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

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

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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 = {459, 329, 275, 205} \begin {gather*} \frac {2 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 \sqrt {a} b^{3/2}}+\frac {2 B x^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 52, normalized size = 0.98 \begin {gather*} \frac {2}{3} \left (\frac {B x^{3/2}}{b}-\frac {(a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 108, normalized size = 2.04 \begin {gather*} \left [\frac {2 \, B a b x^{\frac {3}{2}} + {\left (B a - A b\right )} \sqrt {-a b} \log \left (\frac {b x^{3} - 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right )}{3 \, a b^{2}}, \frac {2 \, {\left (B a b x^{\frac {3}{2}} - {\left (B a - A b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right )\right )}}{3 \, a b^{2}}\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),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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maple [A]  time = 0.05, size = 53, normalized size = 1.00 \begin {gather*} \frac {2 A \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}}-\frac {2 B a \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}\, b}+\frac {2 B \,x^{\frac {3}{2}}}{3 b} \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),x)

[Out]

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

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maxima [A]  time = 1.21, size = 39, normalized size = 0.74 \begin {gather*} \frac {2 \, B x^{\frac {3}{2}}}{3 \, b} - \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} 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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.60, size = 93, normalized size = 1.75 \begin {gather*} \frac {2\,B\,x^{3/2}}{3\,b}-\frac {2\,\mathrm {atan}\left (\frac {3\,\sqrt {a}\,b^{3/2}\,x^{3/2}\,\left (24\,A^2\,b^3-48\,A\,B\,a\,b^2+24\,B^2\,a^2\,b\right )}{\left (72\,B\,a^2\,b^2-72\,A\,a\,b^3\right )\,\left (A\,b-B\,a\right )}\right )\,\left (A\,b-B\,a\right )}{3\,\sqrt {a}\,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),x)

[Out]

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

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sympy [A]  time = 22.39, size = 537, normalized size = 10.13 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} + \frac {2 B x^{\frac {3}{2}}}{3}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} + \frac {2 B x^{\frac {3}{2}}}{3}}{b} & \text {for}\: a = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {9}{2}}}{9}}{a} & \text {for}\: b = 0 \\- \frac {i A \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 A \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 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 \sqrt {a} b \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 \sqrt {a} b \sqrt {\frac {1}{b}}} + \frac {i B \sqrt {a} \log {\left (- \sqrt [6]{-1} \sqrt [6]{a} \sqrt [6]{\frac {1}{b}} + \sqrt {x} \right )}}{3 b^{2} \sqrt {\frac {1}{b}}} - \frac {i B \sqrt {a} \log {\left (\sqrt [6]{-1} \sqrt [6]{a} \sqrt [6]{\frac {1}{b}} + \sqrt {x} \right )}}{3 b^{2} \sqrt {\frac {1}{b}}} - \frac {i B \sqrt {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 b^{2} \sqrt {\frac {1}{b}}} + \frac {i B \sqrt {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 b^{2} \sqrt {\frac {1}{b}}} + \frac {2 B x^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases} \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),x)

[Out]

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

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