3.129 \(\int \frac{(c+d x^3)^{5/12}}{(a+b x^3)^{7/4}} \, dx\)

Optimal. Leaf size=121 \[ \frac{5 x \left (c+d x^3\right )^{5/12} \left (\frac{c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \, _2F_1\left (\frac{1}{3},\frac{3}{4};\frac{4}{3};-\frac{(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{9 a \left (a+b x^3\right )^{3/4}}+\frac{4 x \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}} \]

[Out]

(4*x*(c + d*x^3)^(5/12))/(9*a*(a + b*x^3)^(3/4)) + (5*x*((c*(a + b*x^3))/(a*(c + d*x^3)))^(3/4)*(c + d*x^3)^(5
/12)*Hypergeometric2F1[1/3, 3/4, 4/3, -(((b*c - a*d)*x^3)/(a*(c + d*x^3)))])/(9*a*(a + b*x^3)^(3/4))

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Rubi [A]  time = 0.0373702, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {378, 380} \[ \frac{5 x \left (c+d x^3\right )^{5/12} \left (\frac{c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \, _2F_1\left (\frac{1}{3},\frac{3}{4};\frac{4}{3};-\frac{(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{9 a \left (a+b x^3\right )^{3/4}}+\frac{4 x \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^(5/12)/(a + b*x^3)^(7/4),x]

[Out]

(4*x*(c + d*x^3)^(5/12))/(9*a*(a + b*x^3)^(3/4)) + (5*x*((c*(a + b*x^3))/(a*(c + d*x^3)))^(3/4)*(c + d*x^3)^(5
/12)*Hypergeometric2F1[1/3, 3/4, 4/3, -(((b*c - a*d)*x^3)/(a*(c + d*x^3)))])/(9*a*(a + b*x^3)^(3/4))

Rule 378

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

Rule 380

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(x*(a + b*x^n)^p*Hypergeome
tric2F1[1/n, -p, 1 + 1/n, -(((b*c - a*d)*x^n)/(a*(c + d*x^n)))])/(c*((c*(a + b*x^n))/(a*(c + d*x^n)))^p*(c + d
*x^n)^(1/n + p)), x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0]

Rubi steps

\begin{align*} \int \frac{\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx &=\frac{4 x \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}}+\frac{(5 c) \int \frac{1}{\left (a+b x^3\right )^{3/4} \left (c+d x^3\right )^{7/12}} \, dx}{9 a}\\ &=\frac{4 x \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}}+\frac{5 x \left (\frac{c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{5/12} \, _2F_1\left (\frac{1}{3},\frac{3}{4};\frac{4}{3};-\frac{(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{9 a \left (a+b x^3\right )^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.0248773, size = 89, normalized size = 0.74 \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{3/4} \left (c+d x^3\right )^{5/12} \, _2F_1\left (\frac{1}{3},\frac{7}{4};\frac{4}{3};\frac{(a d-b c) x^3}{a \left (d x^3+c\right )}\right )}{a \left (a+b x^3\right )^{3/4} \left (\frac{d x^3}{c}+1\right )^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x^3)^(5/12)/(a + b*x^3)^(7/4),x]

[Out]

(x*(1 + (b*x^3)/a)^(3/4)*(c + d*x^3)^(5/12)*Hypergeometric2F1[1/3, 7/4, 4/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^
3))])/(a*(a + b*x^3)^(3/4)*(1 + (d*x^3)/c)^(3/4))

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Maple [F]  time = 0.433, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d{x}^{3}+c \right ) ^{{\frac{5}{12}}} \left ( b{x}^{3}+a \right ) ^{-{\frac{7}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x)

[Out]

int((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{\frac{5}{12}}}{{\left (b x^{3} + a\right )}^{\frac{7}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)^(5/12)/(b*x^3 + a)^(7/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{3} + a\right )}^{\frac{1}{4}}{\left (d x^{3} + c\right )}^{\frac{5}{12}}}{b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^(1/4)*(d*x^3 + c)^(5/12)/(b^2*x^6 + 2*a*b*x^3 + a^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)**(5/12)/(b*x**3+a)**(7/4),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{\frac{5}{12}}}{{\left (b x^{3} + a\right )}^{\frac{7}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^(5/12)/(b*x^3 + a)^(7/4), x)