3.99 \(\int \frac {(a+b x^3)^{5/3}}{(c+d x^3)^2} \, dx\)

Optimal. Leaf size=301 \[ -\frac {b^{5/3} \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 d^2}+\frac {b^{5/3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^2}-\frac {(b c-a d)^{2/3} (2 a d+3 b c) \log \left (c+d x^3\right )}{18 c^{5/3} d^2}+\frac {(b c-a d)^{2/3} (2 a d+3 b c) \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{6 c^{5/3} d^2}-\frac {(b c-a d)^{2/3} (2 a d+3 b c) \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} d^2}-\frac {x \left (a+b x^3\right )^{2/3} (b c-a d)}{3 c d \left (c+d x^3\right )} \]

[Out]

-1/3*(-a*d+b*c)*x*(b*x^3+a)^(2/3)/c/d/(d*x^3+c)-1/18*(-a*d+b*c)^(2/3)*(2*a*d+3*b*c)*ln(d*x^3+c)/c^(5/3)/d^2+1/
6*(-a*d+b*c)^(2/3)*(2*a*d+3*b*c)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(5/3)/d^2-1/2*b^(5/3)*ln(-b^
(1/3)*x+(b*x^3+a)^(1/3))/d^2+1/3*b^(5/3)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/d^2*3^(1/2)-1/9*(
-a*d+b*c)^(2/3)*(2*a*d+3*b*c)*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1/2))/c^(5/3)/d^2
*3^(1/2)

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Rubi [C]  time = 0.03, antiderivative size = 60, normalized size of antiderivative = 0.20, 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} \[ \frac {a x \left (a+b x^3\right )^{2/3} F_1\left (\frac {1}{3};-\frac {5}{3},2;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^2 \left (\frac {b x^3}{a}+1\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*x^3)^(5/3)/(c + d*x^3)^2,x]

[Out]

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

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 (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^2} \, dx &=\frac {\left (a \left (a+b x^3\right )^{2/3}\right ) \int \frac {\left (1+\frac {b x^3}{a}\right )^{5/3}}{\left (c+d x^3\right )^2} \, dx}{\left (1+\frac {b x^3}{a}\right )^{2/3}}\\ &=\frac {a x \left (a+b x^3\right )^{2/3} F_1\left (\frac {1}{3};-\frac {5}{3},2;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^2 \left (1+\frac {b x^3}{a}\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.64, size = 450, normalized size = 1.50 \[ \frac {\frac {4 a^2 \left (\log \left (\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+\frac {x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+c^{2/3}\right )-2 \log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt {3}}\right )\right )}{\sqrt [3]{b c-a d}}+\frac {9 b^2 c^{2/3} x^4 \sqrt [3]{\frac {b x^3}{a}+1} F_1\left (\frac {4}{3};\frac {1}{3},1;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{d \sqrt [3]{a+b x^3}}-\frac {12 c^{2/3} x \left (a+b x^3\right )^{2/3} (b c-a d)}{d \left (c+d x^3\right )}+\frac {2 a b c \left (\log \left (\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+\frac {x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+c^{2/3}\right )-2 \log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt {3}}\right )\right )}{d \sqrt [3]{b c-a d}}}{36 c^{5/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(5/3)/(c + d*x^3)^2,x]

[Out]

((-12*c^(2/3)*(b*c - a*d)*x*(a + b*x^3)^(2/3))/(d*(c + d*x^3)) + (9*b^2*c^(2/3)*x^4*(1 + (b*x^3)/a)^(1/3)*Appe
llF1[4/3, 1/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(d*(a + b*x^3)^(1/3)) + (4*a^2*(2*Sqrt[3]*ArcTan[(1 + (2*(
b*c - a*d)^(1/3)*x)/(c^(1/3)*(b + a*x^3)^(1/3)))/Sqrt[3]] - 2*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^
(1/3)] + Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/3)*(b*c - a*d)^(1/3)*x)/(b + a*x^3)^(
1/3)]))/(b*c - a*d)^(1/3) + (2*a*b*c*(2*Sqrt[3]*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(b + a*x^3)^(1/3)
))/Sqrt[3]] - 2*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)] + Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)
/(b + a*x^3)^(2/3) + (c^(1/3)*(b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)]))/(d*(b*c - a*d)^(1/3)))/(36*c^(5/3))

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fricas [B]  time = 1.34, size = 631, normalized size = 2.10 \[ -\frac {2 \, \sqrt {3} {\left ({\left (3 \, b c d + 2 \, a d^{2}\right )} x^{3} + 3 \, b c^{2} + 2 \, a c d\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} {\left (b c - a d\right )} x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}}}{3 \, {\left (b c - a d\right )} x}\right ) + 6 \, \sqrt {3} {\left (b c d x^{3} + b c^{2}\right )} \left (-b^{2}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}}}{3 \, b x}\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c d - a d^{2}\right )} x - 2 \, {\left ({\left (3 \, b c d + 2 \, a d^{2}\right )} x^{3} + 3 \, b c^{2} + 2 \, a c d\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (\frac {c x \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}}{x}\right ) - 6 \, {\left (b c d x^{3} + b c^{2}\right )} \left (-b^{2}\right )^{\frac {1}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + 3 \, {\left (b c d x^{3} + b c^{2}\right )} \left (-b^{2}\right )^{\frac {1}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + {\left ({\left (3 \, b c d + 2 \, a d^{2}\right )} x^{3} + 3 \, b c^{2} + 2 \, a c d\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (-\frac {{\left (b c - a d\right )} x^{2} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} c x \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )}}{x^{2}}\right )}{18 \, {\left (c d^{3} x^{3} + c^{2} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/3)/(d*x^3+c)^2,x, algorithm="fricas")

[Out]

-1/18*(2*sqrt(3)*((3*b*c*d + 2*a*d^2)*x^3 + 3*b*c^2 + 2*a*c*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*arc
tan(-1/3*(sqrt(3)*(b*c - a*d)*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*c*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3))/(
(b*c - a*d)*x)) + 6*sqrt(3)*(b*c*d*x^3 + b*c^2)*(-b^2)^(1/3)*arctan(-1/3*(sqrt(3)*b*x - 2*sqrt(3)*(b*x^3 + a)^
(1/3)*(-b^2)^(1/3))/(b*x)) + 6*(b*x^3 + a)^(2/3)*(b*c*d - a*d^2)*x - 2*((3*b*c*d + 2*a*d^2)*x^3 + 3*b*c^2 + 2*
a*c*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*log((c*x*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(2/3) - (b*x
^3 + a)^(1/3)*(b*c - a*d))/x) - 6*(b*c*d*x^3 + b*c^2)*(-b^2)^(1/3)*log(-((-b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)
/x) + 3*(b*c*d*x^3 + b*c^2)*(-b^2)^(1/3)*log(-((-b^2)^(1/3)*b*x^2 - (b*x^3 + a)^(1/3)*(-b^2)^(2/3)*x - (b*x^3
+ a)^(2/3)*b)/x^2) + ((3*b*c*d + 2*a*d^2)*x^3 + 3*b*c^2 + 2*a*c*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)
*log(-((b*c - a*d)*x^2*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3) + (b*x^3 + a)^(1/3)*c*x*((b^2*c^2 - 2*a*b*c
*d + a^2*d^2)/c^2)^(2/3) + (b*x^3 + a)^(2/3)*(b*c - a*d))/x^2))/(c*d^3*x^3 + c^2*d^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{{\left (d x^{3} + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/3)/(d*x^3+c)^2,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(5/3)/(d*x^3 + c)^2, x)

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maple [F]  time = 0.58, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {5}{3}}}{\left (d \,x^{3}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(5/3)/(d*x^3+c)^2,x)

[Out]

int((b*x^3+a)^(5/3)/(d*x^3+c)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{{\left (d x^{3} + c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(5/3)/(d*x^3+c)^2,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(5/3)/(d*x^3 + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^3+a\right )}^{5/3}}{{\left (d\,x^3+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(5/3)/(c + d*x^3)^2,x)

[Out]

int((a + b*x^3)^(5/3)/(c + d*x^3)^2, x)

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sympy [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((b*x**3+a)**(5/3)/(d*x**3+c)**2,x)

[Out]

Timed out

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