\(\int \frac {1}{(a+b x)^{11/3} (c+d x)^{2/3}} \, dx\) [1614]

   Optimal result
   Rubi [A] (warning: unable to verify)
   Mathematica [C] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 19, antiderivative size = 656 \[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {21 d \sqrt [3]{c+d x}}{40 (b c-a d)^2 (a+b x)^{5/3}}-\frac {21 d^2 \sqrt [3]{c+d x}}{20 (b c-a d)^3 (a+b x)^{2/3}}-\frac {7\ 3^{3/4} \sqrt {2+\sqrt {3}} d^{8/3} ((a+b x) (c+d x))^{2/3} \sqrt {(b c+a d+2 b d x)^2} \left ((b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right ) \sqrt {\frac {(b c-a d)^{4/3}-2^{2/3} \sqrt [3]{b} \sqrt [3]{d} (b c-a d)^{2/3} \sqrt [3]{(a+b x) (c+d x)}+2 \sqrt [3]{2} b^{2/3} d^{2/3} ((a+b x) (c+d x))^{2/3}}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}{\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}\right ),-7-4 \sqrt {3}\right )}{10\ 2^{2/3} \sqrt [3]{b} (b c-a d)^3 (a+b x)^{2/3} (c+d x)^{2/3} (b c+a d+2 b d x) \sqrt {\frac {(b c-a d)^{2/3} \left ((b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right )^2}} \sqrt {(a d+b (c+2 d x))^2}} \]

[Out]

-3/8*(d*x+c)^(1/3)/(-a*d+b*c)/(b*x+a)^(8/3)+21/40*d*(d*x+c)^(1/3)/(-a*d+b*c)^2/(b*x+a)^(5/3)-21/20*d^2*(d*x+c)
^(1/3)/(-a*d+b*c)^3/(b*x+a)^(2/3)-7/20*3^(3/4)*d^(8/3)*((b*x+a)*(d*x+c))^(2/3)*((-a*d+b*c)^(2/3)+2^(2/3)*b^(1/
3)*d^(1/3)*((b*x+a)*(d*x+c))^(1/3))*EllipticF((2^(2/3)*b^(1/3)*d^(1/3)*((b*x+a)*(d*x+c))^(1/3)+(-a*d+b*c)^(2/3
)*(1-3^(1/2)))/(2^(2/3)*b^(1/3)*d^(1/3)*((b*x+a)*(d*x+c))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(
(2*b*d*x+a*d+b*c)^2)^(1/2)*(1/2*6^(1/2)+1/2*2^(1/2))*(((-a*d+b*c)^(4/3)-2^(2/3)*b^(1/3)*d^(1/3)*(-a*d+b*c)^(2/
3)*((b*x+a)*(d*x+c))^(1/3)+2*2^(1/3)*b^(2/3)*d^(2/3)*((b*x+a)*(d*x+c))^(2/3))/(2^(2/3)*b^(1/3)*d^(1/3)*((b*x+a
)*(d*x+c))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2)))^2)^(1/2)*2^(1/3)/b^(1/3)/(-a*d+b*c)^3/(b*x+a)^(2/3)/(d*x+c)^(2/
3)/(2*b*d*x+a*d+b*c)/((a*d+b*(2*d*x+c))^2)^(1/2)/((-a*d+b*c)^(2/3)*((-a*d+b*c)^(2/3)+2^(2/3)*b^(1/3)*d^(1/3)*(
(b*x+a)*(d*x+c))^(1/3))/(2^(2/3)*b^(1/3)*d^(1/3)*((b*x+a)*(d*x+c))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2)))^2)^(1/2
)

Rubi [A] (warning: unable to verify)

Time = 0.74 (sec) , antiderivative size = 656, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {53, 64, 637, 224} \[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{2/3}} \, dx=-\frac {7\ 3^{3/4} \sqrt {2+\sqrt {3}} d^{8/3} ((a+b x) (c+d x))^{2/3} \sqrt {(a d+b c+2 b d x)^2} \left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+(b c-a d)^{2/3}\right ) \sqrt {\frac {2 \sqrt [3]{2} b^{2/3} d^{2/3} ((a+b x) (c+d x))^{2/3}-2^{2/3} \sqrt [3]{b} \sqrt [3]{d} (b c-a d)^{2/3} \sqrt [3]{(a+b x) (c+d x)}+(b c-a d)^{4/3}}{\left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}{\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}\right ),-7-4 \sqrt {3}\right )}{10\ 2^{2/3} \sqrt [3]{b} (a+b x)^{2/3} (c+d x)^{2/3} (b c-a d)^3 (a d+b c+2 b d x) \sqrt {\frac {(b c-a d)^{2/3} \left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+(b c-a d)^{2/3}\right )}{\left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}\right )^2}} \sqrt {(a d+b (c+2 d x))^2}}-\frac {21 d^2 \sqrt [3]{c+d x}}{20 (a+b x)^{2/3} (b c-a d)^3}+\frac {21 d \sqrt [3]{c+d x}}{40 (a+b x)^{5/3} (b c-a d)^2}-\frac {3 \sqrt [3]{c+d x}}{8 (a+b x)^{8/3} (b c-a d)} \]

[In]

Int[1/((a + b*x)^(11/3)*(c + d*x)^(2/3)),x]

[Out]

(-3*(c + d*x)^(1/3))/(8*(b*c - a*d)*(a + b*x)^(8/3)) + (21*d*(c + d*x)^(1/3))/(40*(b*c - a*d)^2*(a + b*x)^(5/3
)) - (21*d^2*(c + d*x)^(1/3))/(20*(b*c - a*d)^3*(a + b*x)^(2/3)) - (7*3^(3/4)*Sqrt[2 + Sqrt[3]]*d^(8/3)*((a +
b*x)*(c + d*x))^(2/3)*Sqrt[(b*c + a*d + 2*b*d*x)^2]*((b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c
 + d*x))^(1/3))*Sqrt[((b*c - a*d)^(4/3) - 2^(2/3)*b^(1/3)*d^(1/3)*(b*c - a*d)^(2/3)*((a + b*x)*(c + d*x))^(1/3
) + 2*2^(1/3)*b^(2/3)*d^(2/3)*((a + b*x)*(c + d*x))^(2/3))/((1 + Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*
d^(1/3)*((a + b*x)*(c + d*x))^(1/3))^2]*EllipticF[ArcSin[((1 - Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^
(1/3)*((a + b*x)*(c + d*x))^(1/3))/((1 + Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c +
d*x))^(1/3))], -7 - 4*Sqrt[3]])/(10*2^(2/3)*b^(1/3)*(b*c - a*d)^3*(a + b*x)^(2/3)*(c + d*x)^(2/3)*(b*c + a*d +
 2*b*d*x)*Sqrt[((b*c - a*d)^(2/3)*((b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3)))/(
(1 + Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3))^2]*Sqrt[(a*d + b*(c + 2
*d*x))^2])

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 64

Int[((a_.) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[(a + b*x)^m*((c + d*x)^m/((a + b*x)*
(c + d*x))^m), Int[(a*c + (b*c + a*d)*x + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] &&
 LtQ[-1, m, 0] && LeQ[3, Denominator[m], 4]

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt
[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq
rt[s*((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)
], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 637

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[d*(Sqrt[(b + 2*c*x)
^2]/(b + 2*c*x)), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
 /; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \sqrt [3]{c+d x}}{8 (b c-a d) (a+b x)^{8/3}}-\frac {(7 d) \int \frac {1}{(a+b x)^{8/3} (c+d x)^{2/3}} \, dx}{8 (b c-a d)} \\ & = -\frac {3 \sqrt [3]{c+d x}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {21 d \sqrt [3]{c+d x}}{40 (b c-a d)^2 (a+b x)^{5/3}}+\frac {\left (7 d^2\right ) \int \frac {1}{(a+b x)^{5/3} (c+d x)^{2/3}} \, dx}{10 (b c-a d)^2} \\ & = -\frac {3 \sqrt [3]{c+d x}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {21 d \sqrt [3]{c+d x}}{40 (b c-a d)^2 (a+b x)^{5/3}}-\frac {21 d^2 \sqrt [3]{c+d x}}{20 (b c-a d)^3 (a+b x)^{2/3}}-\frac {\left (7 d^3\right ) \int \frac {1}{(a+b x)^{2/3} (c+d x)^{2/3}} \, dx}{20 (b c-a d)^3} \\ & = -\frac {3 \sqrt [3]{c+d x}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {21 d \sqrt [3]{c+d x}}{40 (b c-a d)^2 (a+b x)^{5/3}}-\frac {21 d^2 \sqrt [3]{c+d x}}{20 (b c-a d)^3 (a+b x)^{2/3}}-\frac {\left (7 d^3 ((a+b x) (c+d x))^{2/3}\right ) \int \frac {1}{\left (a c+(b c+a d) x+b d x^2\right )^{2/3}} \, dx}{20 (b c-a d)^3 (a+b x)^{2/3} (c+d x)^{2/3}} \\ & = -\frac {3 \sqrt [3]{c+d x}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {21 d \sqrt [3]{c+d x}}{40 (b c-a d)^2 (a+b x)^{5/3}}-\frac {21 d^2 \sqrt [3]{c+d x}}{20 (b c-a d)^3 (a+b x)^{2/3}}-\frac {\left (21 d^3 ((a+b x) (c+d x))^{2/3} \sqrt {(b c+a d+2 b d x)^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-4 a b c d+(b c+a d)^2+4 b d x^3}} \, dx,x,\sqrt [3]{(a+b x) (c+d x)}\right )}{20 (b c-a d)^3 (a+b x)^{2/3} (c+d x)^{2/3} (b c+a d+2 b d x)} \\ & = -\frac {3 \sqrt [3]{c+d x}}{8 (b c-a d) (a+b x)^{8/3}}+\frac {21 d \sqrt [3]{c+d x}}{40 (b c-a d)^2 (a+b x)^{5/3}}-\frac {21 d^2 \sqrt [3]{c+d x}}{20 (b c-a d)^3 (a+b x)^{2/3}}-\frac {7\ 3^{3/4} \sqrt {2+\sqrt {3}} d^{8/3} ((a+b x) (c+d x))^{2/3} \sqrt {(b c+a d+2 b d x)^2} \left ((b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right ) \sqrt {\frac {(b c-a d)^{4/3}-2^{2/3} \sqrt [3]{b} \sqrt [3]{d} (b c-a d)^{2/3} \sqrt [3]{(a+b x) (c+d x)}+2 \sqrt [3]{2} b^{2/3} d^{2/3} ((a+b x) (c+d x))^{2/3}}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}{\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}\right )|-7-4 \sqrt {3}\right )}{10\ 2^{2/3} \sqrt [3]{b} (b c-a d)^3 (a+b x)^{2/3} (c+d x)^{2/3} (b c+a d+2 b d x) \sqrt {\frac {(b c-a d)^{2/3} \left ((b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right )^2}} \sqrt {(a d+b (c+2 d x))^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.11 \[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{2/3}} \, dx=-\frac {3 \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {8}{3},\frac {2}{3},-\frac {5}{3},\frac {d (a+b x)}{-b c+a d}\right )}{8 b (a+b x)^{8/3} (c+d x)^{2/3}} \]

[In]

Integrate[1/((a + b*x)^(11/3)*(c + d*x)^(2/3)),x]

[Out]

(-3*((b*(c + d*x))/(b*c - a*d))^(2/3)*Hypergeometric2F1[-8/3, 2/3, -5/3, (d*(a + b*x))/(-(b*c) + a*d)])/(8*b*(
a + b*x)^(8/3)*(c + d*x)^(2/3))

Maple [F]

\[\int \frac {1}{\left (b x +a \right )^{\frac {11}{3}} \left (d x +c \right )^{\frac {2}{3}}}d x\]

[In]

int(1/(b*x+a)^(11/3)/(d*x+c)^(2/3),x)

[Out]

int(1/(b*x+a)^(11/3)/(d*x+c)^(2/3),x)

Fricas [F]

\[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(2/3),x, algorithm="fricas")

[Out]

integral((b*x + a)^(1/3)*(d*x + c)^(1/3)/(b^4*d*x^5 + a^4*c + (b^4*c + 4*a*b^3*d)*x^4 + 2*(2*a*b^3*c + 3*a^2*b
^2*d)*x^3 + 2*(3*a^2*b^2*c + 2*a^3*b*d)*x^2 + (4*a^3*b*c + a^4*d)*x), x)

Sympy [F]

\[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {11}{3}} \left (c + d x\right )^{\frac {2}{3}}}\, dx \]

[In]

integrate(1/(b*x+a)**(11/3)/(d*x+c)**(2/3),x)

[Out]

Integral(1/((a + b*x)**(11/3)*(c + d*x)**(2/3)), x)

Maxima [F]

\[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(2/3)), x)

Giac [F]

\[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(2/3)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{11/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{11/3}\,{\left (c+d\,x\right )}^{2/3}} \,d x \]

[In]

int(1/((a + b*x)^(11/3)*(c + d*x)^(2/3)),x)

[Out]

int(1/((a + b*x)^(11/3)*(c + d*x)^(2/3)), x)