3.360 \(\int \frac{x^2 \sqrt{c+d x^3}}{a+b x^3} \, dx\)

Optimal. Leaf size=70 \[ \frac{2 \sqrt{c+d x^3}}{3 b}-\frac{2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{3/2}} \]

[Out]

(2*Sqrt[c + d*x^3])/(3*b) - (2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(3/2))

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Rubi [A]  time = 0.0585363, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {444, 50, 63, 208} \[ \frac{2 \sqrt{c+d x^3}}{3 b}-\frac{2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sqrt[c + d*x^3])/(a + b*x^3),x]

[Out]

(2*Sqrt[c + d*x^3])/(3*b) - (2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(3*b^(3/2))

Rule 444

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{x^2 \sqrt{c+d x^3}}{a+b x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{a+b x} \, dx,x,x^3\right )\\ &=\frac{2 \sqrt{c+d x^3}}{3 b}+\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{3 b}\\ &=\frac{2 \sqrt{c+d x^3}}{3 b}+\frac{(2 (b c-a d)) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{3 b d}\\ &=\frac{2 \sqrt{c+d x^3}}{3 b}-\frac{2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0303183, size = 70, normalized size = 1. \[ \frac{1}{3} \left (\frac{2 \sqrt{c+d x^3}}{b}-\frac{2 \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{b^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sqrt[c + d*x^3])/(a + b*x^3),x]

[Out]

((2*Sqrt[c + d*x^3])/b - (2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/b^(3/2))/3

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Maple [C]  time = 0.006, size = 434, normalized size = 6.2 \begin{align*}{\frac{2}{3\,b}\sqrt{d{x}^{3}+c}}+{\frac{{\frac{i}{3}}\sqrt{2}}{b{d}^{2}}\sum _{{\it \_alpha}={\it RootOf} \left ( b{{\it \_Z}}^{3}+a \right ) }{\sqrt [3]{-{d}^{2}c}\sqrt{{{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( -i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}\sqrt{{d \left ( x-{\frac{1}{d}\sqrt [3]{-{d}^{2}c}} \right ) \left ( -3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c} \right ) ^{-1}}}\sqrt{{-{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i\sqrt{3} \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{{i\sqrt{3}d \left ( x+{\frac{1}{2\,d}\sqrt [3]{-{d}^{2}c}}-{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}},{\frac{b}{2\,d \left ( ad-bc \right ) } \left ( 2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd \right ) },\sqrt{{\frac{i\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c} \left ( -{\frac{3}{2\,d}\sqrt [3]{-{d}^{2}c}}+{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{d{x}^{3}+c}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(d*x^3+c)^(1/2)/(b*x^3+a),x)

[Out]

2/3*(d*x^3+c)^(1/2)/b+1/3*I/b/d^2*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^
2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/
2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^
2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*El
lipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2
),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3
)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^
(1/2)),_alpha=RootOf(_Z^3*b+a))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.29459, size = 331, normalized size = 4.73 \begin{align*} \left [\frac{\sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x^{3} + 2 \, b c - a d - 2 \, \sqrt{d x^{3} + c} b \sqrt{\frac{b c - a d}{b}}}{b x^{3} + a}\right ) + 2 \, \sqrt{d x^{3} + c}}{3 \, b}, -\frac{2 \,{\left (\sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x^{3} + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) - \sqrt{d x^{3} + c}\right )}}{3 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 5.5318, size = 68, normalized size = 0.97 \begin{align*} \frac{2 \left (\frac{d \sqrt{c + d x^{3}}}{3 b} - \frac{d \left (a d - b c\right ) \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{3 b^{2} \sqrt{\frac{a d - b c}{b}}}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(d*sqrt(c + d*x**3)/(3*b) - d*(a*d - b*c)*atan(sqrt(c + d*x**3)/sqrt((a*d - b*c)/b))/(3*b**2*sqrt((a*d - b*c
)/b)))/d

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Giac [A]  time = 1.10274, size = 89, normalized size = 1.27 \begin{align*} \frac{2 \,{\left (b c - a d\right )} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{3 \, \sqrt{-b^{2} c + a b d} b} + \frac{2 \, \sqrt{d x^{3} + c}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*c - a*d)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b) + 2/3*sqrt(d*x^3 + c)/
b