Integrand size = 26, antiderivative size = 369 \[ \int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx=\frac {f (9 b d e-5 b c f-4 a d f) (a+b x)^{2/3} \sqrt [3]{c+d x}}{6 b^2 d^2}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{3 \sqrt {3} b^{7/3} d^{8/3}}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log (c+d x)}{18 b^{7/3} d^{8/3}}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log \left (-1+\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{6 b^{7/3} d^{8/3}} \]
1/6*f*(-4*a*d*f-5*b*c*f+9*b*d*e)*(b*x+a)^(2/3)*(d*x+c)^(1/3)/b^2/d^2+1/2*f *(b*x+a)^(2/3)*(d*x+c)^(1/3)*(f*x+e)/b/d-1/18*(2*a^2*d^2*f^2-2*a*b*d*f*(-c *f+3*d*e)+b^2*(5*c^2*f^2-12*c*d*e*f+9*d^2*e^2))*ln(d*x+c)/b^(7/3)/d^(8/3)- 1/6*(2*a^2*d^2*f^2-2*a*b*d*f*(-c*f+3*d*e)+b^2*(5*c^2*f^2-12*c*d*e*f+9*d^2* e^2))*ln(-1+d^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(d*x+c)^(1/3))/b^(7/3)/d^(8/3)-1 /9*(2*a^2*d^2*f^2-2*a*b*d*f*(-c*f+3*d*e)+b^2*(5*c^2*f^2-12*c*d*e*f+9*d^2*e ^2))*arctan(1/3*3^(1/2)+2/3*d^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(d*x+c)^(1/3)*3^ (1/2))/b^(7/3)/d^(8/3)*3^(1/2)
Time = 0.70 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.00 \[ \int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx=\frac {3 \sqrt [3]{b} d^{2/3} f (a+b x)^{2/3} \sqrt [3]{c+d x} (-5 b c f-4 a d f+3 b d (4 e+f x))+2 \sqrt {3} \left (2 a^2 d^2 f^2+2 a b d f (-3 d e+c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}}{\sqrt {3}}\right )-2 \left (2 a^2 d^2 f^2+2 a b d f (-3 d e+c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log \left (\sqrt [3]{d}-\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{a+b x}}\right )+\left (2 a^2 d^2 f^2+2 a b d f (-3 d e+c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log \left (d^{2/3}+\frac {\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{c+d x}}{\sqrt [3]{a+b x}}+\frac {b^{2/3} (c+d x)^{2/3}}{(a+b x)^{2/3}}\right )}{18 b^{7/3} d^{8/3}} \]
(3*b^(1/3)*d^(2/3)*f*(a + b*x)^(2/3)*(c + d*x)^(1/3)*(-5*b*c*f - 4*a*d*f + 3*b*d*(4*e + f*x)) + 2*Sqrt[3]*(2*a^2*d^2*f^2 + 2*a*b*d*f*(-3*d*e + c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*ArcTan[(1 + (2*b^(1/3)*(c + d* x)^(1/3))/(d^(1/3)*(a + b*x)^(1/3)))/Sqrt[3]] - 2*(2*a^2*d^2*f^2 + 2*a*b*d *f*(-3*d*e + c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*Log[d^(1/3) - (b^(1/3)*(c + d*x)^(1/3))/(a + b*x)^(1/3)] + (2*a^2*d^2*f^2 + 2*a*b*d*f* (-3*d*e + c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*Log[d^(2/3) + ( b^(1/3)*d^(1/3)*(c + d*x)^(1/3))/(a + b*x)^(1/3) + (b^(2/3)*(c + d*x)^(2/3 ))/(a + b*x)^(2/3)])/(18*b^(7/3)*d^(8/3))
Time = 0.35 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.73, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {101, 27, 90, 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.
| \(\displaystyle \int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx\) |
| \(\Big \downarrow \) 101 |
| \(\displaystyle \frac {\int \frac {6 b d e^2-f (2 b c e+a d e+3 a c f)+f (9 b d e-5 b c f-4 a d f) x}{3 \sqrt [3]{a+b x} (c+d x)^{2/3}}dx}{2 b d}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {6 b d e^2-f (2 b c e+a d e+3 a c f)+f (9 b d e-5 b c f-4 a d f) x}{\sqrt [3]{a+b x} (c+d x)^{2/3}}dx}{6 b d}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {\frac {2}{3} \left (\frac {2 a^2 d f^2}{b}-2 a f (3 d e-c f)+b \left (\frac {5 c^2 f^2}{d}-12 c e f+9 d e^2\right )\right ) \int \frac {1}{\sqrt [3]{a+b x} (c+d x)^{2/3}}dx+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (-4 a d f-5 b c f+9 b d e)}{b d}}{6 b d}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}\) |
| \(\Big \downarrow \) 71 |
| \(\displaystyle \frac {\frac {2}{3} \left (\frac {2 a^2 d f^2}{b}-2 a f (3 d e-c f)+b \left (\frac {5 c^2 f^2}{d}-12 c e f+9 d e^2\right )\right ) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{b} d^{2/3}}-\frac {3 \log \left (\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{2 \sqrt [3]{b} d^{2/3}}-\frac {\log (c+d x)}{2 \sqrt [3]{b} d^{2/3}}\right )+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (-4 a d f-5 b c f+9 b d e)}{b d}}{6 b d}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}\) |
(f*(a + b*x)^(2/3)*(c + d*x)^(1/3)*(e + f*x))/(2*b*d) + ((f*(9*b*d*e - 5*b *c*f - 4*a*d*f)*(a + b*x)^(2/3)*(c + d*x)^(1/3))/(b*d) + (2*((2*a^2*d*f^2) /b - 2*a*f*(3*d*e - c*f) + b*(9*d*e^2 - 12*c*e*f + (5*c^2*f^2)/d))*(-((Sqr t[3]*ArcTan[1/Sqrt[3] + (2*d^(1/3)*(a + b*x)^(1/3))/(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))])/(b^(1/3)*d^(2/3))) - Log[c + d*x]/(2*b^(1/3)*d^(2/3)) - (3*L og[-1 + (d^(1/3)*(a + b*x)^(1/3))/(b^(1/3)*(c + d*x)^(1/3))])/(2*b^(1/3)*d ^(2/3))))/3)/(6*b*d)
3.31.16.3.1 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[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]
\[\int \frac {\left (f x +e \right )^{2}}{\left (b x +a \right )^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}}}d x\]
Time = 0.29 (sec) , antiderivative size = 1001, normalized size of antiderivative = 2.71 \[ \int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx=\text {Too large to display} \]
[1/18*(3*sqrt(1/3)*(9*b^3*d^3*e^2 - 6*(2*b^3*c*d^2 + a*b^2*d^3)*e*f + (5*b ^3*c^2*d + 2*a*b^2*c*d^2 + 2*a^2*b*d^3)*f^2)*sqrt((-b*d^2)^(1/3)/b)*log(-3 *b*d^2*x - 2*b*c*d - a*d^2 - 3*(-b*d^2)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^(1 /3)*d - 3*sqrt(1/3)*(2*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (-b*d^2)^(2/3 )*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (-b*d^2)^(1/3)*(b*d*x + a*d))*sqrt((-b *d^2)^(1/3)/b)) - 2*(9*b^2*d^2*e^2 - 6*(2*b^2*c*d + a*b*d^2)*e*f + (5*b^2* c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2)*(-b*d^2)^(2/3)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (-b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + (9*b^2*d^2*e^2 - 6*(2*b^2*c*d + a*b*d^2)*e*f + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2)*(-b *d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) + 3 *(3*b^2*d^3*f^2*x + 12*b^2*d^3*e*f - (5*b^2*c*d^2 + 4*a*b*d^3)*f^2)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^3*d^4), 1/18*(6*sqrt(1/3)*(9*b^3*d^3*e^2 - 6 *(2*b^3*c*d^2 + a*b^2*d^3)*e*f + (5*b^3*c^2*d + 2*a*b^2*c*d^2 + 2*a^2*b*d^ 3)*f^2)*sqrt(-(-b*d^2)^(1/3)/b)*arctan(sqrt(1/3)*(2*(-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2)^(1/3)*(b*d*x + a*d))*sqrt(-(-b*d^2)^(1 /3)/b)/(b*d^2*x + a*d^2)) - 2*(9*b^2*d^2*e^2 - 6*(2*b^2*c*d + a*b*d^2)*e*f + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2)*(-b*d^2)^(2/3)*log(((b*x + a)^ (2/3)*(d*x + c)^(1/3)*b*d - (-b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + (9*b^2* d^2*e^2 - 6*(2*b^2*c*d + a*b*d^2)*e*f + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*...
\[ \int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx=\int \frac {\left (e + f x\right )^{2}}{\sqrt [3]{a + b x} \left (c + d x\right )^{\frac {2}{3}}}\, dx \]
\[ \int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx=\int { \frac {{\left (f x + e\right )}^{2}}{{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx=\int { \frac {{\left (f x + e\right )}^{2}}{{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx=\int \frac {{\left (e+f\,x\right )}^2}{{\left (a+b\,x\right )}^{1/3}\,{\left (c+d\,x\right )}^{2/3}} \,d x \]