3.2.61 \(\int \frac {A+B x^3}{x^{5/2} (a+b x^3)} \, dx\) [161]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {464, 335, 281, 211} \begin {gather*} -\frac {2 (A b-a B) \text {ArcTan}\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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 281

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 335

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 464

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 ) \text {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)) \text {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}{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 {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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Maple [A]
time = 0.31, size = 40, normalized size = 0.75

method	result	size
derivativedivides	\(-\frac {2 A}{3 a \,x^{\frac {3}{2}}}-\frac {2 \left (A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 a \sqrt {a b}}\)	\(40\)
default	\(-\frac {2 A}{3 a \,x^{\frac {3}{2}}}-\frac {2 \left (A b -B a \right ) \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 a \sqrt {a b}}\)	\(40\)
risch	\(-\frac {2 A}{3 a \,x^{\frac {3}{2}}}-\frac {2 \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right ) A b}{3 a \sqrt {a b}}+\frac {2 \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right ) B}{3 \sqrt {a b}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^(5/2)/(b*x^3+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, 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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Fricas [A]
time = 1.87, 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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (53) = 106\).
time = 20.93, size = 371, normalized size = 7.00 \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 {A \log {\left (\sqrt {x} - \sqrt [6]{- \frac {a}{b}} \right )}}{3 a \sqrt {- \frac {a}{b}}} + \frac {A \log {\left (\sqrt {x} + \sqrt [6]{- \frac {a}{b}} \right )}}{3 a \sqrt {- \frac {a}{b}}} + \frac {A \log {\left (- 4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 a \sqrt {- \frac {a}{b}}} - \frac {A \log {\left (4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 a \sqrt {- \frac {a}{b}}} - \frac {2 A}{3 a x^{\frac {3}{2}}} + \frac {B \log {\left (\sqrt {x} - \sqrt [6]{- \frac {a}{b}} \right )}}{3 b \sqrt {- \frac {a}{b}}} - \frac {B \log {\left (\sqrt {x} + \sqrt [6]{- \frac {a}{b}} \right )}}{3 b \sqrt {- \frac {a}{b}}} - \frac {B \log {\left (- 4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b \sqrt {- \frac {a}{b}}} + \frac {B \log {\left (4 \sqrt {x} \sqrt [6]{- \frac {a}{b}} + 4 x + 4 \sqrt [3]{- \frac {a}{b}} \right )}}{3 b \sqrt {- \frac {a}{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)), (-A*log(sqrt(x) - (-a/b)**(1/6))/(3*a
*sqrt(-a/b)) + A*log(sqrt(x) + (-a/b)**(1/6))/(3*a*sqrt(-a/b)) + A*log(-4*sqrt(x)*(-a/b)**(1/6) + 4*x + 4*(-a/
b)**(1/3))/(3*a*sqrt(-a/b)) - A*log(4*sqrt(x)*(-a/b)**(1/6) + 4*x + 4*(-a/b)**(1/3))/(3*a*sqrt(-a/b)) - 2*A/(3
*a*x**(3/2)) + B*log(sqrt(x) - (-a/b)**(1/6))/(3*b*sqrt(-a/b)) - B*log(sqrt(x) + (-a/b)**(1/6))/(3*b*sqrt(-a/b
)) - B*log(-4*sqrt(x)*(-a/b)**(1/6) + 4*x + 4*(-a/b)**(1/3))/(3*b*sqrt(-a/b)) + B*log(4*sqrt(x)*(-a/b)**(1/6)
+ 4*x + 4*(-a/b)**(1/3))/(3*b*sqrt(-a/b)), True))

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Giac [A]
time = 0.57, 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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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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