\(\int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx\) [529]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

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)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

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))
 

Rubi [A] (verified)

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)
 

Defintions of rubi rules used

rule 60
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]
 

rule 71
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]
 
Maple [F]

\[\int \frac {\left (b x +a \right )^{\frac {4}{3}}}{\left (x d +c \right )^{\frac {1}{3}}}d x\]

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)
 

Fricas [B] (verification not implemented)

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)]
 

Sympy [F]

\[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\int \frac {\left (a + b x\right )^{\frac {4}{3}}}{\sqrt [3]{c + d x}}\, dx \] 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)
 

Maxima [F]

\[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] 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)
 

Giac [F]

\[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {4}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] 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)
 

Mupad [F(-1)]

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)
 

Reduce [F]

\[ \int \frac {(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx=\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {1}{3}}}d x \right ) a +\left (\int \frac {\left (b x +a \right )^{\frac {1}{3}} x}{\left (d x +c \right )^{\frac {1}{3}}}d x \right ) b \] 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