\(\int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx\) [288]

Optimal result

Integrand size = 26, antiderivative size = 268 \[ \int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx=-\frac {4 c \left (c^2-d^2 x^2\right )^{2/3}}{3 d \sqrt [3]{c+d x}}-\frac {(c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3}}{2 d}+\frac {8 c^2 \sqrt [3]{c-d x} \sqrt [3]{c+d x} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c-d x}}{\sqrt {3} \sqrt [3]{c+d x}}\right )}{3 \sqrt {3} d \sqrt [3]{c^2-d^2 x^2}}+\frac {4 c^2 \sqrt [3]{c-d x} \sqrt [3]{c+d x} \log (c+d x)}{9 d \sqrt [3]{c^2-d^2 x^2}}+\frac {4 c^2 \sqrt [3]{c-d x} \sqrt [3]{c+d x} \log \left (1+\frac {\sqrt [3]{c-d x}}{\sqrt [3]{c+d x}}\right )}{3 d \sqrt [3]{c^2-d^2 x^2}} \] Output:

-4/3*c*(-d^2*x^2+c^2)^(2/3)/d/(d*x+c)^(1/3)-1/2*(d*x+c)^(2/3)*(-d^2*x^2+c^ 
2)^(2/3)/d-8/9*c^2*(-d*x+c)^(1/3)*(d*x+c)^(1/3)*arctan(-1/3*3^(1/2)+2/3*(- 
d*x+c)^(1/3)*3^(1/2)/(d*x+c)^(1/3))*3^(1/2)/d/(-d^2*x^2+c^2)^(1/3)+4/9*c^2 
*(-d*x+c)^(1/3)*(d*x+c)^(1/3)*ln(d*x+c)/d/(-d^2*x^2+c^2)^(1/3)+4/3*c^2*(-d 
*x+c)^(1/3)*(d*x+c)^(1/3)*ln(1+(-d*x+c)^(1/3)/(d*x+c)^(1/3))/d/(-d^2*x^2+c 
^2)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.68 \[ \int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx=\frac {-\frac {3 (11 c+3 d x) \left (c^2-d^2 x^2\right )^{2/3}}{\sqrt [3]{c+d x}}+16 \sqrt {3} c^2 \arctan \left (\frac {1-\frac {2 \sqrt [3]{c^2-d^2 x^2}}{(c+d x)^{2/3}}}{\sqrt {3}}\right )+16 c^2 \log \left (d \left ((c+d x)^{2/3}+\sqrt [3]{c^2-d^2 x^2}\right )\right )-8 c^2 \log \left ((c+d x)^{4/3}-(c+d x)^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}\right )}{18 d} \] Input:

Integrate[(c + d*x)^(5/3)/(c^2 - d^2*x^2)^(1/3),x]
 

Output:

((-3*(11*c + 3*d*x)*(c^2 - d^2*x^2)^(2/3))/(c + d*x)^(1/3) + 16*Sqrt[3]*c^ 
2*ArcTan[(1 - (2*(c^2 - d^2*x^2)^(1/3))/(c + d*x)^(2/3))/Sqrt[3]] + 16*c^2 
*Log[d*((c + d*x)^(2/3) + (c^2 - d^2*x^2)^(1/3))] - 8*c^2*Log[(c + d*x)^(4 
/3) - (c + d*x)^(2/3)*(c^2 - d^2*x^2)^(1/3) + (c^2 - d^2*x^2)^(2/3)])/(18* 
d)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {474, 473, 60, 60, 72}

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.

\(\displaystyle \int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx\)

\(\Big \downarrow \) 474

\(\displaystyle \frac {c (c+d x)^{2/3} \int \frac {\left (\frac {d x}{c}+1\right )^{5/3}}{\sqrt [3]{c^2-d^2 x^2}}dx}{\left (\frac {d x}{c}+1\right )^{2/3}}\)

\(\Big \downarrow \) 473

\(\displaystyle \frac {c (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \int \frac {\left (\frac {d x}{c}+1\right )^{4/3}}{\sqrt [3]{c^2-c d x}}dx}{\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {c (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \left (\frac {4}{3} \int \frac {\sqrt [3]{\frac {d x}{c}+1}}{\sqrt [3]{c^2-c d x}}dx-\frac {\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}{2 c d}\right )}{\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {c (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \left (\frac {4}{3} \left (\frac {2}{3} \int \frac {1}{\left (\frac {d x}{c}+1\right )^{2/3} \sqrt [3]{c^2-c d x}}dx-\frac {\sqrt [3]{\frac {d x}{c}+1} \left (c^2-c d x\right )^{2/3}}{c d}\right )-\frac {\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}{2 c d}\right )}{\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}\)

\(\Big \downarrow \) 72

\(\displaystyle \frac {c (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \left (\frac {4}{3} \left (\frac {2}{3} \left (\frac {\sqrt {3} \sqrt [3]{c} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c^2-c d x}}{\sqrt {3} c^{2/3} \sqrt [3]{\frac {d x}{c}+1}}\right )}{d}+\frac {3 \sqrt [3]{c} \log \left (\frac {\sqrt [3]{c^2-c d x}}{c^{2/3} \sqrt [3]{\frac {d x}{c}+1}}+1\right )}{2 d}+\frac {\sqrt [3]{c} \log \left (\frac {d x}{c}+1\right )}{2 d}\right )-\frac {\sqrt [3]{\frac {d x}{c}+1} \left (c^2-c d x\right )^{2/3}}{c d}\right )-\frac {\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}{2 c d}\right )}{\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}\)

Input:

Int[(c + d*x)^(5/3)/(c^2 - d^2*x^2)^(1/3),x]
 

Output:

(c*(c + d*x)^(2/3)*(c^2 - d^2*x^2)^(2/3)*(-1/2*((1 + (d*x)/c)^(4/3)*(c^2 - 
 c*d*x)^(2/3))/(c*d) + (4*(-(((1 + (d*x)/c)^(1/3)*(c^2 - c*d*x)^(2/3))/(c* 
d)) + (2*((Sqrt[3]*c^(1/3)*ArcTan[1/Sqrt[3] - (2*(c^2 - c*d*x)^(1/3))/(Sqr 
t[3]*c^(2/3)*(1 + (d*x)/c)^(1/3))])/d + (c^(1/3)*Log[1 + (d*x)/c])/(2*d) + 
 (3*c^(1/3)*Log[1 + (c^2 - c*d*x)^(1/3)/(c^(2/3)*(1 + (d*x)/c)^(1/3))])/(2 
*d)))/3))/3))/((1 + (d*x)/c)^(4/3)*(c^2 - c*d*x)^(2/3))
 

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 72

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[1/Sqrt[3] - 2*q*((a + b* 
x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))], 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])] /; F 
reeQ[{a, b, c, d}, x] && NegQ[d/b]
 

rule 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

rule 474

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n])   Int[(1 + d 
*(x/c))^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + 
 a*d^2, 0] &&  !(IntegerQ[n] || GtQ[c, 0])
 

Maple [F]

\[\int \frac {\left (d x +c \right )^{\frac {5}{3}}}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}d x\]

Input:

int((d*x+c)^(5/3)/(-d^2*x^2+c^2)^(1/3),x)
 

Output:

int((d*x+c)^(5/3)/(-d^2*x^2+c^2)^(1/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.03 \[ \int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx=-\frac {16 \, \sqrt {3} {\left (c^{2} d x + c^{3}\right )} \arctan \left (\frac {2 \, \sqrt {3} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}} + \sqrt {3} {\left (d^{2} x^{2} - c^{2}\right )}}{3 \, {\left (d^{2} x^{2} - c^{2}\right )}}\right ) + 3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (3 \, d x + 11 \, c\right )} {\left (d x + c\right )}^{\frac {2}{3}} + 8 \, {\left (c^{2} d x + c^{3}\right )} \log \left (\frac {d^{2} x^{2} - c^{2} - {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {4}{3}} + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{d^{2} x^{2} - c^{2}}\right ) - 16 \, {\left (c^{2} d x + c^{3}\right )} \log \left (-\frac {d^{2} x^{2} - c^{2} - {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{d^{2} x^{2} - c^{2}}\right )}{18 \, {\left (d^{2} x + c d\right )}} \] Input:

integrate((d*x+c)^(5/3)/(-d^2*x^2+c^2)^(1/3),x, algorithm="fricas")
 

Output:

-1/18*(16*sqrt(3)*(c^2*d*x + c^3)*arctan(1/3*(2*sqrt(3)*(-d^2*x^2 + c^2)^( 
2/3)*(d*x + c)^(2/3) + sqrt(3)*(d^2*x^2 - c^2))/(d^2*x^2 - c^2)) + 3*(-d^2 
*x^2 + c^2)^(2/3)*(3*d*x + 11*c)*(d*x + c)^(2/3) + 8*(c^2*d*x + c^3)*log(( 
d^2*x^2 - c^2 - (-d^2*x^2 + c^2)^(1/3)*(d*x + c)^(4/3) + (-d^2*x^2 + c^2)^ 
(2/3)*(d*x + c)^(2/3))/(d^2*x^2 - c^2)) - 16*(c^2*d*x + c^3)*log(-(d^2*x^2 
 - c^2 - (-d^2*x^2 + c^2)^(2/3)*(d*x + c)^(2/3))/(d^2*x^2 - c^2)))/(d^2*x 
+ c*d)
 

Sympy [F]

\[ \int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{3}}}{\sqrt [3]{- \left (- c + d x\right ) \left (c + d x\right )}}\, dx \] Input:

integrate((d*x+c)**(5/3)/(-d**2*x**2+c**2)**(1/3),x)
 

Output:

Integral((c + d*x)**(5/3)/(-(-c + d*x)*(c + d*x))**(1/3), x)
 

Maxima [F]

\[ \int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{3}}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((d*x+c)^(5/3)/(-d^2*x^2+c^2)^(1/3),x, algorithm="maxima")
 

Output:

integrate((d*x + c)^(5/3)/(-d^2*x^2 + c^2)^(1/3), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.57 \[ \int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx=-\frac {16 \, \sqrt {3} c^{3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) + 8 \, c^{3} \log \left ({\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {2}{3}} - {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {1}{3}} + 1\right ) - 16 \, c^{3} \log \left ({\left | {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {1}{3}} + 1 \right |}\right ) + \frac {3 \, {\left (4 \, c^{3} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{3}} + 7 \, c^{3} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {2}{3}}\right )} {\left (d x + c\right )}^{2}}{c^{2}}}{18 \, c d} \] Input:

integrate((d*x+c)^(5/3)/(-d^2*x^2+c^2)^(1/3),x, algorithm="giac")
 

Output:

-1/18*(16*sqrt(3)*c^3*arctan(1/3*sqrt(3)*(2*(2*c/(d*x + c) - 1)^(1/3) - 1) 
) + 8*c^3*log((2*c/(d*x + c) - 1)^(2/3) - (2*c/(d*x + c) - 1)^(1/3) + 1) - 
 16*c^3*log(abs((2*c/(d*x + c) - 1)^(1/3) + 1)) + 3*(4*c^3*(2*c/(d*x + c) 
- 1)^(5/3) + 7*c^3*(2*c/(d*x + c) - 1)^(2/3))*(d*x + c)^2/c^2)/(c*d)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/3}}{{\left (c^2-d^2\,x^2\right )}^{1/3}} \,d x \] Input:

int((c + d*x)^(5/3)/(c^2 - d^2*x^2)^(1/3),x)
 

Output:

int((c + d*x)^(5/3)/(c^2 - d^2*x^2)^(1/3), x)
 

Reduce [F]

\[ \int \frac {(c+d x)^{5/3}}{\sqrt [3]{c^2-d^2 x^2}} \, dx=\left (\int \frac {\left (d x +c \right )^{\frac {2}{3}}}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}d x \right ) c +\left (\int \frac {\left (d x +c \right )^{\frac {2}{3}} x}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}d x \right ) d \] Input:

int((d*x+c)^(5/3)/(-d^2*x^2+c^2)^(1/3),x)
 

Output:

int((c + d*x)**(2/3)/(c**2 - d**2*x**2)**(1/3),x)*c + int(((c + d*x)**(2/3 
)*x)/(c**2 - d**2*x**2)**(1/3),x)*d