Integrand size = 19, antiderivative size = 196 \[ \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx=-\frac {3 (a+b x)^{4/3}}{d \sqrt [3]{c+d x}}+\frac {4 b \sqrt [3]{a+b x} (c+d x)^{2/3}}{d^2}+\frac {4 \sqrt [3]{b} (b c-a d) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{\sqrt {3} d^{7/3}}+\frac {2 \sqrt [3]{b} (b c-a d) \log (a+b x)}{3 d^{7/3}}+\frac {2 \sqrt [3]{b} (b c-a d) \log \left (1-\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{d^{7/3}} \] Output:
-3*(b*x+a)^(4/3)/d/(d*x+c)^(1/3)+4*b*(b*x+a)^(1/3)*(d*x+c)^(2/3)/d^2+4/3*b ^(1/3)*(-a*d+b*c)*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)/d^(7/3)+2/3*b^(1/3)*(-a*d+b*c)*ln(b*x+a)/d^(7/ 3)+2*b^(1/3)*(-a*d+b*c)*ln(1-b^(1/3)*(d*x+c)^(1/3)/d^(1/3)/(b*x+a)^(1/3))/ d^(7/3)
Time = 0.18 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx=\frac {\frac {3 \sqrt [3]{d} \sqrt [3]{a+b x} (4 b c-3 a d+b d x)}{\sqrt [3]{c+d x}}+4 \sqrt {3} \sqrt [3]{b} (b c-a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{d} \sqrt [3]{a+b x}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )+4 \sqrt [3]{b} (b c-a d) \log \left (\sqrt [3]{d} \sqrt [3]{a+b x}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )+2 \sqrt [3]{b} (-b c+a d) \log \left (d^{2/3} (a+b x)^{2/3}+\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )}{3 d^{7/3}} \] Input:
Integrate[(a + b*x)^(4/3)/(c + d*x)^(4/3),x]
Output:
((3*d^(1/3)*(a + b*x)^(1/3)*(4*b*c - 3*a*d + b*d*x))/(c + d*x)^(1/3) + 4*S qrt[3]*b^(1/3)*(b*c - a*d)*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^(1/3)*(b*c - a*d)*L og[d^(1/3)*(a + b*x)^(1/3) - b^(1/3)*(c + d*x)^(1/3)] + 2*b^(1/3)*(-(b*c) + a*d)*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)])/(3*d^(7/3))
Time = 0.23 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {57, 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.
\(\displaystyle \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {4 b \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}dx}{d}-\frac {3 (a+b x)^{4/3}}{d \sqrt [3]{c+d x}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {4 b \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}}dx}{3 d}\right )}{d}-\frac {3 (a+b x)^{4/3}}{d \sqrt [3]{c+d x}}\) |
\(\Big \downarrow \) 71 |
\(\displaystyle \frac {4 b \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{d}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{b^{2/3} \sqrt [3]{d}}-\frac {3 \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{2 b^{2/3} \sqrt [3]{d}}-\frac {\log (a+b x)}{2 b^{2/3} \sqrt [3]{d}}\right )}{3 d}\right )}{d}-\frac {3 (a+b x)^{4/3}}{d \sqrt [3]{c+d x}}\) |
Input:
Int[(a + b*x)^(4/3)/(c + d*x)^(4/3),x]
Output:
(-3*(a + b*x)^(4/3))/(d*(c + d*x)^(1/3)) + (4*b*(((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)))/d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, 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]
\[\int \frac {\left (b x +a \right )^{\frac {4}{3}}}{\left (x d +c \right )^{\frac {4}{3}}}d x\]
Input:
int((b*x+a)^(4/3)/(d*x+c)^(4/3),x)
Output:
int((b*x+a)^(4/3)/(d*x+c)^(4/3),x)
Time = 0.11 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.56 \[ \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx=\frac {4 \, \sqrt {3} {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} d \left (-\frac {b}{d}\right )^{\frac {2}{3}} + \sqrt {3} {\left (b d x + b c\right )}}{3 \, {\left (b d x + b c\right )}}\right ) + 2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \log \left (\frac {{\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {2}{3}} - {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} \left (-\frac {b}{d}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{d x + c}\right ) - 4 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \log \left (\frac {{\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{d x + c}\right ) + 3 \, {\left (b d x + 4 \, b c - 3 \, a d\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{3 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:
integrate((b*x+a)^(4/3)/(d*x+c)^(4/3),x, algorithm="fricas")
Output:
1/3*(4*sqrt(3)*(b*c^2 - a*c*d + (b*c*d - a*d^2)*x)*(-b/d)^(1/3)*arctan(1/3 *(2*sqrt(3)*(b*x + a)^(1/3)*(d*x + c)^(2/3)*d*(-b/d)^(2/3) + sqrt(3)*(b*d* x + b*c))/(b*d*x + b*c)) + 2*(b*c^2 - a*c*d + (b*c*d - a*d^2)*x)*(-b/d)^(1 /3)*log(((d*x + c)*(-b/d)^(2/3) - (b*x + a)^(1/3)*(d*x + c)^(2/3)*(-b/d)^( 1/3) + (b*x + a)^(2/3)*(d*x + c)^(1/3))/(d*x + c)) - 4*(b*c^2 - a*c*d + (b *c*d - a*d^2)*x)*(-b/d)^(1/3)*log(((d*x + c)*(-b/d)^(1/3) + (b*x + a)^(1/3 )*(d*x + c)^(2/3))/(d*x + c)) + 3*(b*d*x + 4*b*c - 3*a*d)*(b*x + a)^(1/3)* (d*x + c)^(2/3))/(d^3*x + c*d^2)
\[ \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx=\int \frac {\left (a + b x\right )^{\frac {4}{3}}}{\left (c + d x\right )^{\frac {4}{3}}}\, dx \] Input:
integrate((b*x+a)**(4/3)/(d*x+c)**(4/3),x)
Output:
Integral((a + b*x)**(4/3)/(c + d*x)**(4/3), x)
\[ \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate((b*x+a)^(4/3)/(d*x+c)^(4/3),x, algorithm="maxima")
Output:
integrate((b*x + a)^(4/3)/(d*x + c)^(4/3), x)
\[ \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate((b*x+a)^(4/3)/(d*x+c)^(4/3),x, algorithm="giac")
Output:
integrate((b*x + a)^(4/3)/(d*x + c)^(4/3), x)
Timed out. \[ \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx=\int \frac {{\left (a+b\,x\right )}^{4/3}}{{\left (c+d\,x\right )}^{4/3}} \,d x \] Input:
int((a + b*x)^(4/3)/(c + d*x)^(4/3),x)
Output:
int((a + b*x)^(4/3)/(c + d*x)^(4/3), x)
\[ \int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx=\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {1}{3}} c +\left (d x +c \right )^{\frac {1}{3}} d x}d x \right ) a +\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}} x}{\left (d x +c \right )^{\frac {1}{3}} c +\left (d x +c \right )^{\frac {1}{3}} d x}d x \right ) b \] Input:
int((b*x+a)^(4/3)/(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 + in t(((a + b*x)**(1/3)*x)/((c + d*x)**(1/3)*c + (c + d*x)**(1/3)*d*x),x)*b