Integrand size = 26, antiderivative size = 562 \[ \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx=\frac {3 (d e-c f)^2 (a+b x)^{7/3}}{d^2 (b c-a d) \sqrt [3]{c+d x}}-\frac {4 \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{27 b d^4}+\frac {\left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) (a+b x)^{4/3} (c+d x)^{2/3}}{9 b d^3 (b c-a d)}+\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3}}{3 b d^2}-\frac {4 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{27 \sqrt {3} b^{5/3} d^{13/3}}-\frac {2 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) \log (a+b x)}{81 b^{5/3} d^{13/3}}-\frac {2 (b c-a d) \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{27 b^{5/3} d^{13/3}} \]
3*(-c*f+d*e)^2*(b*x+a)^(7/3)/d^2/(-a*d+b*c)/(d*x+c)^(1/3)-4/27*(a^2*d^2*f^ 2-a*b*d*f*(-7*c*f+9*d*e)-b^2*(35*c^2*f^2-63*c*d*e*f+27*d^2*e^2))*(b*x+a)^( 1/3)*(d*x+c)^(2/3)/b/d^4+1/9*(a^2*d^2*f^2-a*b*d*f*(-7*c*f+9*d*e)-b^2*(35*c ^2*f^2-63*c*d*e*f+27*d^2*e^2))*(b*x+a)^(4/3)*(d*x+c)^(2/3)/b/d^3/(-a*d+b*c )+1/3*f^2*(b*x+a)^(7/3)*(d*x+c)^(2/3)/b/d^2-2/81*(-a*d+b*c)*(a^2*d^2*f^2-a *b*d*f*(-7*c*f+9*d*e)-b^2*(35*c^2*f^2-63*c*d*e*f+27*d^2*e^2))*ln(b*x+a)/b^ (5/3)/d^(13/3)-2/27*(-a*d+b*c)*(a^2*d^2*f^2-a*b*d*f*(-7*c*f+9*d*e)-b^2*(35 *c^2*f^2-63*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^(5/3)/d^(13/3)-4/81*(-a*d+b*c)*(a^2*d^2*f^2-a*b*d*f*(-7*c*f+9* d*e)-b^2*(35*c^2*f^2-63*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^(5/3)/d^(13/3)*3^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.16 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.31 \[ \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx=\frac {(a+b x)^{7/3} \left (-63 (d e-c f)^2-\frac {7 (b c-a d) f^2 (c+d x)}{b}+\frac {2 \left (-a^2 d^2 f^2+a b d f (9 d e-7 c f)+b^2 \left (27 d^2 e^2-63 c d e f+35 c^2 f^2\right )\right ) \sqrt [3]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {7}{3},\frac {10}{3},\frac {d (a+b x)}{-b c+a d}\right )}{b^2}\right )}{21 d^2 (-b c+a d) \sqrt [3]{c+d x}} \]
((a + b*x)^(7/3)*(-63*(d*e - c*f)^2 - (7*(b*c - a*d)*f^2*(c + d*x))/b + (2 *(-(a^2*d^2*f^2) + a*b*d*f*(9*d*e - 7*c*f) + b^2*(27*d^2*e^2 - 63*c*d*e*f + 35*c^2*f^2))*((b*(c + d*x))/(b*c - a*d))^(1/3)*Hypergeometric2F1[1/3, 7/ 3, 10/3, (d*(a + b*x))/(-(b*c) + a*d)])/b^2))/(21*d^2*(-(b*c) + a*d)*(c + d*x)^(1/3))
Time = 0.40 (sec) , antiderivative size = 362, normalized size of antiderivative = 0.64, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {100, 27, 90, 60, 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.
| \(\displaystyle \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)^2}{d^2 \sqrt [3]{c+d x} (b c-a d)}-\frac {3 \int \frac {(a+b x)^{4/3} \left (-d (b c-a d) x f^2+a d (2 d e-c f) f+b \left (6 d^2 e^2-14 c d f e+7 c^2 f^2\right )\right )}{3 \sqrt [3]{c+d x}}dx}{d^2 (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)^2}{d^2 \sqrt [3]{c+d x} (b c-a d)}-\frac {\int \frac {(a+b x)^{4/3} \left (-d (b c-a d) x f^2+a d (2 d e-c f) f+b \left (6 d^2 e^2-14 c d f e+7 c^2 f^2\right )\right )}{\sqrt [3]{c+d x}}dx}{d^2 (b c-a d)}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)^2}{d^2 \sqrt [3]{c+d x} (b c-a d)}-\frac {-\frac {2 \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-\left (b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right ) \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}}dx}{9 b}-\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3} (b c-a d)}{3 b}}{d^2 (b c-a d)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)^2}{d^2 \sqrt [3]{c+d x} (b c-a d)}-\frac {-\frac {2 \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-\left (b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right ) \left (\frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}dx}{3 d}\right )}{9 b}-\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3} (b c-a d)}{3 b}}{d^2 (b c-a d)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)^2}{d^2 \sqrt [3]{c+d x} (b c-a d)}-\frac {-\frac {2 \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-\left (b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right ) \left (\frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}}dx}{3 d}\right )}{3 d}\right )}{9 b}-\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3} (b c-a d)}{3 b}}{d^2 (b c-a d)}\) |
| \(\Big \downarrow \) 71 |
| \(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)^2}{d^2 \sqrt [3]{c+d x} (b c-a d)}-\frac {-\frac {2 \left (a^2 d^2 f^2-a b d f (9 d e-7 c f)-\left (b^2 \left (35 c^2 f^2-63 c d e f+27 d^2 e^2\right )\right )\right ) \left (\frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{b^{2/3} \sqrt [3]{d}}-\frac {3 \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{2 b^{2/3} \sqrt [3]{d}}-\frac {\log (a+b x)}{2 b^{2/3} \sqrt [3]{d}}\right )}{3 d}\right )}{3 d}\right )}{9 b}-\frac {f^2 (a+b x)^{7/3} (c+d x)^{2/3} (b c-a d)}{3 b}}{d^2 (b c-a d)}\) |
(3*(d*e - c*f)^2*(a + b*x)^(7/3))/(d^2*(b*c - a*d)*(c + d*x)^(1/3)) - (-1/ 3*((b*c - a*d)*f^2*(a + b*x)^(7/3)*(c + d*x)^(2/3))/b - (2*(a^2*d^2*f^2 - a*b*d*f*(9*d*e - 7*c*f) - b^2*(27*d^2*e^2 - 63*c*d*e*f + 35*c^2*f^2))*(((a + b*x)^(4/3)*(c + d*x)^(2/3))/(2*d) - (2*(b*c - a*d)*(((a + b*x)^(1/3)*(c + d*x)^(2/3))/d - ((b*c - a*d)*(-((Sqrt[3]*ArcTan[1/Sqrt[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*d)))/(9*b))/(d ^2*(b*c - a*d))
3.31.37.3.1 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*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
\[\int \frac {\left (b x +a \right )^{\frac {4}{3}} \left (f x +e \right )^{2}}{\left (d x +c \right )^{\frac {4}{3}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 1108 vs. \(2 (502) = 1004\).
Time = 0.36 (sec) , antiderivative size = 2267, normalized size of antiderivative = 4.03 \[ \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx=\text {Too large to display} \]
[1/81*(6*sqrt(1/3)*(27*(b^4*c^2*d^3 - a*b^3*c*d^4)*e^2 - 9*(7*b^4*c^3*d^2 - 8*a*b^3*c^2*d^3 + a^2*b^2*c*d^4)*e*f + (35*b^4*c^4*d - 42*a*b^3*c^3*d^2 + 6*a^2*b^2*c^2*d^3 + a^3*b*c*d^4)*f^2 + (27*(b^4*c*d^4 - a*b^3*d^5)*e^2 - 9*(7*b^4*c^2*d^3 - 8*a*b^3*c*d^4 + a^2*b^2*d^5)*e*f + (35*b^4*c^3*d^2 - 4 2*a*b^3*c^2*d^3 + 6*a^2*b^2*c*d^4 + a^3*b*d^5)*f^2)*x)*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))*sq rt(-(b^2*d)^(1/3)/d)) - 2*(b^2*d)^(2/3)*(27*(b^3*c^2*d^2 - a*b^2*c*d^3)*e^ 2 - 9*(7*b^3*c^3*d - 8*a*b^2*c^2*d^2 + a^2*b*c*d^3)*e*f + (35*b^3*c^4 - 42 *a*b^2*c^3*d + 6*a^2*b*c^2*d^2 + a^3*c*d^3)*f^2 + (27*(b^3*c*d^3 - a*b^2*d ^4)*e^2 - 9*(7*b^3*c^2*d^2 - 8*a*b^2*c*d^3 + a^2*b*d^4)*e*f + (35*b^3*c^3* d - 42*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3 + a^3*d^4)*f^2)*x)*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)) + 4*(b^2*d)^(2/3)*(27*(b^3*c^2*d^2 - a*b^2*c*d^3)*e^2 - 9*(7*b^3*c^3*d - 8*a*b^2*c^2*d^2 + a^2*b*c*d^3)*e*f + (35*b^3*c^4 - 42*a*b^2*c^3*d + 6*a^2*b*c^2*d^2 + a^3*c*d^3)*f^2 + (27*(b^ 3*c*d^3 - a*b^2*d^4)*e^2 - 9*(7*b^3*c^2*d^2 - 8*a*b^2*c*d^3 + a^2*b*d^4)*e *f + (35*b^3*c^3*d - 42*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3 + a^3*d^4)*f^2)*x)*l og(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (b^2*d)^(2/3)*(d*x + c))/(d*x...
\[ \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx=\int \frac {\left (a + b x\right )^{\frac {4}{3}} \left (e + f x\right )^{2}}{\left (c + d x\right )^{\frac {4}{3}}}\, dx \]
\[ \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}} {\left (f x + e\right )}^{2}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \]
\[ \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}} {\left (f x + e\right )}^{2}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^{4/3} (e+f x)^2}{(c+d x)^{4/3}} \, dx=\int \frac {{\left (e+f\,x\right )}^2\,{\left (a+b\,x\right )}^{4/3}}{{\left (c+d\,x\right )}^{4/3}} \,d x \]