Integrand size = 26, antiderivative size = 475 \[ \int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx=\frac {\left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (27 d^2 e^2-36 c d e f+14 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{27 b^2 d^3}+\frac {f (12 b d e-7 b c f-5 a d f) (a+b x)^{4/3} (c+d x)^{2/3}}{18 b^2 d^2}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}+\frac {(b c-a d) \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (27 d^2 e^2-36 c d e f+14 c^2 f^2\right )\right ) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{27 \sqrt {3} b^{8/3} d^{10/3}}+\frac {(b c-a d) \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (27 d^2 e^2-36 c d e f+14 c^2 f^2\right )\right ) \log (a+b x)}{162 b^{8/3} d^{10/3}}+\frac {(b c-a d) \left (5 a^2 d^2 f^2-2 a b d f (9 d e-4 c f)+b^2 \left (27 d^2 e^2-36 c d e f+14 c^2 f^2\right )\right ) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{54 b^{8/3} d^{10/3}} \]
1/27*(5*a^2*d^2*f^2-2*a*b*d*f*(-4*c*f+9*d*e)+b^2*(14*c^2*f^2-36*c*d*e*f+27 *d^2*e^2))*(b*x+a)^(1/3)*(d*x+c)^(2/3)/b^2/d^3+1/18*f*(-5*a*d*f-7*b*c*f+12 *b*d*e)*(b*x+a)^(4/3)*(d*x+c)^(2/3)/b^2/d^2+1/3*f*(b*x+a)^(4/3)*(d*x+c)^(2 /3)*(f*x+e)/b/d+1/162*(-a*d+b*c)*(5*a^2*d^2*f^2-2*a*b*d*f*(-4*c*f+9*d*e)+b ^2*(14*c^2*f^2-36*c*d*e*f+27*d^2*e^2))*ln(b*x+a)/b^(8/3)/d^(10/3)+1/54*(-a *d+b*c)*(5*a^2*d^2*f^2-2*a*b*d*f*(-4*c*f+9*d*e)+b^2*(14*c^2*f^2-36*c*d*e*f +27*d^2*e^2))*ln(-1+b^(1/3)*(d*x+c)^(1/3)/d^(1/3)/(b*x+a)^(1/3))/b^(8/3)/d ^(10/3)+1/81*(-a*d+b*c)*(5*a^2*d^2*f^2-2*a*b*d*f*(-4*c*f+9*d*e)+b^2*(14*c^ 2*f^2-36*c*d*e*f+27*d^2*e^2))*arctan(1/3*3^(1/2)+2/3*b^(1/3)*(d*x+c)^(1/3) /d^(1/3)/(b*x+a)^(1/3)*3^(1/2))/b^(8/3)/d^(10/3)*3^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.15 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.37 \[ \int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx=\frac {(a+b x)^{4/3} \left (-4 b f (-12 b d e+7 b c f+5 a d f) (c+d x)+24 b^2 d f (c+d x) (e+f x)+2 \left (5 a^2 d^2 f^2+2 a b d f (-9 d e+4 c f)+b^2 \left (27 d^2 e^2-36 c d e f+14 c^2 f^2\right )\right ) \sqrt [3]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {4}{3},\frac {7}{3},\frac {d (a+b x)}{-b c+a d}\right )\right )}{72 b^3 d^2 \sqrt [3]{c+d x}} \]
((a + b*x)^(4/3)*(-4*b*f*(-12*b*d*e + 7*b*c*f + 5*a*d*f)*(c + d*x) + 24*b^ 2*d*f*(c + d*x)*(e + f*x) + 2*(5*a^2*d^2*f^2 + 2*a*b*d*f*(-9*d*e + 4*c*f) + b^2*(27*d^2*e^2 - 36*c*d*e*f + 14*c^2*f^2))*((b*(c + d*x))/(b*c - a*d))^ (1/3)*Hypergeometric2F1[1/3, 4/3, 7/3, (d*(a + b*x))/(-(b*c) + a*d)]))/(72 *b^3*d^2*(c + d*x)^(1/3))
Time = 0.38 (sec) , antiderivative size = 310, normalized size of antiderivative = 0.65, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {101, 27, 90, 60, 71}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx\) |
| \(\Big \downarrow \) 101 |
| \(\displaystyle \frac {\int \frac {\sqrt [3]{a+b x} \left (9 b d e^2-f (4 b c e+2 a d e+3 a c f)+f (12 b d e-7 b c f-5 a d f) x\right )}{3 \sqrt [3]{c+d x}}dx}{3 b d}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt [3]{a+b x} (b e (9 d e-4 c f)-a f (2 d e+3 c f)+f (12 b d e-7 b c f-5 a d f) x)}{\sqrt [3]{c+d x}}dx}{9 b d}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (\frac {14 c^2 f^2}{d}-36 c e f+27 d e^2\right )\right ) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}dx+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (-5 a d f-7 b c f+12 b d e)}{2 b d}}{9 b d}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (\frac {14 c^2 f^2}{d}-36 c e f+27 d e^2\right )\right ) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}}dx}{3 d}\right )+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (-5 a d f-7 b c f+12 b d e)}{2 b d}}{9 b d}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}\) |
| \(\Big \downarrow \) 71 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {5 a^2 d f^2}{b}-2 a f (9 d e-4 c f)+b \left (\frac {14 c^2 f^2}{d}-36 c e f+27 d e^2\right )\right ) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{b^{2/3} \sqrt [3]{d}}-\frac {3 \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{2 b^{2/3} \sqrt [3]{d}}-\frac {\log (a+b x)}{2 b^{2/3} \sqrt [3]{d}}\right )}{3 d}\right )+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (-5 a d f-7 b c f+12 b d e)}{2 b d}}{9 b d}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3} (e+f x)}{3 b d}\) |
(f*(a + b*x)^(4/3)*(c + d*x)^(2/3)*(e + f*x))/(3*b*d) + ((f*(12*b*d*e - 7* b*c*f - 5*a*d*f)*(a + b*x)^(4/3)*(c + d*x)^(2/3))/(2*b*d) + (((5*a^2*d*f^2 )/b - 2*a*f*(9*d*e - 4*c*f) + b*(27*d*e^2 - 36*c*e*f + (14*c^2*f^2)/d))*(( (a + b*x)^(1/3)*(c + d*x)^(2/3))/d - ((b*c - a*d)*(-((Sqrt[3]*ArcTan[1/Sqr t[3] + (2*b^(1/3)*(c + d*x)^(1/3))/(Sqrt[3]*d^(1/3)*(a + b*x)^(1/3))])/(b^ (2/3)*d^(1/3))) - Log[a + b*x]/(2*b^(2/3)*d^(1/3)) - (3*Log[-1 + (b^(1/3)* (c + d*x)^(1/3))/(d^(1/3)*(a + b*x)^(1/3))])/(2*b^(2/3)*d^(1/3))))/(3*d))) /3)/(9*b*d)
3.31.8.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, Simp[(-Sqrt[3])*(q/d)*ArcTan[2*q*((a + b*x)^(1/3)/( Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[3]], x] + (-Simp[3*(q/(2*d))*Log[q*((a + b*x)^(1/3)/(c + d*x)^(1/3)) - 1], x] - Simp[(q/(2*d))*Log[c + d*x], x])] / ; FreeQ[{a, b, c, d}, x] && PosQ[d/b]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
\[\int \frac {\left (b x +a \right )^{\frac {1}{3}} \left (f x +e \right )^{2}}{\left (d x +c \right )^{\frac {1}{3}}}d x\]
Time = 0.30 (sec) , antiderivative size = 1400, normalized size of antiderivative = 2.95 \[ \int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx=\text {Too large to display} \]
[-1/162*(3*sqrt(1/3)*(27*(b^4*c*d^3 - a*b^3*d^4)*e^2 - 18*(2*b^4*c^2*d^2 - a*b^3*c*d^3 - a^2*b^2*d^4)*e*f + (14*b^4*c^3*d - 6*a*b^3*c^2*d^2 - 3*a^2* b^2*c*d^3 - 5*a^3*b*d^4)*f^2)*sqrt((-b^2*d)^(1/3)/d)*log(3*b^2*d*x + b^2*c + 2*a*b*d + 3*(-b^2*d)^(1/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b + 3*sqrt(1 /3)*(2*(b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (-b^2*d)^(2/3)*(b*x + a)^(1/3 )*(d*x + c)^(2/3) + (-b^2*d)^(1/3)*(b*d*x + b*c))*sqrt((-b^2*d)^(1/3)/d)) + (-b^2*d)^(2/3)*(27*(b^3*c*d^2 - a*b^2*d^3)*e^2 - 18*(2*b^3*c^2*d - a*b^2 *c*d^2 - a^2*b*d^3)*e*f + (14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5* a^3*d^3)*f^2)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d + (-b^2*d)^(2/3)*(b *x + a)^(1/3)*(d*x + c)^(2/3) - (-b^2*d)^(1/3)*(b*d*x + b*c))/(d*x + c)) - 2*(-b^2*d)^(2/3)*(27*(b^3*c*d^2 - a*b^2*d^3)*e^2 - 18*(2*b^3*c^2*d - a*b^ 2*c*d^2 - a^2*b*d^3)*e*f + (14*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5 *a^3*d^3)*f^2)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (-b^2*d)^(2/3)*( d*x + c))/(d*x + c)) - 3*(18*b^4*d^3*f^2*x^2 + 54*b^4*d^3*e^2 - 18*(4*b^4* c*d^2 - a*b^3*d^3)*e*f + (28*b^4*c^2*d - 5*a*b^3*c*d^2 - 5*a^2*b^2*d^3)*f^ 2 + 3*(18*b^4*d^3*e*f - (7*b^4*c*d^2 - a*b^3*d^3)*f^2)*x)*(b*x + a)^(1/3)* (d*x + c)^(2/3))/(b^4*d^4), -1/162*(6*sqrt(1/3)*(27*(b^4*c*d^3 - a*b^3*d^4 )*e^2 - 18*(2*b^4*c^2*d^2 - a*b^3*c*d^3 - a^2*b^2*d^4)*e*f + (14*b^4*c^3*d - 6*a*b^3*c^2*d^2 - 3*a^2*b^2*c*d^3 - 5*a^3*b*d^4)*f^2)*sqrt(-(-b^2*d)^(1 /3)/d)*arctan(sqrt(1/3)*(2*(-b^2*d)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/...
\[ \int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx=\int \frac {\sqrt [3]{a + b x} \left (e + f x\right )^{2}}{\sqrt [3]{c + d x}}\, dx \]
\[ \int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (f x + e\right )}^{2}}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (f x + e\right )}^{2}}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{a+b x} (e+f x)^2}{\sqrt [3]{c+d x}} \, dx=\int \frac {{\left (e+f\,x\right )}^2\,{\left (a+b\,x\right )}^{1/3}}{{\left (c+d\,x\right )}^{1/3}} \,d x \]