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

Optimal. Leaf size=53 \[ -\frac {2 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{3/2} \sqrt {b}}-\frac {2 A}{3 a x^{3/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {453, 329, 275, 205} \begin {gather*} -\frac {2 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{3/2} \sqrt {b}}-\frac {2 A}{3 a x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(3*a*x^(3/2)) - (2*(A*b - a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(3*a^(3/2)*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 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 453

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 53, normalized size = 1.00 \begin {gather*} \frac {2 (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{3/2} \sqrt {b}}-\frac {2 A}{3 a x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.05, size = 53, normalized size = 1.00 \begin {gather*} \frac {2 (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{3/2} \sqrt {b}}-\frac {2 A}{3 a x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.78, size = 120, normalized size = 2.26 \begin {gather*} \left [\frac {{\left (B a - A b\right )} \sqrt {-a b} x^{2} \log \left (\frac {b x^{3} + 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right ) - 2 \, A a b \sqrt {x}}{3 \, a^{2} b x^{2}}, \frac {2 \, {\left ({\left (B a - A b\right )} \sqrt {a b} x^{2} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right ) - A a b \sqrt {x}\right )}}{3 \, a^{2} b x^{2}}\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),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 39, normalized size = 0.74 \begin {gather*} \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} a} - \frac {2 \, A}{3 \, a x^{\frac {3}{2}}} \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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 53, normalized size = 1.00 \begin {gather*} -\frac {2 A b \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}\, a}+\frac {2 B \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}}-\frac {2 A}{3 a \,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),x)

[Out]

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

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maxima [A]  time = 1.40, size = 39, normalized size = 0.74 \begin {gather*} \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} a} - \frac {2 \, A}{3 \, a x^{\frac {3}{2}}} \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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.10, size = 102, normalized size = 1.92 \begin {gather*} -\frac {2\,A}{3\,a\,x^{3/2}}-\frac {2\,\mathrm {atan}\left (\frac {3\,a^{3/2}\,\sqrt {b}\,x^{3/2}\,\left (24\,A^2\,a^3\,b^5-48\,A\,B\,a^4\,b^4+24\,B^2\,a^5\,b^3\right )}{\left (A\,b-B\,a\right )\,\left (72\,A\,a^5\,b^4-72\,B\,a^6\,b^3\right )}\right )\,\left (A\,b-B\,a\right )}{3\,a^{3/2}\,\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)),x)

[Out]

- (2*A)/(3*a*x^(3/2)) - (2*atan((3*a^(3/2)*b^(1/2)*x^(3/2)*(24*A^2*a^3*b^5 + 24*B^2*a^5*b^3 - 48*A*B*a^4*b^4))
/((A*b - B*a)*(72*A*a^5*b^4 - 72*B*a^6*b^3)))*(A*b - B*a))/(3*a^(3/2)*b^(1/2))

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sympy [A]  time = 156.58, size = 527, normalized size = 9.94 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{9 x^{\frac {9}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{b} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} + \frac {2 B x^{\frac {3}{2}}}{3}}{a} & \text {for}\: b = 0 \\- \frac {2 A}{3 a x^{\frac {3}{2}}} + \frac {i A \log {\left (- \sqrt [6]{-1} \sqrt [6]{a} \sqrt [6]{\frac {1}{b}} + \sqrt {x} \right )}}{3 a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {i A \log {\left (\sqrt [6]{-1} \sqrt [6]{a} \sqrt [6]{\frac {1}{b}} + \sqrt {x} \right )}}{3 a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {i A \log {\left (- 4 \sqrt [6]{-1} \sqrt [6]{a} \sqrt {x} \sqrt [6]{\frac {1}{b}} + 4 \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac {1}{b}} + 4 x \right )}}{3 a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} + \frac {i A \log {\left (4 \sqrt [6]{-1} \sqrt [6]{a} \sqrt {x} \sqrt [6]{\frac {1}{b}} + 4 \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac {1}{b}} + 4 x \right )}}{3 a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {i B \log {\left (- \sqrt [6]{-1} \sqrt [6]{a} \sqrt [6]{\frac {1}{b}} + \sqrt {x} \right )}}{3 \sqrt {a} b \sqrt {\frac {1}{b}}} + \frac {i B \log {\left (\sqrt [6]{-1} \sqrt [6]{a} \sqrt [6]{\frac {1}{b}} + \sqrt {x} \right )}}{3 \sqrt {a} b \sqrt {\frac {1}{b}}} + \frac {i B \log {\left (- 4 \sqrt [6]{-1} \sqrt [6]{a} \sqrt {x} \sqrt [6]{\frac {1}{b}} + 4 \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac {1}{b}} + 4 x \right )}}{3 \sqrt {a} b \sqrt {\frac {1}{b}}} - \frac {i B \log {\left (4 \sqrt [6]{-1} \sqrt [6]{a} \sqrt {x} \sqrt [6]{\frac {1}{b}} + 4 \sqrt [3]{-1} \sqrt [3]{a} \sqrt [3]{\frac {1}{b}} + 4 x \right )}}{3 \sqrt {a} b \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \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),x)

[Out]

Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(3*x**(3/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(3*x**(
3/2)))/b, Eq(a, 0)), ((-2*A/(3*x**(3/2)) + 2*B*x**(3/2)/3)/a, Eq(b, 0)), (-2*A/(3*a*x**(3/2)) + I*A*log(-(-1)*
*(1/6)*a**(1/6)*(1/b)**(1/6) + sqrt(x))/(3*a**(3/2)*sqrt(1/b)) - I*A*log((-1)**(1/6)*a**(1/6)*(1/b)**(1/6) + s
qrt(x))/(3*a**(3/2)*sqrt(1/b)) - I*A*log(-4*(-1)**(1/6)*a**(1/6)*sqrt(x)*(1/b)**(1/6) + 4*(-1)**(1/3)*a**(1/3)
*(1/b)**(1/3) + 4*x)/(3*a**(3/2)*sqrt(1/b)) + I*A*log(4*(-1)**(1/6)*a**(1/6)*sqrt(x)*(1/b)**(1/6) + 4*(-1)**(1
/3)*a**(1/3)*(1/b)**(1/3) + 4*x)/(3*a**(3/2)*sqrt(1/b)) - I*B*log(-(-1)**(1/6)*a**(1/6)*(1/b)**(1/6) + sqrt(x)
)/(3*sqrt(a)*b*sqrt(1/b)) + I*B*log((-1)**(1/6)*a**(1/6)*(1/b)**(1/6) + sqrt(x))/(3*sqrt(a)*b*sqrt(1/b)) + I*B
*log(-4*(-1)**(1/6)*a**(1/6)*sqrt(x)*(1/b)**(1/6) + 4*(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + 4*x)/(3*sqrt(a)*b*sq
rt(1/b)) - I*B*log(4*(-1)**(1/6)*a**(1/6)*sqrt(x)*(1/b)**(1/6) + 4*(-1)**(1/3)*a**(1/3)*(1/b)**(1/3) + 4*x)/(3
*sqrt(a)*b*sqrt(1/b)), True))

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