3.2.31 \(\int \frac {(c+d x^3)^{17/12}}{(a+b x^3)^{11/4}} \, dx\) [131]

Optimal. Leaf size=153 \[ \frac {68 c x \left (c+d x^3\right )^{5/12}}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}+\frac {85 c x \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{5/12} \, _2F_1\left (\frac {1}{3},\frac {3}{4};\frac {4}{3};-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{189 a^2 \left (a+b x^3\right )^{3/4}} \]

[Out]

68/189*c*x*(d*x^3+c)^(5/12)/a^2/(b*x^3+a)^(3/4)+4/21*x*(d*x^3+c)^(17/12)/a/(b*x^3+a)^(7/4)+85/189*c*x*(c*(b*x^
3+a)/a/(d*x^3+c))^(3/4)*(d*x^3+c)^(5/12)*hypergeom([1/3, 3/4],[4/3],-(-a*d+b*c)*x^3/a/(d*x^3+c))/a^2/(b*x^3+a)
^(3/4)

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Rubi [A]
time = 0.04, antiderivative size = 153, normalized size of antiderivative = 1.00, 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 = {386, 388} \begin {gather*} \frac {85 c 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 )}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {68 c x \left (c+d x^3\right )^{5/12}}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^(17/12)/(a + b*x^3)^(11/4),x]

[Out]

(68*c*x*(c + d*x^3)^(5/12))/(189*a^2*(a + b*x^3)^(3/4)) + (4*x*(c + d*x^3)^(17/12))/(21*a*(a + b*x^3)^(7/4)) +
(85*c*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)))])/(189*a^2*(a + b*x^3)^(3/4))

Rule 386

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(c*(c*((a
+ b*x^n)/(a*(c + d*x^n))))^p*(c + d*x^n)^(1/n + p)))*Hypergeometric2F1[1/n, -p, 1 + 1/n, (-(b*c - a*d))*(x^n/(
a*(c + d*x^n)))], 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 )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx &=\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}+\frac {(17 c) \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx}{21 a}\\ &=\frac {68 c x \left (c+d x^3\right )^{5/12}}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}+\frac {\left (85 c^2\right ) \int \frac {1}{\left (a+b x^3\right )^{3/4} \left (c+d x^3\right )^{7/12}} \, dx}{189 a^2}\\ &=\frac {68 c x \left (c+d x^3\right )^{5/12}}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}+\frac {85 c 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 )}{189 a^2 \left (a+b x^3\right )^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 5.85, size = 90, normalized size = 0.59 \begin {gather*} \frac {c x \left (1+\frac {b x^3}{a}\right )^{3/4} \left (c+d x^3\right )^{5/12} \, _2F_1\left (\frac {1}{3},\frac {11}{4};\frac {4}{3};\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{a^2 \left (a+b x^3\right )^{3/4} \left (1+\frac {d x^3}{c}\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x^3)^(17/12)/(a + b*x^3)^(11/4),x]

[Out]

(c*x*(1 + (b*x^3)/a)^(3/4)*(c + d*x^3)^(5/12)*Hypergeometric2F1[1/3, 11/4, 4/3, ((-(b*c) + a*d)*x^3)/(a*(c + d
*x^3))])/(a^2*(a + b*x^3)^(3/4)*(1 + (d*x^3)/c)^(3/4))

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (d \,x^{3}+c \right )^{\frac {17}{12}}}{\left (b \,x^{3}+a \right )^{\frac {11}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^(17/12)/(b*x^3+a)^(11/4),x)

[Out]

int((d*x^3+c)^(17/12)/(b*x^3+a)^(11/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(17/12)/(b*x^3+a)^(11/4),x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)^(17/12)/(b*x^3 + a)^(11/4), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(17/12)/(b*x^3+a)^(11/4),x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^(1/4)*(d*x^3 + c)^(17/12)/(b^3*x^9 + 3*a*b^2*x^6 + 3*a^2*b*x^3 + a^3), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)**(17/12)/(b*x**3+a)**(11/4),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(17/12)/(b*x^3+a)^(11/4),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^(17/12)/(b*x^3 + a)^(11/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x^3+c\right )}^{17/12}}{{\left (b\,x^3+a\right )}^{11/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)^(17/12)/(a + b*x^3)^(11/4),x)

[Out]

int((c + d*x^3)^(17/12)/(a + b*x^3)^(11/4), x)

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