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

Optimal. Leaf size=60 \[ \frac {c x \sqrt {c+d x^3} F_1\left (\frac {1}{3};2,-\frac {3}{2};\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 \sqrt {\frac {d x^3}{c}+1}} \]

[Out]

c*x*AppellF1(1/3,2,-3/2,4/3,-b*x^3/a,-d*x^3/c)*(d*x^3+c)^(1/2)/a^2/(1+d*x^3/c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {430, 429} \[ \frac {c x \sqrt {c+d x^3} F_1\left (\frac {1}{3};2,-\frac {3}{2};\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 \sqrt {\frac {d x^3}{c}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 429

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

Rule 430

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

Rubi steps

\begin {align*} \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx &=\frac {\left (c \sqrt {c+d x^3}\right ) \int \frac {\left (1+\frac {d x^3}{c}\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx}{\sqrt {1+\frac {d x^3}{c}}}\\ &=\frac {c x \sqrt {c+d x^3} F_1\left (\frac {1}{3};2,-\frac {3}{2};\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 \sqrt {1+\frac {d x^3}{c}}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.43, size = 339, normalized size = 5.65 \[ \frac {x \left (\frac {a \left (24 x^3 \left (c+d x^3\right ) (b c-a d) \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )-64 a c \left (b c \left (3 c+d x^3\right )-a d^2 x^3\right ) F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}{\left (a+b x^3\right ) \left (3 x^3 \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )-8 a c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}+d x^3 \sqrt {\frac {d x^3}{c}+1} (5 a d+b c) F_1\left (\frac {4}{3};\frac {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}{24 a^2 b \sqrt {c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(d*(b*c + 5*a*d)*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + (a*(-64*a
*c*(-(a*d^2*x^3) + b*c*(3*c + d*x^3))*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] + 24*(b*c - a*d)*
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)])))))/(24*a^2*b*Sqrt[c + d*x^3])

________________________________________________________________________________________

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)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.26, size = 801, normalized size = 13.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a*d-b*c)*(d*x^3+c)^(1/2)/(b*x^3+a)/a/b*x-2/3*I*(1/b^2*d^2-1/6*(a*d-b*c)/a/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))+1/18*I/a/b^2/d^2*2^(1/2)*sum((5*a^2*d^2-a*b*c*d-4*b^2*c^2)/_a
lpha^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))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________