3.2.59 \(\int \frac {x^{7/2} (A+B x^3)}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=95 \[ \frac {(A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 \sqrt {a} b^{5/2}}-\frac {x^{3/2} (A b-3 a B)}{3 a b^2}+\frac {x^{9/2} (A b-a B)}{3 a b \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {457, 321, 329, 275, 205} \begin {gather*} -\frac {x^{3/2} (A b-3 a B)}{3 a b^2}+\frac {(A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 \sqrt {a} b^{5/2}}+\frac {x^{9/2} (A b-a B)}{3 a b \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

-((A*b - 3*a*B)*x^(3/2))/(3*a*b^2) + ((A*b - a*B)*x^(9/2))/(3*a*b*(a + b*x^3)) + ((A*b - 3*a*B)*ArcTan[(Sqrt[b
]*x^(3/2))/Sqrt[a]])/(3*Sqrt[a]*b^(5/2))

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 {x^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx &=\frac {(A b-a B) x^{9/2}}{3 a b \left (a+b x^3\right )}+\frac {\left (-\frac {3 A b}{2}+\frac {9 a B}{2}\right ) \int \frac {x^{7/2}}{a+b x^3} \, dx}{3 a b}\\ &=-\frac {(A b-3 a B) x^{3/2}}{3 a b^2}+\frac {(A b-a B) x^{9/2}}{3 a b \left (a+b x^3\right )}+\frac {(A b-3 a B) \int \frac {\sqrt {x}}{a+b x^3} \, dx}{2 b^2}\\ &=-\frac {(A b-3 a B) x^{3/2}}{3 a b^2}+\frac {(A b-a B) x^{9/2}}{3 a b \left (a+b x^3\right )}+\frac {(A b-3 a B) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^6} \, dx,x,\sqrt {x}\right )}{b^2}\\ &=-\frac {(A b-3 a B) x^{3/2}}{3 a b^2}+\frac {(A b-a B) x^{9/2}}{3 a b \left (a+b x^3\right )}+\frac {(A b-3 a B) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^{3/2}\right )}{3 b^2}\\ &=-\frac {(A b-3 a B) x^{3/2}}{3 a b^2}+\frac {(A b-a B) x^{9/2}}{3 a b \left (a+b x^3\right )}+\frac {(A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 \sqrt {a} b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 77, normalized size = 0.81 \begin {gather*} \frac {\frac {(A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\sqrt {b} x^{3/2} \left (3 a B-A b+2 b B x^3\right )}{a+b x^3}}{3 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.11, size = 77, normalized size = 0.81 \begin {gather*} \frac {(A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 \sqrt {a} b^{5/2}}+\frac {x^{3/2} \left (3 a B-A b+2 b B x^3\right )}{3 b^2 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(7/2)*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

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

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fricas [A]  time = 0.72, size = 222, normalized size = 2.34 \begin {gather*} \left [\frac {{\left ({\left (3 \, B a b - A b^{2}\right )} x^{3} + 3 \, B a^{2} - A a b\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 (2 \, B a b^{2} x^{4} + {\left (3 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {x}}{6 \, {\left (a b^{4} x^{3} + a^{2} b^{3}\right )}}, -\frac {{\left ({\left (3 \, B a b - A b^{2}\right )} x^{3} + 3 \, B a^{2} - A a b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right ) - {\left (2 \, B a b^{2} x^{4} + {\left (3 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {x}}{3 \, {\left (a b^{4} x^{3} + a^{2} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 68, normalized size = 0.72 \begin {gather*} \frac {2 \, B x^{\frac {3}{2}}}{3 \, b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b^{2}} + \frac {B a x^{\frac {3}{2}} - A b x^{\frac {3}{2}}}{3 \, {\left (b x^{3} + a\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

2/3*B*x^(3/2)/b^2 - 1/3*(3*B*a - A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*b^2) + 1/3*(B*a*x^(3/2) - A*b*x^(
3/2))/((b*x^3 + a)*b^2)

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maple [A]  time = 0.06, size = 93, normalized size = 0.98 \begin {gather*} -\frac {A \,x^{\frac {3}{2}}}{3 \left (b \,x^{3}+a \right ) b}+\frac {B a \,x^{\frac {3}{2}}}{3 \left (b \,x^{3}+a \right ) b^{2}}+\frac {A \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}\, b}-\frac {B a \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{2}}+\frac {2 B \,x^{\frac {3}{2}}}{3 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x^3+A)/(b*x^3+a)^2,x)

[Out]

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

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maxima [A]  time = 1.24, size = 68, normalized size = 0.72 \begin {gather*} \frac {{\left (B a - A b\right )} x^{\frac {3}{2}}}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}} + \frac {2 \, B x^{\frac {3}{2}}}{3 \, b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

1/3*(B*a - A*b)*x^(3/2)/(b^3*x^3 + a*b^2) + 2/3*B*x^(3/2)/b^2 - 1/3*(3*B*a - A*b)*arctan(b*x^(3/2)/sqrt(a*b))/
(sqrt(a*b)*b^2)

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mupad [B]  time = 2.65, size = 116, normalized size = 1.22 \begin {gather*} \frac {2\,B\,x^{3/2}}{3\,b^2}-\frac {x^{3/2}\,\left (\frac {A\,b}{3}-\frac {B\,a}{3}\right )}{b^3\,x^3+a\,b^2}+\frac {\mathrm {atan}\left (\frac {36\,\sqrt {a}\,b^{3/2}\,x^{3/2}\,\left (A^2\,b^2-6\,A\,B\,a\,b+9\,B^2\,a^2\right )}{\left (A\,b-3\,B\,a\right )\,\left (36\,A\,a\,b^2-108\,B\,a^2\,b\right )}\right )\,\left (A\,b-3\,B\,a\right )}{3\,\sqrt {a}\,b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(7/2)*(A + B*x^3))/(a + b*x^3)^2,x)

[Out]

(2*B*x^(3/2))/(3*b^2) - (x^(3/2)*((A*b)/3 - (B*a)/3))/(a*b^2 + b^3*x^3) + (atan((36*a^(1/2)*b^(3/2)*x^(3/2)*(A
^2*b^2 + 9*B^2*a^2 - 6*A*B*a*b))/((A*b - 3*B*a)*(36*A*a*b^2 - 108*B*a^2*b)))*(A*b - 3*B*a))/(3*a^(1/2)*b^(5/2)
)

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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(x**(7/2)*(B*x**3+A)/(b*x**3+a)**2,x)

[Out]

Timed out

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