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

Optimal. Leaf size=73 \[ -\frac {2 \sqrt {a} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 b^{5/2}}+\frac {2 x^{3/2} (A b-a B)}{3 b^2}+\frac {2 B x^{9/2}}{9 b} \]

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Rubi [A]  time = 0.05, antiderivative size = 73, 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 = {459, 321, 329, 275, 205} \begin {gather*} \frac {2 x^{3/2} (A b-a B)}{3 b^2}-\frac {2 \sqrt {a} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 b^{5/2}}+\frac {2 B x^{9/2}}{9 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*x^(3/2))/(3*b^2) + (2*B*x^(9/2))/(9*b) - (2*Sqrt[a]*(A*b - a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a
]])/(3*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 459

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

Rubi steps

\begin {align*} \int \frac {x^{7/2} \left (A+B x^3\right )}{a+b x^3} \, dx &=\frac {2 B x^{9/2}}{9 b}-\frac {\left (2 \left (-\frac {9 A b}{2}+\frac {9 a B}{2}\right )\right ) \int \frac {x^{7/2}}{a+b x^3} \, dx}{9 b}\\ &=\frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{9/2}}{9 b}-\frac {(a (A b-a B)) \int \frac {\sqrt {x}}{a+b x^3} \, dx}{b^2}\\ &=\frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{9/2}}{9 b}-\frac {(2 a (A b-a B)) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^6} \, dx,x,\sqrt {x}\right )}{b^2}\\ &=\frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{9/2}}{9 b}-\frac {(2 a (A b-a B)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^{3/2}\right )}{3 b^2}\\ &=\frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{9/2}}{9 b}-\frac {2 \sqrt {a} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 67, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {a} (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 b^{5/2}}+\frac {2 x^{3/2} \left (-3 a B+3 A b+b B x^3\right )}{9 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 71, normalized size = 0.97 \begin {gather*} \frac {2 \left (a^{3/2} B-\sqrt {a} A b\right ) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 b^{5/2}}+\frac {2 x^{3/2} \left (-3 a B+3 A b+b B x^3\right )}{9 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 143, normalized size = 1.96 \begin {gather*} \left [-\frac {3 \, {\left (B a - A b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{3} - 2 \, b x^{\frac {3}{2}} \sqrt {-\frac {a}{b}} - a}{b x^{3} + a}\right ) - 2 \, {\left (B b x^{4} - 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}}{9 \, b^{2}}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{\frac {3}{2}} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (B b x^{4} - 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}\right )}}{9 \, b^{2}}\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),x, algorithm="fricas")

[Out]

[-1/9*(3*(B*a - A*b)*sqrt(-a/b)*log((b*x^3 - 2*b*x^(3/2)*sqrt(-a/b) - a)/(b*x^3 + a)) - 2*(B*b*x^4 - 3*(B*a -
A*b)*x)*sqrt(x))/b^2, 2/9*(3*(B*a - A*b)*sqrt(a/b)*arctan(b*x^(3/2)*sqrt(a/b)/a) + (B*b*x^4 - 3*(B*a - A*b)*x)
*sqrt(x))/b^2]

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giac [A]  time = 0.17, size = 64, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (B a^{2} - A a b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b^{2}} + \frac {2 \, {\left (B b^{2} x^{\frac {9}{2}} - 3 \, B a b x^{\frac {3}{2}} + 3 \, A b^{2} x^{\frac {3}{2}}\right )}}{9 \, b^{3}} \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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 78, normalized size = 1.07 \begin {gather*} \frac {2 B \,x^{\frac {9}{2}}}{9 b}-\frac {2 A a \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}\, b}+\frac {2 B \,a^{2} \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}\, b^{2}}+\frac {2 A \,x^{\frac {3}{2}}}{3 b}-\frac {2 B a \,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),x)

[Out]

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

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maxima [A]  time = 1.32, size = 58, normalized size = 0.79 \begin {gather*} \frac {2 \, {\left (B a^{2} - A a b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b^{2}} + \frac {2 \, {\left (B b x^{\frac {9}{2}} - 3 \, {\left (B a - A b\right )} x^{\frac {3}{2}}\right )}}{9 \, 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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.61, size = 111, normalized size = 1.52 \begin {gather*} x^{3/2}\,\left (\frac {2\,A}{3\,b}-\frac {2\,B\,a}{3\,b^2}\right )+\frac {2\,B\,x^{9/2}}{9\,b}-\frac {2\,\sqrt {a}\,\mathrm {atan}\left (\frac {72\,b^{3/2}\,x^{3/2}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{\sqrt {a}\,\left (72\,A\,a^2\,b^2-72\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}\right )\,\left (A\,b-B\,a\right )}{3\,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),x)

[Out]

x^(3/2)*((2*A)/(3*b) - (2*B*a)/(3*b^2)) + (2*B*x^(9/2))/(9*b) - (2*a^(1/2)*atan((72*b^(3/2)*x^(3/2)*(B^2*a^4 +
A^2*a^2*b^2 - 2*A*B*a^3*b))/(a^(1/2)*(72*A*a^2*b^2 - 72*B*a^3*b)*(A*b - B*a)))*(A*b - B*a))/(3*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),x)

[Out]

Timed out

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