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

Optimal. Leaf size=104 \[ \frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{3/2} b^{5/2}}-\frac {x^{3/2} (3 a B+A b)}{12 a b^2 \left (a+b x^3\right )}+\frac {x^{9/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 104, 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, 288, 329, 275, 205} \begin {gather*} \frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{3/2} b^{5/2}}-\frac {x^{3/2} (3 a B+A b)}{12 a b^2 \left (a+b x^3\right )}+\frac {x^{9/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^(9/2))/(6*a*b*(a + b*x^3)^2) - ((A*b + 3*a*B)*x^(3/2))/(12*a*b^2*(a + b*x^3)) + ((A*b + 3*a*B)*
ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(12*a^(3/2)*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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 )^3} \, dx &=\frac {(A b-a B) x^{9/2}}{6 a b \left (a+b x^3\right )^2}+\frac {\left (\frac {3 A b}{2}+\frac {9 a B}{2}\right ) \int \frac {x^{7/2}}{\left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac {(A b-a B) x^{9/2}}{6 a b \left (a+b x^3\right )^2}-\frac {(A b+3 a B) x^{3/2}}{12 a b^2 \left (a+b x^3\right )}+\frac {(A b+3 a B) \int \frac {\sqrt {x}}{a+b x^3} \, dx}{8 a b^2}\\ &=\frac {(A b-a B) x^{9/2}}{6 a b \left (a+b x^3\right )^2}-\frac {(A b+3 a B) x^{3/2}}{12 a b^2 \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 )}{4 a b^2}\\ &=\frac {(A b-a B) x^{9/2}}{6 a b \left (a+b x^3\right )^2}-\frac {(A b+3 a B) x^{3/2}}{12 a b^2 \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 )}{12 a b^2}\\ &=\frac {(A b-a B) x^{9/2}}{6 a b \left (a+b x^3\right )^2}-\frac {(A b+3 a B) x^{3/2}}{12 a b^2 \left (a+b x^3\right )}+\frac {(A b+3 a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{3/2} b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 93, normalized size = 0.89 \begin {gather*} \frac {\frac {\sqrt {a} \sqrt {b} x^{3/2} \left (-3 a^2 B-a b \left (A+5 B x^3\right )+A b^2 x^3\right )}{\left (a+b x^3\right )^2}+(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{3/2} b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.17, size = 92, normalized size = 0.88 \begin {gather*} \frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{3/2} b^{5/2}}-\frac {x^{3/2} \left (3 a^2 B+a A b+5 a b B x^3-A b^2 x^3\right )}{12 a b^2 \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(x^(3/2)*(a*A*b + 3*a^2*B - A*b^2*x^3 + 5*a*b*B*x^3))/(a*b^2*(a + b*x^3)^2) + ((A*b + 3*a*B)*ArcTan[(Sqr
t[b]*x^(3/2))/Sqrt[a]])/(12*a^(3/2)*b^(5/2))

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fricas [A]  time = 0.77, size = 314, normalized size = 3.02 \begin {gather*} \left [-\frac {{\left ({\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + 3 \, B a^{3} + A a^{2} b + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x^{3}\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 ({\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{4} + {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{2} b^{5} x^{6} + 2 \, a^{3} b^{4} x^{3} + a^{4} b^{3}\right )}}, \frac {{\left ({\left (3 \, B a b^{2} + A b^{3}\right )} x^{6} + 3 \, B a^{3} + A a^{2} b + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right ) - {\left ({\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{4} + {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (a^{2} b^{5} x^{6} + 2 \, a^{3} b^{4} x^{3} + a^{4} 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)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 84, normalized size = 0.81 \begin {gather*} \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \, \sqrt {a b} a b^{2}} - \frac {5 \, B a b x^{\frac {9}{2}} - A b^{2} x^{\frac {9}{2}} + 3 \, B a^{2} x^{\frac {3}{2}} + A a b x^{\frac {3}{2}}}{12 \, {\left (b x^{3} + a\right )}^{2} a 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)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 96, normalized size = 0.92 \begin {gather*} \frac {A \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \sqrt {a b}\, a b}+\frac {B \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, b^{2}}+\frac {\frac {\left (A b -5 B a \right ) x^{\frac {9}{2}}}{12 a b}-\frac {\left (A b +3 B a \right ) x^{\frac {3}{2}}}{12 b^{2}}}{\left (b \,x^{3}+a \right )^{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)^3,x)

[Out]

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

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maxima [A]  time = 1.43, size = 96, normalized size = 0.92 \begin {gather*} -\frac {{\left (5 \, B a b - A b^{2}\right )} x^{\frac {9}{2}} + {\left (3 \, B a^{2} + A a b\right )} x^{\frac {3}{2}}}{12 \, {\left (a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}} + \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \, \sqrt {a b} a 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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.76, size = 133, normalized size = 1.28 \begin {gather*} \frac {\mathrm {atan}\left (\frac {9\,b^{3/2}\,x^{3/2}\,\left (A^2\,b^2+6\,A\,B\,a\,b+9\,B^2\,a^2\right )}{\sqrt {a}\,\left (9\,A\,b^2+27\,B\,a\,b\right )\,\left (A\,b+3\,B\,a\right )}\right )\,\left (A\,b+3\,B\,a\right )}{12\,a^{3/2}\,b^{5/2}}-\frac {\frac {x^{3/2}\,\left (A\,b+3\,B\,a\right )}{12\,b^2}-\frac {x^{9/2}\,\left (A\,b-5\,B\,a\right )}{12\,a\,b}}{a^2+2\,a\,b\,x^3+b^2\,x^6} \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)^3,x)

[Out]

(atan((9*b^(3/2)*x^(3/2)*(A^2*b^2 + 9*B^2*a^2 + 6*A*B*a*b))/(a^(1/2)*(9*A*b^2 + 27*B*a*b)*(A*b + 3*B*a)))*(A*b
+ 3*B*a))/(12*a^(3/2)*b^(5/2)) - ((x^(3/2)*(A*b + 3*B*a))/(12*b^2) - (x^(9/2)*(A*b - 5*B*a))/(12*a*b))/(a^2 +
b^2*x^6 + 2*a*b*x^3)

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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)**3,x)

[Out]

Timed out

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