3.124 \(\int \frac {(a+b x^3)^{3/4}}{(c+d x^3)^{25/12}} \, dx\)

Optimal. Leaf size=122 \[ \frac {9 a x \sqrt [4]{\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}} \, _2F_1\left (\frac {1}{4},\frac {1}{3};\frac {4}{3};-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{13 c^2 \sqrt [4]{a+b x^3} \sqrt [12]{c+d x^3}}+\frac {4 x \left (a+b x^3\right )^{3/4}}{13 c \left (c+d x^3\right )^{13/12}} \]

[Out]

4/13*x*(b*x^3+a)^(3/4)/c/(d*x^3+c)^(13/12)+9/13*a*x*(c*(b*x^3+a)/a/(d*x^3+c))^(1/4)*hypergeom([1/4, 1/3],[4/3]
,-(-a*d+b*c)*x^3/a/(d*x^3+c))/c^2/(b*x^3+a)^(1/4)/(d*x^3+c)^(1/12)

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Rubi [A]  time = 0.04, antiderivative size = 122, normalized size of antiderivative = 1.00, 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} \[ \frac {9 a x \sqrt [4]{\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}} \, _2F_1\left (\frac {1}{4},\frac {1}{3};\frac {4}{3};-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{13 c^2 \sqrt [4]{a+b x^3} \sqrt [12]{c+d x^3}}+\frac {4 x \left (a+b x^3\right )^{3/4}}{13 c \left (c+d x^3\right )^{13/12}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(3/4)/(c + d*x^3)^(25/12),x]

[Out]

(4*x*(a + b*x^3)^(3/4))/(13*c*(c + d*x^3)^(13/12)) + (9*a*x*((c*(a + b*x^3))/(a*(c + d*x^3)))^(1/4)*Hypergeome
tric2F1[1/4, 1/3, 4/3, -(((b*c - a*d)*x^3)/(a*(c + d*x^3)))])/(13*c^2*(a + b*x^3)^(1/4)*(c + d*x^3)^(1/12))

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 )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx &=\frac {4 x \left (a+b x^3\right )^{3/4}}{13 c \left (c+d x^3\right )^{13/12}}+\frac {(9 a) \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx}{13 c}\\ &=\frac {4 x \left (a+b x^3\right )^{3/4}}{13 c \left (c+d x^3\right )^{13/12}}+\frac {9 a x \sqrt [4]{\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}} \, _2F_1\left (\frac {1}{4},\frac {1}{3};\frac {4}{3};-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{13 c^2 \sqrt [4]{a+b x^3} \sqrt [12]{c+d x^3}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 89, normalized size = 0.73 \[ \frac {x \left (a+b x^3\right )^{3/4} \, _2F_1\left (-\frac {3}{4},\frac {1}{3};\frac {4}{3};\frac {(a d-b c) x^3}{a \left (d x^3+c\right )}\right )}{c^2 \left (\frac {b x^3}{a}+1\right )^{3/4} \sqrt [12]{c+d x^3} \sqrt [4]{\frac {d x^3}{c}+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(3/4)/(c + d*x^3)^(25/12),x]

[Out]

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

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fricas [F]  time = 47.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{3} + a\right )}^{\frac {3}{4}} {\left (d x^{3} + c\right )}^{\frac {11}{12}}}{d^{3} x^{9} + 3 \, c d^{2} x^{6} + 3 \, c^{2} d x^{3} + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^(3/4)*(d*x^3 + c)^(11/12)/(d^3*x^9 + 3*c*d^2*x^6 + 3*c^2*d*x^3 + c^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {3}{4}}}{{\left (d x^{3} + c\right )}^{\frac {25}{12}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(3/4)/(d*x^3 + c)^(25/12), x)

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maple [F]  time = 0.55, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {3}{4}}}{\left (d \,x^{3}+c \right )^{\frac {25}{12}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x)

[Out]

int((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {3}{4}}}{{\left (d x^{3} + c\right )}^{\frac {25}{12}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(3/4)/(d*x^3 + c)^(25/12), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^3+a\right )}^{3/4}}{{\left (d\,x^3+c\right )}^{25/12}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(3/4)/(c + d*x^3)^(25/12),x)

[Out]

int((a + b*x^3)^(3/4)/(c + d*x^3)^(25/12), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(3/4)/(d*x**3+c)**(25/12),x)

[Out]

Timed out

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