*e^2) + (x^6*(16*A*b^2 - 4*B*a
*b))/(9*a^3*b^2*e^2)))/(x^7*(e*x)^(1/2) + (a^2*x*(e*x)^(1/2))/b^2 + (2*a*x^4*(e*x)^(1/2))/b)

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