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

Optimal. Leaf size=132 \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2-5 a b c d+5 b^2 c^2\right ) \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{5 b^2 \left (a+b x^3\right )^{2/3}}+\frac{2 d x \sqrt [3]{a+b x^3} (2 b c-a d)}{5 b^2}+\frac{d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b} \]

[Out]

(2*d*(2*b*c - a*d)*x*(a + b*x^3)^(1/3))/(5*b^2) + (d*x*(a + b*x^3)^(1/3)*(c + d*x^3))/(5*b) + ((5*b^2*c^2 - 5*
a*b*c*d + 2*a^2*d^2)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(5*b^2*(a + b*x^3
)^(2/3))

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Rubi [A]  time = 0.0754575, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {416, 388, 246, 245} \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2-5 a b c d+5 b^2 c^2\right ) \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{5 b^2 \left (a+b x^3\right )^{2/3}}+\frac{2 d x \sqrt [3]{a+b x^3} (2 b c-a d)}{5 b^2}+\frac{d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(2*b*c - a*d)*x*(a + b*x^3)^(1/3))/(5*b^2) + (d*x*(a + b*x^3)^(1/3)*(c + d*x^3))/(5*b) + ((5*b^2*c^2 - 5*
a*b*c*d + 2*a^2*d^2)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(5*b^2*(a + b*x^3
)^(2/3))

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 388

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{2/3}} \, dx &=\frac{d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}+\frac{\int \frac{c (5 b c-a d)+4 d (2 b c-a d) x^3}{\left (a+b x^3\right )^{2/3}} \, dx}{5 b}\\ &=\frac{2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac{d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}-\frac{(4 a d (2 b c-a d)-2 b c (5 b c-a d)) \int \frac{1}{\left (a+b x^3\right )^{2/3}} \, dx}{10 b^2}\\ &=\frac{2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac{d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}-\frac{\left ((4 a d (2 b c-a d)-2 b c (5 b c-a d)) \left (1+\frac{b x^3}{a}\right )^{2/3}\right ) \int \frac{1}{\left (1+\frac{b x^3}{a}\right )^{2/3}} \, dx}{10 b^2 \left (a+b x^3\right )^{2/3}}\\ &=\frac{2 d (2 b c-a d) x \sqrt [3]{a+b x^3}}{5 b^2}+\frac{d x \sqrt [3]{a+b x^3} \left (c+d x^3\right )}{5 b}+\frac{\left (5 b^2 c^2-5 a b c d+2 a^2 d^2\right ) x \left (1+\frac{b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{5 b^2 \left (a+b x^3\right )^{2/3}}\\ \end{align*}

Mathematica [B]  time = 4.10862, size = 304, normalized size = 2.3 \[ -\frac{x \text{Gamma}\left (\frac{4}{3}\right ) \left (\frac{b x^3}{a}+1\right )^{2/3} \left (81 b x^3 \text{Gamma}\left (\frac{10}{3}\right ) \left (c+d x^3\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{4}{3},\frac{5}{3},2\right \},\left \{1,\frac{13}{3}\right \},-\frac{b x^3}{a}\right )-270 a \text{Gamma}\left (\frac{10}{3}\right ) \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{10}{3};-\frac{b x^3}{a}\right )+297 b c^2 x^3 \text{Gamma}\left (\frac{10}{3}\right ) \, _2F_1\left (\frac{4}{3},\frac{5}{3};\frac{13}{3};-\frac{b x^3}{a}\right )+432 b c d x^6 \text{Gamma}\left (\frac{10}{3}\right ) \, _2F_1\left (\frac{4}{3},\frac{5}{3};\frac{13}{3};-\frac{b x^3}{a}\right )+135 b d^2 x^9 \text{Gamma}\left (\frac{10}{3}\right ) \, _2F_1\left (\frac{4}{3},\frac{5}{3};\frac{13}{3};-\frac{b x^3}{a}\right )+3780 a c^2 \text{Gamma}\left (\frac{10}{3}\right )-3920 a c^2 \text{Gamma}\left (\frac{1}{3}\right )+1890 a c d x^3 \text{Gamma}\left (\frac{10}{3}\right )-1960 a c d x^3 \text{Gamma}\left (\frac{1}{3}\right )+540 a d^2 x^6 \text{Gamma}\left (\frac{10}{3}\right )-560 a d^2 x^6 \text{Gamma}\left (\frac{1}{3}\right )\right )}{1260 a \text{Gamma}\left (\frac{1}{3}\right ) \text{Gamma}\left (\frac{10}{3}\right ) \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(x*(1 + (b*x^3)/a)^(2/3)*Gamma[4/3]*(-3920*a*c^2*Gamma[1/3] - 1960*a*c*d*x^3*Gamma[1/3] - 560*a*d^2*x^6*Gamma
[1/3] + 3780*a*c^2*Gamma[10/3] + 1890*a*c*d*x^3*Gamma[10/3] + 540*a*d^2*x^6*Gamma[10/3] - 270*a*(14*c^2 + 7*c*
d*x^3 + 2*d^2*x^6)*Gamma[10/3]*Hypergeometric2F1[1/3, 2/3, 10/3, -((b*x^3)/a)] + 297*b*c^2*x^3*Gamma[10/3]*Hyp
ergeometric2F1[4/3, 5/3, 13/3, -((b*x^3)/a)] + 432*b*c*d*x^6*Gamma[10/3]*Hypergeometric2F1[4/3, 5/3, 13/3, -((
b*x^3)/a)] + 135*b*d^2*x^9*Gamma[10/3]*Hypergeometric2F1[4/3, 5/3, 13/3, -((b*x^3)/a)] + 81*b*x^3*(c + d*x^3)^
2*Gamma[10/3]*HypergeometricPFQ[{4/3, 5/3, 2}, {1, 13/3}, -((b*x^3)/a)]))/(1260*a*(a + b*x^3)^(2/3)*Gamma[1/3]
*Gamma[10/3])

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Maple [F]  time = 0.217, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d{x}^{3}+c \right ) ^{2} \left ( b{x}^{3}+a \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d^{2} x^{6} + 2 \, c d x^{3} + c^{2}}{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d^2*x^6 + 2*c*d*x^3 + c^2)/(b*x^3 + a)^(2/3), x)

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Sympy [C]  time = 2.88902, size = 126, normalized size = 0.95 \begin{align*} \frac{c^{2} x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{2}{3} \\ \frac{4}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{2}{3}} \Gamma \left (\frac{4}{3}\right )} + \frac{2 c d x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{2}{3}, \frac{4}{3} \\ \frac{7}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{2}{3}} \Gamma \left (\frac{7}{3}\right )} + \frac{d^{2} x^{7} \Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{2}{3}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{2}{3}} \Gamma \left (\frac{10}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*x*gamma(1/3)*hyper((1/3, 2/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(4/3)) + 2*c*d*x**4*gam
ma(4/3)*hyper((2/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(7/3)) + d**2*x**7*gamma(7/3)*hyp
er((2/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(10/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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