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

Optimal. Leaf size=129 \[ -\frac {(5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}}+\frac {a B-5 A b}{4 a^3 b x^{3/2}}+\frac {5 A b-a B}{12 a^2 b x^{3/2} \left (a+b x^3\right )}+\frac {A b-a B}{6 a b x^{3/2} \left (a+b x^3\right )^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 130, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {457, 290, 325, 329, 275, 205} \begin {gather*} \frac {5 A b-a B}{12 a^2 b x^{3/2} \left (a+b x^3\right )}-\frac {5 A b-a B}{4 a^3 b x^{3/2}}-\frac {(5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}}+\frac {A b-a B}{6 a b x^{3/2} \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(5*A*b - a*B)/(4*a^3*b*x^(3/2)) + (A*b - a*B)/(6*a*b*x^(3/2)*(a + b*x^3)^2) + (5*A*b - a*B)/(12*a^2*b*x^(3/2)
*(a + b*x^3)) - ((5*A*b - a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(4*a^(7/2)*Sqrt[b])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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))]))

Rubi steps

\begin {align*} \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^3} \, dx &=\frac {A b-a B}{6 a b x^{3/2} \left (a+b x^3\right )^2}+\frac {\left (\frac {15 A b}{2}-\frac {3 a B}{2}\right ) \int \frac {1}{x^{5/2} \left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac {A b-a B}{6 a b x^{3/2} \left (a+b x^3\right )^2}+\frac {5 A b-a B}{12 a^2 b x^{3/2} \left (a+b x^3\right )}+\frac {(3 (5 A b-a B)) \int \frac {1}{x^{5/2} \left (a+b x^3\right )} \, dx}{8 a^2 b}\\ &=-\frac {5 A b-a B}{4 a^3 b x^{3/2}}+\frac {A b-a B}{6 a b x^{3/2} \left (a+b x^3\right )^2}+\frac {5 A b-a B}{12 a^2 b x^{3/2} \left (a+b x^3\right )}-\frac {(3 (5 A b-a B)) \int \frac {\sqrt {x}}{a+b x^3} \, dx}{8 a^3}\\ &=-\frac {5 A b-a B}{4 a^3 b x^{3/2}}+\frac {A b-a B}{6 a b x^{3/2} \left (a+b x^3\right )^2}+\frac {5 A b-a B}{12 a^2 b x^{3/2} \left (a+b x^3\right )}-\frac {(3 (5 A b-a B)) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^6} \, dx,x,\sqrt {x}\right )}{4 a^3}\\ &=-\frac {5 A b-a B}{4 a^3 b x^{3/2}}+\frac {A b-a B}{6 a b x^{3/2} \left (a+b x^3\right )^2}+\frac {5 A b-a B}{12 a^2 b x^{3/2} \left (a+b x^3\right )}-\frac {(5 A b-a B) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^{3/2}\right )}{4 a^3}\\ &=-\frac {5 A b-a B}{4 a^3 b x^{3/2}}+\frac {A b-a B}{6 a b x^{3/2} \left (a+b x^3\right )^2}+\frac {5 A b-a B}{12 a^2 b x^{3/2} \left (a+b x^3\right )}-\frac {(5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 102, normalized size = 0.79 \begin {gather*} \frac {(a B-5 A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}}+\frac {a^2 \left (5 B x^3-8 A\right )+a \left (3 b B x^6-25 A b x^3\right )-15 A b^2 x^6}{12 a^3 x^{3/2} \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*A*b^2*x^6 + a^2*(-8*A + 5*B*x^3) + a*(-25*A*b*x^3 + 3*b*B*x^6))/(12*a^3*x^(3/2)*(a + b*x^3)^2) + ((-5*A*b
+ a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(4*a^(7/2)*Sqrt[b])

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IntegrateAlgebraic [A]  time = 0.17, size = 102, normalized size = 0.79 \begin {gather*} \frac {(a B-5 A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}}+\frac {-8 a^2 A+5 a^2 B x^3-25 a A b x^3+3 a b B x^6-15 A b^2 x^6}{12 a^3 x^{3/2} \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2*A - 25*a*A*b*x^3 + 5*a^2*B*x^3 - 15*A*b^2*x^6 + 3*a*b*B*x^6)/(12*a^3*x^(3/2)*(a + b*x^3)^2) + ((-5*A*b
+ a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(4*a^(7/2)*Sqrt[b])

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fricas [A]  time = 0.83, size = 347, normalized size = 2.69 \begin {gather*} \left [\frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{8} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{5} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{3} + 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right ) + 2 \, {\left (3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{6} - 8 \, A a^{3} b + 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{3} x^{8} + 2 \, a^{5} b^{2} x^{5} + a^{6} b x^{2}\right )}}, \frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{8} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{5} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right ) + {\left (3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{6} - 8 \, A a^{3} b + 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3}\right )} \sqrt {x}}{12 \, {\left (a^{4} b^{3} x^{8} + 2 \, a^{5} b^{2} x^{5} + a^{6} b x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(3*((B*a*b^2 - 5*A*b^3)*x^8 + 2*(B*a^2*b - 5*A*a*b^2)*x^5 + (B*a^3 - 5*A*a^2*b)*x^2)*sqrt(-a*b)*log((b*x
^3 + 2*sqrt(-a*b)*x^(3/2) - a)/(b*x^3 + a)) + 2*(3*(B*a^2*b^2 - 5*A*a*b^3)*x^6 - 8*A*a^3*b + 5*(B*a^3*b - 5*A*
a^2*b^2)*x^3)*sqrt(x))/(a^4*b^3*x^8 + 2*a^5*b^2*x^5 + a^6*b*x^2), 1/12*(3*((B*a*b^2 - 5*A*b^3)*x^8 + 2*(B*a^2*
b - 5*A*a*b^2)*x^5 + (B*a^3 - 5*A*a^2*b)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x^(3/2)/a) + (3*(B*a^2*b^2 - 5*A*a*b^
3)*x^6 - 8*A*a^3*b + 5*(B*a^3*b - 5*A*a^2*b^2)*x^3)*sqrt(x))/(a^4*b^3*x^8 + 2*a^5*b^2*x^5 + a^6*b*x^2)]

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giac [A]  time = 0.18, size = 88, normalized size = 0.68 \begin {gather*} \frac {{\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} - \frac {2 \, A}{3 \, a^{3} x^{\frac {3}{2}}} + \frac {3 \, B a b x^{\frac {9}{2}} - 7 \, A b^{2} x^{\frac {9}{2}} + 5 \, B a^{2} x^{\frac {3}{2}} - 9 \, A a b x^{\frac {3}{2}}}{12 \, {\left (b x^{3} + a\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(B*a - 5*A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/3*A/(a^3*x^(3/2)) + 1/12*(3*B*a*b*x^(9/2) -
7*A*b^2*x^(9/2) + 5*B*a^2*x^(3/2) - 9*A*a*b*x^(3/2))/((b*x^3 + a)^2*a^3)

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maple [A]  time = 0.07, size = 133, normalized size = 1.03 \begin {gather*} -\frac {7 A \,b^{2} x^{\frac {9}{2}}}{12 \left (b \,x^{3}+a \right )^{2} a^{3}}+\frac {B b \,x^{\frac {9}{2}}}{4 \left (b \,x^{3}+a \right )^{2} a^{2}}-\frac {3 A b \,x^{\frac {3}{2}}}{4 \left (b \,x^{3}+a \right )^{2} a^{2}}+\frac {5 B \,x^{\frac {3}{2}}}{12 \left (b \,x^{3}+a \right )^{2} a}-\frac {5 A b \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{3}}+\frac {B \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{2}}-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/12/a^3/(b*x^3+a)^2*x^(9/2)*b^2*A+1/4/a^2/(b*x^3+a)^2*x^(9/2)*B*b-3/4/a^2/(b*x^3+a)^2*A*x^(3/2)*b+5/12/a/(b*
x^3+a)^2*B*x^(3/2)-5/4/a^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(3/2))*A*b+1/4/a^2/(a*b)^(1/2)*arctan(1/(a*b)^
(1/2)*b*x^(3/2))*B-2/3/a^3*A/x^(3/2)

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maxima [A]  time = 1.35, size = 100, normalized size = 0.78 \begin {gather*} \frac {3 \, {\left (B a b - 5 \, A b^{2}\right )} x^{6} + 5 \, {\left (B a^{2} - 5 \, A a b\right )} x^{3} - 8 \, A a^{2}}{12 \, {\left (a^{3} b^{2} x^{\frac {15}{2}} + 2 \, a^{4} b x^{\frac {9}{2}} + a^{5} x^{\frac {3}{2}}\right )}} + \frac {{\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(B*a*b - 5*A*b^2)*x^6 + 5*(B*a^2 - 5*A*a*b)*x^3 - 8*A*a^2)/(a^3*b^2*x^(15/2) + 2*a^4*b*x^(9/2) + a^5*x
^(3/2)) + 1/4*(B*a - 5*A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*a^3)

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mupad [B]  time = 2.73, size = 163, normalized size = 1.26 \begin {gather*} -\frac {\frac {2\,A}{3\,a}+\frac {5\,x^3\,\left (5\,A\,b-B\,a\right )}{12\,a^2}+\frac {b\,x^6\,\left (5\,A\,b-B\,a\right )}{4\,a^3}}{a^2\,x^{3/2}+b^2\,x^{15/2}+2\,a\,b\,x^{9/2}}-\frac {\mathrm {atan}\left (\frac {8\,a^{7/2}\,\sqrt {b}\,x^{3/2}\,\left (86400\,A^2\,a^9\,b^5-34560\,A\,B\,a^{10}\,b^4+3456\,B^2\,a^{11}\,b^3\right )}{\left (5\,A\,b-B\,a\right )\,\left (138240\,A\,a^{13}\,b^4-27648\,B\,a^{14}\,b^3\right )}\right )\,\left (5\,A\,b-B\,a\right )}{4\,a^{7/2}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*A)/(3*a) + (5*x^3*(5*A*b - B*a))/(12*a^2) + (b*x^6*(5*A*b - B*a))/(4*a^3))/(a^2*x^(3/2) + b^2*x^(15/2) +
2*a*b*x^(9/2)) - (atan((8*a^(7/2)*b^(1/2)*x^(3/2)*(86400*A^2*a^9*b^5 + 3456*B^2*a^11*b^3 - 34560*A*B*a^10*b^4
))/((5*A*b - B*a)*(138240*A*a^13*b^4 - 27648*B*a^14*b^3)))*(5*A*b - B*a))/(4*a^(7/2)*b^(1/2))

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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)/x**(5/2)/(b*x**3+a)**3,x)

[Out]

Timed out

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