3.132 \(\int \frac{(c+d x^3)^{23/12}}{(a+b x^3)^{13/4}} \, dx\)

Optimal. Leaf size=153 \[ \frac{253 c x \left (c+d x^3\right )^{11/12} \left (\frac{c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{5/4} \, _2F_1\left (\frac{1}{3},\frac{5}{4};\frac{4}{3};-\frac{(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{405 a^2 \left (a+b x^3\right )^{5/4}}+\frac{92 c x \left (c+d x^3\right )^{11/12}}{405 a^2 \left (a+b x^3\right )^{5/4}}+\frac{4 x \left (c+d x^3\right )^{23/12}}{27 a \left (a+b x^3\right )^{9/4}} \]

[Out]

(92*c*x*(c + d*x^3)^(11/12))/(405*a^2*(a + b*x^3)^(5/4)) + (4*x*(c + d*x^3)^(23/12))/(27*a*(a + b*x^3)^(9/4))
+ (253*c*x*((c*(a + b*x^3))/(a*(c + d*x^3)))^(5/4)*(c + d*x^3)^(11/12)*Hypergeometric2F1[1/3, 5/4, 4/3, -(((b*
c - a*d)*x^3)/(a*(c + d*x^3)))])/(405*a^2*(a + b*x^3)^(5/4))

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Rubi [A]  time = 0.0565592, antiderivative size = 153, 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} \[ \frac{253 c x \left (c+d x^3\right )^{11/12} \left (\frac{c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{5/4} \, _2F_1\left (\frac{1}{3},\frac{5}{4};\frac{4}{3};-\frac{(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{405 a^2 \left (a+b x^3\right )^{5/4}}+\frac{92 c x \left (c+d x^3\right )^{11/12}}{405 a^2 \left (a+b x^3\right )^{5/4}}+\frac{4 x \left (c+d x^3\right )^{23/12}}{27 a \left (a+b x^3\right )^{9/4}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^(23/12)/(a + b*x^3)^(13/4),x]

[Out]

(92*c*x*(c + d*x^3)^(11/12))/(405*a^2*(a + b*x^3)^(5/4)) + (4*x*(c + d*x^3)^(23/12))/(27*a*(a + b*x^3)^(9/4))
+ (253*c*x*((c*(a + b*x^3))/(a*(c + d*x^3)))^(5/4)*(c + d*x^3)^(11/12)*Hypergeometric2F1[1/3, 5/4, 4/3, -(((b*
c - a*d)*x^3)/(a*(c + d*x^3)))])/(405*a^2*(a + b*x^3)^(5/4))

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 )^{23/12}}{\left (a+b x^3\right )^{13/4}} \, dx &=\frac{4 x \left (c+d x^3\right )^{23/12}}{27 a \left (a+b x^3\right )^{9/4}}+\frac{(23 c) \int \frac{\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx}{27 a}\\ &=\frac{92 c x \left (c+d x^3\right )^{11/12}}{405 a^2 \left (a+b x^3\right )^{5/4}}+\frac{4 x \left (c+d x^3\right )^{23/12}}{27 a \left (a+b x^3\right )^{9/4}}+\frac{\left (253 c^2\right ) \int \frac{1}{\left (a+b x^3\right )^{5/4} \sqrt [12]{c+d x^3}} \, dx}{405 a^2}\\ &=\frac{92 c x \left (c+d x^3\right )^{11/12}}{405 a^2 \left (a+b x^3\right )^{5/4}}+\frac{4 x \left (c+d x^3\right )^{23/12}}{27 a \left (a+b x^3\right )^{9/4}}+\frac{253 c x \left (\frac{c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{5/4} \left (c+d x^3\right )^{11/12} \, _2F_1\left (\frac{1}{3},\frac{5}{4};\frac{4}{3};-\frac{(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{405 a^2 \left (a+b x^3\right )^{5/4}}\\ \end{align*}

Mathematica [A]  time = 0.0255563, size = 90, normalized size = 0.59 \[ \frac{c x \sqrt [4]{\frac{b x^3}{a}+1} \left (c+d x^3\right )^{11/12} \, _2F_1\left (\frac{1}{3},\frac{13}{4};\frac{4}{3};\frac{(a d-b c) x^3}{a \left (d x^3+c\right )}\right )}{a^3 \sqrt [4]{a+b x^3} \left (\frac{d x^3}{c}+1\right )^{5/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x^3)^(23/12)/(a + b*x^3)^(13/4),x]

[Out]

(c*x*(1 + (b*x^3)/a)^(1/4)*(c + d*x^3)^(11/12)*Hypergeometric2F1[1/3, 13/4, 4/3, ((-(b*c) + a*d)*x^3)/(a*(c +
d*x^3))])/(a^3*(a + b*x^3)^(1/4)*(1 + (d*x^3)/c)^(5/4))

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Maple [F]  time = 0.439, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d{x}^{3}+c \right ) ^{{\frac{23}{12}}} \left ( b{x}^{3}+a \right ) ^{-{\frac{13}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^(23/12)/(b*x^3+a)^(13/4),x)

[Out]

int((d*x^3+c)^(23/12)/(b*x^3+a)^(13/4),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{23}{12}}}{{\left (b x^{3} + a\right )}^{\frac{13}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(23/12)/(b*x^3+a)^(13/4),x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)^(23/12)/(b*x^3 + a)^(13/4), 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{3}{4}}{\left (d x^{3} + c\right )}^{\frac{23}{12}}}{b^{4} x^{12} + 4 \, a b^{3} x^{9} + 6 \, a^{2} b^{2} x^{6} + 4 \, a^{3} b x^{3} + a^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(23/12)/(b*x^3+a)^(13/4),x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^(3/4)*(d*x^3 + c)^(23/12)/(b^4*x^12 + 4*a*b^3*x^9 + 6*a^2*b^2*x^6 + 4*a^3*b*x^3 + a^4), 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)**(23/12)/(b*x**3+a)**(13/4),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{23}{12}}}{{\left (b x^{3} + a\right )}^{\frac{13}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(23/12)/(b*x^3+a)^(13/4),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^(23/12)/(b*x^3 + a)^(13/4), x)