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

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

[Out]

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

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Rubi [A]  time = 0.0275824, antiderivative size = 60, normalized size of antiderivative = 1., 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} \[ \frac{c x \sqrt{c+d x^3} F_1\left (\frac{1}{3};1,-\frac{3}{2};\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a \sqrt{\frac{d x^3}{c}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^(3/2)/(a + b*x^3),x]

[Out]

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

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])

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])

Rubi steps

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

Mathematica [B]  time = 0.32665, size = 351, normalized size = 5.85 \[ \frac{x \left (\frac{8 \left (3 d x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (2 b c F_1\left (\frac{4}{3};\frac{1}{2},2;\frac{7}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )+a d F_1\left (\frac{4}{3};\frac{3}{2},1;\frac{7}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )\right )-4 a c \left (2 a d^2 x^3+b \left (5 c^2+2 c d x^3+2 d^2 x^6\right )\right ) F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )\right )}{\left (a+b x^3\right ) \left (3 x^3 \left (2 b c F_1\left (\frac{4}{3};\frac{1}{2},2;\frac{7}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )+a d F_1\left (\frac{4}{3};\frac{3}{2},1;\frac{7}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )\right )-8 a c F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )\right )}+\frac{d x^3 \sqrt{\frac{d x^3}{c}+1} (8 b c-5 a d) F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )}{a}\right )}{20 b \sqrt{c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*((d*(8*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 + (8*
(-4*a*c*(2*a*d^2*x^3 + b*(5*c^2 + 2*c*d*x^3 + 2*d^2*x^6))*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a
)] + 3*d*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*Appel
lF1[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)])))))/(20*b*Sqrt[c + d*x^3])

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Maple [C]  time = 0.004, size = 776, normalized size = 12.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{\frac{3}{2}}}{b x^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x^{3}\right )^{\frac{3}{2}}}{a + b x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{\frac{3}{2}}}{b x^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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