Integrand size = 24, antiderivative size = 295 \[ \int \frac {(a+b x)^{4/3} (e+f x)}{(c+d x)^{4/3}} \, dx=-\frac {3 (d e-c f) (a+b x)^{4/3}}{d^2 \sqrt [3]{c+d x}}+\frac {2 (6 b d e-7 b c f+a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 d^3}+\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}+\frac {2 (b c-a d) (6 b d e-7 b c f+a d f) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} b^{2/3} d^{10/3}}+\frac {(b c-a d) (6 b d e-7 b c f+a d f) \log (a+b x)}{9 b^{2/3} d^{10/3}}+\frac {(b c-a d) (6 b d e-7 b c f+a d f) \log \left (1-\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 b^{2/3} d^{10/3}} \] Output:
-3*(-c*f+d*e)*(b*x+a)^(4/3)/d^2/(d*x+c)^(1/3)+2/3*(a*d*f-7*b*c*f+6*b*d*e)* (b*x+a)^(1/3)*(d*x+c)^(2/3)/d^3+1/2*f*(b*x+a)^(4/3)*(d*x+c)^(2/3)/d^2+2/9* (-a*d+b*c)*(a*d*f-7*b*c*f+6*b*d*e)*arctan(1/3*3^(1/2)+2/3*b^(1/3)*(d*x+c)^ (1/3)*3^(1/2)/d^(1/3)/(b*x+a)^(1/3))*3^(1/2)/b^(2/3)/d^(10/3)+1/9*(-a*d+b* c)*(a*d*f-7*b*c*f+6*b*d*e)*ln(b*x+a)/b^(2/3)/d^(10/3)+1/3*(-a*d+b*c)*(a*d* f-7*b*c*f+6*b*d*e)*ln(1-b^(1/3)*(d*x+c)^(1/3)/d^(1/3)/(b*x+a)^(1/3))/b^(2/ 3)/d^(10/3)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.39 \[ \int \frac {(a+b x)^{4/3} (e+f x)}{(c+d x)^{4/3}} \, dx=\frac {3 (a+b x)^{7/3} \left (b (-7 d e+7 c f)+(6 b d e-7 b c f+a d f) \sqrt [3]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {7}{3},\frac {10}{3},\frac {d (a+b x)}{-b c+a d}\right )\right )}{7 b d (-b c+a d) \sqrt [3]{c+d x}} \] Input:
Integrate[((a + b*x)^(4/3)*(e + f*x))/(c + d*x)^(4/3),x]
Output:
(3*(a + b*x)^(7/3)*(b*(-7*d*e + 7*c*f) + (6*b*d*e - 7*b*c*f + a*d*f)*((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)]))/(7*b*d*(-(b*c) + a*d)*(c + d*x)^(1/3))
Time = 0.31 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 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.
\(\displaystyle \int \frac {(a+b x)^{4/3} (e+f x)}{(c+d x)^{4/3}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)}{d \sqrt [3]{c+d x} (b c-a d)}-\frac {(a d f-7 b c f+6 b d e) \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}}dx}{d (b c-a d)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)}{d \sqrt [3]{c+d x} (b c-a d)}-\frac {(a d f-7 b c f+6 b d e) \left (\frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}dx}{3 d}\right )}{d (b c-a d)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)}{d \sqrt [3]{c+d x} (b c-a d)}-\frac {(a d f-7 b c f+6 b d e) \left (\frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}}dx}{3 d}\right )}{3 d}\right )}{d (b c-a d)}\) |
\(\Big \downarrow \) 71 |
\(\displaystyle \frac {3 (a+b x)^{7/3} (d e-c f)}{d \sqrt [3]{c+d x} (b c-a d)}-\frac {(a d f-7 b c f+6 b d e) \left (\frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{b^{2/3} \sqrt [3]{d}}-\frac {3 \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{2 b^{2/3} \sqrt [3]{d}}-\frac {\log (a+b x)}{2 b^{2/3} \sqrt [3]{d}}\right )}{3 d}\right )}{3 d}\right )}{d (b c-a d)}\) |
Input:
Int[((a + b*x)^(4/3)*(e + f*x))/(c + d*x)^(4/3),x]
Output:
(3*(d*e - c*f)*(a + b*x)^(7/3))/(d*(b*c - a*d)*(c + d*x)^(1/3)) - ((6*b*d* e - 7*b*c*f + a*d*f)*(((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]*ArcTa n[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)))/(d*(b*c - a*d))
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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
\[\int \frac {\left (b x +a \right )^{\frac {4}{3}} \left (f x +e \right )}{\left (x d +c \right )^{\frac {4}{3}}}d x\]
Input:
int((b*x+a)^(4/3)*(f*x+e)/(d*x+c)^(4/3),x)
Output:
int((b*x+a)^(4/3)*(f*x+e)/(d*x+c)^(4/3),x)
Leaf count of result is larger than twice the leaf count of optimal. 670 vs. \(2 (241) = 482\).
Time = 0.15 (sec) , antiderivative size = 1386, normalized size of antiderivative = 4.70 \[ \int \frac {(a+b x)^{4/3} (e+f x)}{(c+d x)^{4/3}} \, dx=\text {Too large to display} \] Input:
integrate((b*x+a)^(4/3)*(f*x+e)/(d*x+c)^(4/3),x, algorithm="fricas")
Output:
[-1/18*(6*sqrt(1/3)*(6*(b^3*c^2*d^2 - a*b^2*c*d^3)*e - (7*b^3*c^3*d - 8*a* b^2*c^2*d^2 + a^2*b*c*d^3)*f + (6*(b^3*c*d^3 - a*b^2*d^4)*e - (7*b^3*c^2*d ^2 - 8*a*b^2*c*d^3 + a^2*b*d^4)*f)*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))*sqrt((-b^2*d)^(1 /3)/d)) + 2*(-b^2*d)^(2/3)*(6*(b^2*c^2*d - a*b*c*d^2)*e - (7*b^2*c^3 - 8*a *b*c^2*d + a^2*c*d^2)*f + (6*(b^2*c*d^2 - a*b*d^3)*e - (7*b^2*c^2*d - 8*a* b*c*d^2 + a^2*d^3)*f)*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)*(6*(b^2*c^2*d - a*b*c*d^2)*e - (7*b^2*c^3 - 8 *a*b*c^2*d + a^2*c*d^2)*f + (6*(b^2*c*d^2 - a*b*d^3)*e - (7*b^2*c^2*d - 8* a*b*c*d^2 + a^2*d^3)*f)*x)*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*(3*b^3*d^3*f*x^2 + 6*(4*b^3*c*d^2 - 3 *a*b^2*d^3)*e - (28*b^3*c^2*d - 25*a*b^2*c*d^2)*f + (6*b^3*d^3*e - 7*(b^3* c*d^2 - a*b^2*d^3)*f)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^2*d^5*x + b^2 *c*d^4), -1/18*(12*sqrt(1/3)*(6*(b^3*c^2*d^2 - a*b^2*c*d^3)*e - (7*b^3*c^3 *d - 8*a*b^2*c^2*d^2 + a^2*b*c*d^3)*f + (6*(b^3*c*d^3 - a*b^2*d^4)*e - (7* b^3*c^2*d^2 - 8*a*b^2*c*d^3 + a^2*b*d^4)*f)*x)*sqrt(-(-b^2*d)^(1/3)/d)*arc tan(sqrt(1/3)*(2*(-b^2*d)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) - (-b^2...
\[ \int \frac {(a+b x)^{4/3} (e+f x)}{(c+d x)^{4/3}} \, dx=\int \frac {\left (a + b x\right )^{\frac {4}{3}} \left (e + f x\right )}{\left (c + d x\right )^{\frac {4}{3}}}\, dx \] Input:
integrate((b*x+a)**(4/3)*(f*x+e)/(d*x+c)**(4/3),x)
Output:
Integral((a + b*x)**(4/3)*(e + f*x)/(c + d*x)**(4/3), x)
\[ \int \frac {(a+b x)^{4/3} (e+f x)}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}} {\left (f x + e\right )}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate((b*x+a)^(4/3)*(f*x+e)/(d*x+c)^(4/3),x, algorithm="maxima")
Output:
integrate((b*x + a)^(4/3)*(f*x + e)/(d*x + c)^(4/3), x)
\[ \int \frac {(a+b x)^{4/3} (e+f x)}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}} {\left (f x + e\right )}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate((b*x+a)^(4/3)*(f*x+e)/(d*x+c)^(4/3),x, algorithm="giac")
Output:
integrate((b*x + a)^(4/3)*(f*x + e)/(d*x + c)^(4/3), x)
Timed out. \[ \int \frac {(a+b x)^{4/3} (e+f x)}{(c+d x)^{4/3}} \, dx=\int \frac {\left (e+f\,x\right )\,{\left (a+b\,x\right )}^{4/3}}{{\left (c+d\,x\right )}^{4/3}} \,d x \] Input:
int(((e + f*x)*(a + b*x)^(4/3))/(c + d*x)^(4/3),x)
Output:
int(((e + f*x)*(a + b*x)^(4/3))/(c + d*x)^(4/3), x)
\[ \int \frac {(a+b x)^{4/3} (e+f x)}{(c+d x)^{4/3}} \, dx=\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {1}{3}} c +\left (d x +c \right )^{\frac {1}{3}} d x}d x \right ) a e +\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}} x^{2}}{\left (d x +c \right )^{\frac {1}{3}} c +\left (d x +c \right )^{\frac {1}{3}} d x}d x \right ) b f +\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}} x}{\left (d x +c \right )^{\frac {1}{3}} c +\left (d x +c \right )^{\frac {1}{3}} d x}d x \right ) a f +\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}} x}{\left (d x +c \right )^{\frac {1}{3}} c +\left (d x +c \right )^{\frac {1}{3}} d x}d x \right ) b e \] Input:
int((b*x+a)^(4/3)*(f*x+e)/(d*x+c)^(4/3),x)
Output:
int((a + b*x)**(1/3)/((c + d*x)**(1/3)*c + (c + d*x)**(1/3)*d*x),x)*a*e + int(((a + b*x)**(1/3)*x**2)/((c + d*x)**(1/3)*c + (c + d*x)**(1/3)*d*x),x) *b*f + int(((a + b*x)**(1/3)*x)/((c + d*x)**(1/3)*c + (c + d*x)**(1/3)*d*x ),x)*a*f + int(((a + b*x)**(1/3)*x)/((c + d*x)**(1/3)*c + (c + d*x)**(1/3) *d*x),x)*b*e