3.110 \(\int \frac {(a+b x^3)^{11/3}}{(c+d x^3)^3} \, dx\)

Optimal. Leaf size=458 \[ \frac {b x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2-7 a b c d+18 b^2 c^2\right )}{18 c^2 d^3}+\frac {(b c-a d)^{5/3} \left (5 a^2 d^2+12 a b c d+27 b^2 c^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} d^4}-\frac {(b c-a d)^{5/3} \left (5 a^2 d^2+12 a b c d+27 b^2 c^2\right ) \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} d^4}+\frac {(b c-a d)^{5/3} \left (5 a^2 d^2+12 a b c d+27 b^2 c^2\right ) \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} c^{8/3} d^4}+\frac {b^{8/3} (9 b c-11 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 d^4}-\frac {b^{8/3} (9 b c-11 a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} d^4}-\frac {x \left (a+b x^3\right )^{5/3} (b c-a d) (5 a d+9 b c)}{18 c^2 d^2 \left (c+d x^3\right )}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2} \]

[Out]

1/18*b*(-5*a^2*d^2-7*a*b*c*d+18*b^2*c^2)*x*(b*x^3+a)^(2/3)/c^2/d^3-1/6*(-a*d+b*c)*x*(b*x^3+a)^(8/3)/c/d/(d*x^3
+c)^2-1/18*(-a*d+b*c)*(5*a*d+9*b*c)*x*(b*x^3+a)^(5/3)/c^2/d^2/(d*x^3+c)+1/54*(-a*d+b*c)^(5/3)*(5*a^2*d^2+12*a*
b*c*d+27*b^2*c^2)*ln(d*x^3+c)/c^(8/3)/d^4-1/18*(-a*d+b*c)^(5/3)*(5*a^2*d^2+12*a*b*c*d+27*b^2*c^2)*ln((-a*d+b*c
)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(8/3)/d^4+1/6*b^(8/3)*(-11*a*d+9*b*c)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/d^4-
1/9*b^(8/3)*(-11*a*d+9*b*c)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/d^4*3^(1/2)+1/27*(-a*d+b*c)^(5
/3)*(5*a^2*d^2+12*a*b*c*d+27*b^2*c^2)*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^(
8/3)/d^4*3^(1/2)

________________________________________________________________________________________

Rubi [C]  time = 0.03, antiderivative size = 62, normalized size of antiderivative = 0.14, 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} \[ \frac {a^3 x \left (a+b x^3\right )^{2/3} F_1\left (\frac {1}{3};-\frac {11}{3},3;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^3 \left (\frac {b x^3}{a}+1\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^3*x*(a + b*x^3)^(2/3)*AppellF1[1/3, -11/3, 3, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(c^3*(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 )^{11/3}}{\left (c+d x^3\right )^3} \, dx &=\frac {\left (a^3 \left (a+b x^3\right )^{2/3}\right ) \int \frac {\left (1+\frac {b x^3}{a}\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx}{\left (1+\frac {b x^3}{a}\right )^{2/3}}\\ &=\frac {a^3 x \left (a+b x^3\right )^{2/3} F_1\left (\frac {1}{3};-\frac {11}{3},3;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^3 \left (1+\frac {b x^3}{a}\right )^{2/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.84, size = 908, normalized size = 1.98 \[ \frac {1}{108} \left (\frac {10 \left (2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a x^3+b}}\right )+\log \left (\frac {(b c-a d)^{2/3} x^2}{\left (a x^3+b\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )\right ) a^4}{c^{8/3} \sqrt [3]{b c-a d}}+\frac {4 b \left (2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a x^3+b}}\right )+\log \left (\frac {(b c-a d)^{2/3} x^2}{\left (a x^3+b\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )\right ) a^3}{c^{5/3} d \sqrt [3]{b c-a d}}+\frac {16 b^2 \left (2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a x^3+b}}\right )+\log \left (\frac {(b c-a d)^{2/3} x^2}{\left (a x^3+b\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )\right ) a^2}{c^{2/3} d^2 \sqrt [3]{b c-a d}}+\frac {99 b^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 ) a}{c d^2 \sqrt [3]{b x^3+a}}-\frac {18 b^3 \sqrt [3]{c} \left (2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a x^3+b}}\right )+\log \left (\frac {(b c-a d)^{2/3} x^2}{\left (a x^3+b\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )\right ) a}{d^3 \sqrt [3]{b c-a d}}+\frac {6 x \left (b x^3+a\right )^{2/3} \left (6 b^3+\frac {5 (b c-a d)^2 (3 b c+a d)}{c^2 \left (d x^3+c\right )}-\frac {3 (b c-a d)^3}{c \left (d x^3+c\right )^2}\right )}{d^3}-\frac {81 b^4 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^3 \sqrt [3]{b x^3+a}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((6*x*(a + b*x^3)^(2/3)*(6*b^3 - (3*(b*c - a*d)^3)/(c*(c + d*x^3)^2) + (5*(b*c - a*d)^2*(3*b*c + a*d))/(c^2*(c
+ d*x^3))))/d^3 - (81*b^4*x^4*(1 + (b*x^3)/a)^(1/3)*AppellF1[4/3, 1/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(
d^3*(a + b*x^3)^(1/3)) + (99*a*b^3*x^4*(1 + (b*x^3)/a)^(1/3)*AppellF1[4/3, 1/3, 1, 7/3, -((b*x^3)/a), -((d*x^3
)/c)])/(c*d^2*(a + b*x^3)^(1/3)) + (10*a^4*(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)]))/(c^(8/3)*(b*c - a*d)^(1/3)) - (1
8*a*b^3*c^(1/3)*(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^3*(b*c - a*d)^(1/3)) + (16*a^2*b^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)]))/(c^(2/3)*d^2*(b*c - a*d)^(1/3)) + (4*a^3*b*(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)]))/(c^(5/3)*d*(b*c -
a*d)^(1/3)))/108

________________________________________________________________________________________

fricas [B]  time = 79.23, size = 1246, normalized size = 2.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(2*sqrt(3)*(27*b^3*c^5 - 15*a*b^2*c^4*d - 7*a^2*b*c^3*d^2 - 5*a^3*c^2*d^3 + (27*b^3*c^3*d^2 - 15*a*b^2*c^
2*d^3 - 7*a^2*b*c*d^4 - 5*a^3*d^5)*x^6 + 2*(27*b^3*c^4*d - 15*a*b^2*c^3*d^2 - 7*a^2*b*c^2*d^3 - 5*a^3*c*d^4)*x
^3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*arctan(-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)*(9*b^3*c^5 - 11*a*b^2*c^4*d + (
9*b^3*c^3*d^2 - 11*a*b^2*c^2*d^3)*x^6 + 2*(9*b^3*c^4*d - 11*a*b^2*c^3*d^2)*x^3)*(-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)) - 2*(27*b^3*c^5 - 15*a*b^2*c^4*d - 7*a^2*b*c^3*d^2
- 5*a^3*c^2*d^3 + (27*b^3*c^3*d^2 - 15*a*b^2*c^2*d^3 - 7*a^2*b*c*d^4 - 5*a^3*d^5)*x^6 + 2*(27*b^3*c^4*d - 15*a
*b^2*c^3*d^2 - 7*a^2*b*c^2*d^3 - 5*a^3*c*d^4)*x^3)*((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*(9*b^3*c^5 - 11*a*b^2*c^4*d + (9
*b^3*c^3*d^2 - 11*a*b^2*c^2*d^3)*x^6 + 2*(9*b^3*c^4*d - 11*a*b^2*c^3*d^2)*x^3)*(-b^2)^(1/3)*log(-((-b^2)^(2/3)
*x - (b*x^3 + a)^(1/3)*b)/x) + 3*(9*b^3*c^5 - 11*a*b^2*c^4*d + (9*b^3*c^3*d^2 - 11*a*b^2*c^2*d^3)*x^6 + 2*(9*b
^3*c^4*d - 11*a*b^2*c^3*d^2)*x^3)*(-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) + (27*b^3*c^5 - 15*a*b^2*c^4*d - 7*a^2*b*c^3*d^2 - 5*a^3*c^2*d^3 + (27*b^3*c^3*d^2 -
15*a*b^2*c^2*d^3 - 7*a^2*b*c*d^4 - 5*a^3*d^5)*x^6 + 2*(27*b^3*c^4*d - 15*a*b^2*c^3*d^2 - 7*a^2*b*c^2*d^3 - 5*a
^3*c*d^4)*x^3)*((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) + 3*(6*b^3*c^2*d^3*x^7 + (27*b^3*c^3*d^2 - 25*a*b^2*c^2*d^3 + 5*a^2*b*c*d^4 + 5*a^3*d^5)*x^4 + 2*(9
*b^3*c^4*d - 8*a*b^2*c^3*d^2 - 2*a^2*b*c^2*d^3 + 4*a^3*c*d^4)*x)*(b*x^3 + a)^(2/3))/(c^2*d^6*x^6 + 2*c^3*d^5*x
^3 + c^4*d^4)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {11}{3}}}{\left (d \,x^{3}+c \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^3+a\right )}^{11/3}}{{\left (d\,x^3+c\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)**(11/3)/(d*x**3+c)**3,x)

[Out]

Timed out

________________________________________________________________________________________