3.2.70 \(\int \frac {\sqrt {x} (A+B x^3)}{(a+b x^3)^3} \, dx\)

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

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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, 290, 329, 275, 205} \begin {gather*} \frac {(a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{5/2} b^{3/2}}+\frac {x^{3/2} (a B+3 A b)}{12 a^2 b \left (a+b x^3\right )}+\frac {x^{3/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 {\sqrt {x} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx &=\frac {(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac {\left (\frac {9 A b}{2}+\frac {3 a B}{2}\right ) \int \frac {\sqrt {x}}{\left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac {(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac {(3 A b+a B) x^{3/2}}{12 a^2 b \left (a+b x^3\right )}+\frac {(3 A b+a B) \int \frac {\sqrt {x}}{a+b x^3} \, dx}{8 a^2 b}\\ &=\frac {(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac {(3 A b+a B) x^{3/2}}{12 a^2 b \left (a+b x^3\right )}+\frac {(3 A b+a B) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^6} \, dx,x,\sqrt {x}\right )}{4 a^2 b}\\ &=\frac {(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac {(3 A b+a B) x^{3/2}}{12 a^2 b \left (a+b x^3\right )}+\frac {(3 A b+a B) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^{3/2}\right )}{12 a^2 b}\\ &=\frac {(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac {(3 A b+a B) x^{3/2}}{12 a^2 b \left (a+b x^3\right )}+\frac {(3 A b+a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{5/2} b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 94, normalized size = 0.90 \begin {gather*} \frac {\frac {\sqrt {a} \sqrt {b} x^{3/2} \left (-a^2 B+a b \left (5 A+B x^3\right )+3 A b^2 x^3\right )}{\left (a+b x^3\right )^2}+(a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{5/2} b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 92, normalized size = 0.88 \begin {gather*} \frac {(a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{12 a^{5/2} b^{3/2}}-\frac {x^{3/2} \left (a^2 B-5 a A b-a b B x^3-3 A b^2 x^3\right )}{12 a^2 b \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 313, normalized size = 3.01 \begin {gather*} \left [-\frac {{\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{6} + B a^{3} + 3 \, A a^{2} b + 2 \, {\left (B a^{2} b + 3 \, 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 (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{4} - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{3} b^{4} x^{6} + 2 \, a^{4} b^{3} x^{3} + a^{5} b^{2}\right )}}, \frac {{\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{6} + B a^{3} + 3 \, A a^{2} b + 2 \, {\left (B a^{2} b + 3 \, 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 (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{4} - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{12 \, {\left (a^{3} b^{4} x^{6} + 2 \, a^{4} b^{3} x^{3} + a^{5} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*x^(1/2)/(b*x^3+a)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 84, normalized size = 0.81 \begin {gather*} \frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \, \sqrt {a b} a^{2} b} + \frac {B a b x^{\frac {9}{2}} + 3 \, A b^{2} x^{\frac {9}{2}} - B a^{2} x^{\frac {3}{2}} + 5 \, A a b x^{\frac {3}{2}}}{12 \, {\left (b x^{3} + a\right )}^{2} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*x^(1/2)/(b*x^3+a)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 97, normalized size = 0.93 \begin {gather*} \frac {A \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{2}}+\frac {B \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \sqrt {a b}\, a b}+\frac {\frac {\left (3 A b +B a \right ) x^{\frac {9}{2}}}{12 a^{2}}+\frac {\left (5 A b -B a \right ) x^{\frac {3}{2}}}{12 a b}}{\left (b \,x^{3}+a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)*x^(1/2)/(b*x^3+a)^3,x)

[Out]

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

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maxima [A]  time = 1.27, size = 96, normalized size = 0.92 \begin {gather*} \frac {{\left (B a b + 3 \, A b^{2}\right )} x^{\frac {9}{2}} - {\left (B a^{2} - 5 \, A a b\right )} x^{\frac {3}{2}}}{12 \, {\left (a^{2} b^{3} x^{6} + 2 \, a^{3} b^{2} x^{3} + a^{4} b\right )}} + \frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{12 \, \sqrt {a b} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*x^(1/2)/(b*x^3+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.71, size = 136, normalized size = 1.31 \begin {gather*} \frac {\frac {x^{9/2}\,\left (3\,A\,b+B\,a\right )}{12\,a^2}+\frac {x^{3/2}\,\left (5\,A\,b-B\,a\right )}{12\,a\,b}}{a^2+2\,a\,b\,x^3+b^2\,x^6}+\frac {\mathrm {atan}\left (\frac {b^{3/2}\,x^{3/2}\,\left (9\,A^2\,b^3+6\,A\,B\,a\,b^2+B^2\,a^2\,b\right )}{\sqrt {a}\,\left (3\,A\,b+B\,a\right )\,\left (3\,A\,b^3+B\,a\,b^2\right )}\right )\,\left (3\,A\,b+B\,a\right )}{12\,a^{5/2}\,b^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/2)*(A + B*x^3))/(a + b*x^3)^3,x)

[Out]

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

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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**(1/2)/(b*x**3+a)**3,x)

[Out]

Timed out

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