Integrand size = 19, antiderivative size = 217 \[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=-\frac {2 (b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 d^2}+\frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d)^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} b^{2/3} d^{7/3}}-\frac {(b c-a d)^2 \log (a+b x)}{9 b^{2/3} d^{7/3}}-\frac {(b c-a d)^2 \log \left (1-\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 b^{2/3} d^{7/3}} \] Output:
-2/3*(-a*d+b*c)*(b*x+a)^(1/3)*(d*x+c)^(2/3)/d^2+1/2*(b*x+a)^(4/3)*(d*x+c)^ (2/3)/d-2/9*(-a*d+b*c)^2*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^(7/3)-1/9*(-a*d+b*c)^2*ln(b*x +a)/b^(2/3)/d^(7/3)-1/3*(-a*d+b*c)^2*ln(1-b^(1/3)*(d*x+c)^(1/3)/d^(1/3)/(b *x+a)^(1/3))/b^(2/3)/d^(7/3)
Time = 0.46 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\frac {3 b^{2/3} \sqrt [3]{d} \sqrt [3]{a+b x} (c+d x)^{2/3} (-4 b c+7 a d+3 b d x)-4 \sqrt {3} (b c-a d)^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{d} \sqrt [3]{a+b x}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )-4 (b c-a d)^2 \log \left (\sqrt [3]{d} \sqrt [3]{a+b x}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )+2 (b c-a d)^2 \log \left (d^{2/3} (a+b x)^{2/3}+\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )}{18 b^{2/3} d^{7/3}} \] Input:
Integrate[(a + b*x)^(4/3)/(c + d*x)^(1/3),x]
Output:
(3*b^(2/3)*d^(1/3)*(a + b*x)^(1/3)*(c + d*x)^(2/3)*(-4*b*c + 7*a*d + 3*b*d *x) - 4*Sqrt[3]*(b*c - a*d)^2*ArcTan[(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))/(2* d^(1/3)*(a + b*x)^(1/3) + b^(1/3)*(c + d*x)^(1/3))] - 4*(b*c - a*d)^2*Log[ d^(1/3)*(a + b*x)^(1/3) - b^(1/3)*(c + d*x)^(1/3)] + 2*(b*c - a*d)^2*Log[d ^(2/3)*(a + b*x)^(2/3) + b^(1/3)*d^(1/3)*(a + b*x)^(1/3)*(c + d*x)^(1/3) + b^(2/3)*(c + d*x)^(2/3)])/(18*b^(2/3)*d^(7/3))
Time = 0.22 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.
\(\displaystyle \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}dx}{3 d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}}dx}{3 d}\right )}{3 d}\) |
\(\Big \downarrow \) 71 |
\(\displaystyle \frac {(a+b x)^{4/3} (c+d x)^{2/3}}{2 d}-\frac {2 (b c-a d) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{b^{2/3} \sqrt [3]{d}}-\frac {3 \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{2 b^{2/3} \sqrt [3]{d}}-\frac {\log (a+b x)}{2 b^{2/3} \sqrt [3]{d}}\right )}{3 d}\right )}{3 d}\) |
Input:
Int[(a + b*x)^(4/3)/(c + d*x)^(1/3),x]
Output:
((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)
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]
\[\int \frac {\left (b x +a \right )^{\frac {4}{3}}}{\left (x d +c \right )^{\frac {1}{3}}}d x\]
Input:
int((b*x+a)^(4/3)/(d*x+c)^(1/3),x)
Output:
int((b*x+a)^(4/3)/(d*x+c)^(1/3),x)
Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (167) = 334\).
Time = 0.12 (sec) , antiderivative size = 740, normalized size of antiderivative = 3.41 \[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx =\text {Too large to display} \] Input:
integrate((b*x+a)^(4/3)/(d*x+c)^(1/3),x, algorithm="fricas")
Output:
[1/18*(6*sqrt(1/3)*(b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*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*c^2 - 2*a*b*c*d + a^2*d^2)*(-b^2*d)^ (2/3)*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* c^2 - 2*a*b*c*d + a^2*d^2)*(-b^2*d)^(2/3)*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^2*x - 4*b^3*c *d + 7*a*b^2*d^2)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^2*d^3), 1/18*(12*sqr t(1/3)*(b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*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*d) ^(1/3)*(b*d*x + b*c))*sqrt(-(-b^2*d)^(1/3)/d)/(b^2*d*x + b^2*c)) + 2*(b^2* c^2 - 2*a*b*c*d + a^2*d^2)*(-b^2*d)^(2/3)*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*c^2 - 2*a*b*c*d + a^2*d^2)*(-b^2*d)^(2 /3)*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^2*x - 4*b^3*c*d + 7*a*b^2*d^2)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^2*d^3)]
\[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\int \frac {\left (a + b x\right )^{\frac {4}{3}}}{\sqrt [3]{c + d x}}\, dx \] Input:
integrate((b*x+a)**(4/3)/(d*x+c)**(1/3),x)
Output:
Integral((a + b*x)**(4/3)/(c + d*x)**(1/3), x)
\[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((b*x+a)^(4/3)/(d*x+c)^(1/3),x, algorithm="maxima")
Output:
integrate((b*x + a)^(4/3)/(d*x + c)^(1/3), x)
\[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((b*x+a)^(4/3)/(d*x+c)^(1/3),x, algorithm="giac")
Output:
integrate((b*x + a)^(4/3)/(d*x + c)^(1/3), x)
Timed out. \[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\int \frac {{\left (a+b\,x\right )}^{4/3}}{{\left (c+d\,x\right )}^{1/3}} \,d x \] Input:
int((a + b*x)^(4/3)/(c + d*x)^(1/3),x)
Output:
int((a + b*x)^(4/3)/(c + d*x)^(1/3), x)
\[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {1}{3}}}d x \right ) a +\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}} x}{\left (d x +c \right )^{\frac {1}{3}}}d x \right ) b \] Input:
int((b*x+a)^(4/3)/(d*x+c)^(1/3),x)
Output:
int((a + b*x)**(1/3)/(c + d*x)**(1/3),x)*a + int(((a + b*x)**(1/3)*x)/(c + d*x)**(1/3),x)*b