3.123 \(\int \frac{(a+b x^3)^{5/4}}{(c+d x^3)^{31/12}} \, dx\)

Optimal. Leaf size=155 \[ \frac{45 a^2 x \left (c+d x^3\right )^{5/12} \left (\frac{c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \, _2F_1\left (\frac{1}{3},\frac{3}{4};\frac{4}{3};-\frac{(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{133 c^3 \left (a+b x^3\right )^{3/4}}+\frac{60 a x \sqrt [4]{a+b x^3}}{133 c^2 \left (c+d x^3\right )^{7/12}}+\frac{4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}} \]

[Out]

(4*x*(a + b*x^3)^(5/4))/(19*c*(c + d*x^3)^(19/12)) + (60*a*x*(a + b*x^3)^(1/4))/(133*c^2*(c + d*x^3)^(7/12)) +
(45*a^2*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)))])/(133*c^3*(a + b*x^3)^(3/4))

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Rubi [A]  time = 0.0577109, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {378, 380} \[ \frac{45 a^2 x \left (c+d x^3\right )^{5/12} \left (\frac{c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \, _2F_1\left (\frac{1}{3},\frac{3}{4};\frac{4}{3};-\frac{(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{133 c^3 \left (a+b x^3\right )^{3/4}}+\frac{60 a x \sqrt [4]{a+b x^3}}{133 c^2 \left (c+d x^3\right )^{7/12}}+\frac{4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(5/4)/(c + d*x^3)^(31/12),x]

[Out]

(4*x*(a + b*x^3)^(5/4))/(19*c*(c + d*x^3)^(19/12)) + (60*a*x*(a + b*x^3)^(1/4))/(133*c^2*(c + d*x^3)^(7/12)) +
(45*a^2*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)))])/(133*c^3*(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 (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx &=\frac{4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}+\frac{(15 a) \int \frac{\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx}{19 c}\\ &=\frac{4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}+\frac{60 a x \sqrt [4]{a+b x^3}}{133 c^2 \left (c+d x^3\right )^{7/12}}+\frac{\left (45 a^2\right ) \int \frac{1}{\left (a+b x^3\right )^{3/4} \left (c+d x^3\right )^{7/12}} \, dx}{133 c^2}\\ &=\frac{4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}+\frac{60 a x \sqrt [4]{a+b x^3}}{133 c^2 \left (c+d x^3\right )^{7/12}}+\frac{45 a^2 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 )}{133 c^3 \left (a+b x^3\right )^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.0315534, size = 90, normalized size = 0.58 \[ \frac{a x \sqrt [4]{a+b x^3} \sqrt [4]{\frac{d x^3}{c}+1} \, _2F_1\left (-\frac{5}{4},\frac{1}{3};\frac{4}{3};\frac{(a d-b c) x^3}{a \left (d x^3+c\right )}\right )}{c^2 \sqrt [4]{\frac{b x^3}{a}+1} \left (c+d x^3\right )^{7/12}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(5/4)/(c + d*x^3)^(31/12),x]

[Out]

(a*x*(a + b*x^3)^(1/4)*(1 + (d*x^3)/c)^(1/4)*Hypergeometric2F1[-5/4, 1/3, 4/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*
x^3))])/(c^2*(1 + (b*x^3)/a)^(1/4)*(c + d*x^3)^(7/12))

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Maple [F]  time = 0.487, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b{x}^{3}+a \right ) ^{{\frac{5}{4}}} \left ( d{x}^{3}+c \right ) ^{-{\frac{31}{12}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x)

[Out]

int((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{5}{4}}}{{\left (d x^{3} + c\right )}^{\frac{31}{12}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(5/4)/(d*x^3 + c)^(31/12), 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{5}{4}}{\left (d x^{3} + c\right )}^{\frac{5}{12}}}{d^{3} x^{9} + 3 \, c d^{2} x^{6} + 3 \, c^{2} d x^{3} + c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^(5/4)*(d*x^3 + c)^(5/12)/(d^3*x^9 + 3*c*d^2*x^6 + 3*c^2*d*x^3 + c^3), 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((b*x**3+a)**(5/4)/(d*x**3+c)**(31/12),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{5}{4}}}{{\left (d x^{3} + c\right )}^{\frac{31}{12}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(5/4)/(d*x^3 + c)^(31/12), x)