3.377 \(\int \frac {(c+d x^3)^{3/2}}{x^3 (a+b x^3)} \, dx\)

Optimal. Leaf size=65 \[ -\frac {c \sqrt {c+d x^3} F_1\left (-\frac {2}{3};1,-\frac {3}{2};\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a x^2 \sqrt {\frac {d x^3}{c}+1}} \]

[Out]

-1/2*c*AppellF1(-2/3,1,-3/2,1/3,-b*x^3/a,-d*x^3/c)*(d*x^3+c)^(1/2)/a/x^2/(1+d*x^3/c)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {511, 510} \[ -\frac {c \sqrt {c+d x^3} F_1\left (-\frac {2}{3};1,-\frac {3}{2};\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a x^2 \sqrt {\frac {d x^3}{c}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^(3/2)/(x^3*(a + b*x^3)),x]

[Out]

-(c*Sqrt[c + d*x^3]*AppellF1[-2/3, 1, -3/2, 1/3, -((b*x^3)/a), -((d*x^3)/c)])/(2*a*x^2*Sqrt[1 + (d*x^3)/c])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rubi steps

\begin {align*} \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx &=\frac {\left (c \sqrt {c+d x^3}\right ) \int \frac {\left (1+\frac {d x^3}{c}\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx}{\sqrt {1+\frac {d x^3}{c}}}\\ &=-\frac {c \sqrt {c+d x^3} F_1\left (-\frac {2}{3};1,-\frac {3}{2};\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a x^2 \sqrt {1+\frac {d x^3}{c}}}\\ \end {align*}

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Mathematica [B]  time = 0.33, size = 343, normalized size = 5.28 \[ -\frac {d x^6 \sqrt {\frac {d x^3}{c}+1} (b c-4 a d) F_1\left (\frac {4}{3};\frac {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+\frac {8 a c \left (3 x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )-4 a c \left (2 a c-5 a d x^3+6 b c x^3+2 b d x^6\right ) F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}{\left (a+b x^3\right ) \left (3 x^3 \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )-8 a c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}}{16 a^2 x^2 \sqrt {c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x^3)^(3/2)/(x^3*(a + b*x^3)),x]

[Out]

-1/16*(d*(b*c - 4*a*d)*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + (8*a*c
*(-4*a*c*(2*a*c + 6*b*c*x^3 - 5*a*d*x^3 + 2*b*d*x^6)*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] +
3*x^3*(a + b*x^3)*(c + d*x^3)*(2*b*c*AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + a*d*AppellF1[4/3
, 3/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)])))/((a + b*x^3)*(-8*a*c*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -(
(b*x^3)/a)] + 3*x^3*(2*b*c*AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + a*d*AppellF1[4/3, 3/2, 1,
7/3, -((d*x^3)/c), -((b*x^3)/a)]))))/(a^2*x^2*Sqrt[c + d*x^3])

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fricas [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((d*x^3+c)^(3/2)/x^3/(b*x^3+a),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(3/2)/x^3/(b*x^3+a),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^(3/2)/((b*x^3 + a)*x^3), x)

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maple [C]  time = 0.30, size = 1096, normalized size = 16.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^(3/2)/x^3/(b*x^3+a),x)

[Out]

-1/a*b*(2/5*(d*x^3+c)^(1/2)/b*d*x-2/3*I*(-2/5/b*c*d-(a*d-2*b*c)/b^2*d)*3^(1/2)*(-c*d^2)^(1/3)/d*(I*(x+1/2*(-c*
d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3/2*(-c*d
^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d))^(1/2)*(-I*(x+1/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)
*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2
)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(
1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2))+1/3*I/b^2/d^2*2^(1/2)*sum((-a^2*d^2+2*a*b*c*d-b^2*c^2)/_alpha^2/(a*d-b*c)*(-c
*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/
3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1
/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/
3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3
^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*
c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3
/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*b+a)))+1/a*(-1/2*c*(d*x^3+c)^
(1/2)/x^2+2/5*d*x*(d*x^3+c)^(1/2)-9/10*I*c*3^(1/2)*(-c*d^2)^(1/3)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c
*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*
d^2)^(1/3)/d))^(1/2)*(-I*(x+1/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/
2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-
c*d^2)^(1/3)*d)^(1/2),(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2
)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(3/2)/x^3/(b*x^3+a),x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)^(3/2)/((b*x^3 + a)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (d\,x^3+c\right )}^{3/2}}{x^3\,\left (b\,x^3+a\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)^(3/2)/(x^3*(a + b*x^3)),x)

[Out]

int((c + d*x^3)^(3/2)/(x^3*(a + b*x^3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{x^{3} \left (a + b x^{3}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)**(3/2)/x**3/(b*x**3+a),x)

[Out]

Integral((c + d*x**3)**(3/2)/(x**3*(a + b*x**3)), x)

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