\(\int (c+d x)^{2/3} (c^2-d^2 x^2)^{2/3} \, dx\) [280]

Optimal result

Integrand size = 26, antiderivative size = 303 \[ \int (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \, dx=\frac {8 c^2 \left (c^2-d^2 x^2\right )^{2/3}}{27 d \sqrt [3]{c+d x}}-\frac {4 c \left (c^2-d^2 x^2\right )^{5/3}}{9 d (c+d x)^{4/3}}-\frac {\left (c^2-d^2 x^2\right )^{5/3}}{3 d \sqrt [3]{c+d x}}+\frac {32 c^3 \left (c^2-d^2 x^2\right )^{2/3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c-d x}}{\sqrt {3} \sqrt [3]{c+d x}}\right )}{27 \sqrt {3} d (c-d x)^{2/3} (c+d x)^{2/3}}+\frac {16 c^3 \left (c^2-d^2 x^2\right )^{2/3} \log (c+d x)}{81 d (c-d x)^{2/3} (c+d x)^{2/3}}+\frac {16 c^3 \left (c^2-d^2 x^2\right )^{2/3} \log \left (1+\frac {\sqrt [3]{c-d x}}{\sqrt [3]{c+d x}}\right )}{27 d (c-d x)^{2/3} (c+d x)^{2/3}} \] Output:

8/27*c^2*(-d^2*x^2+c^2)^(2/3)/d/(d*x+c)^(1/3)-4/9*c*(-d^2*x^2+c^2)^(5/3)/d 
/(d*x+c)^(4/3)-1/3*(-d^2*x^2+c^2)^(5/3)/d/(d*x+c)^(1/3)-32/81*c^3*(-d^2*x^ 
2+c^2)^(2/3)*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*x+c)^(2/3)/(d*x+c)^(2/3)+16/81*c^3*(-d^2*x^2+c^2)^(2/3)*ln( 
d*x+c)/d/(-d*x+c)^(2/3)/(d*x+c)^(2/3)+16/27*c^3*(-d^2*x^2+c^2)^(2/3)*ln(1+ 
(-d*x+c)^(1/3)/(d*x+c)^(1/3))/d/(-d*x+c)^(2/3)/(d*x+c)^(2/3)
 

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.66 \[ \int (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \, dx=-\frac {\frac {3 \left (13 c^2-12 c d x-9 d^2 x^2\right ) \left (c^2-d^2 x^2\right )^{2/3}}{\sqrt [3]{c+d x}}+32 \sqrt {3} c^3 \arctan \left (\frac {\sqrt {3} (c+d x)^{2/3}}{(c+d x)^{2/3}-2 \sqrt [3]{c^2-d^2 x^2}}\right )-32 c^3 \log \left ((c+d x)^{2/3}+\sqrt [3]{c^2-d^2 x^2}\right )+16 c^3 \log \left ((c+d x)^{4/3}-(c+d x)^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}\right )}{81 d} \] Input:

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

Output:

-1/81*((3*(13*c^2 - 12*c*d*x - 9*d^2*x^2)*(c^2 - d^2*x^2)^(2/3))/(c + d*x) 
^(1/3) + 32*Sqrt[3]*c^3*ArcTan[(Sqrt[3]*(c + d*x)^(2/3))/((c + d*x)^(2/3) 
- 2*(c^2 - d^2*x^2)^(1/3))] - 32*c^3*Log[(c + d*x)^(2/3) + (c^2 - d^2*x^2) 
^(1/3)] + 16*c^3*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)])/d
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 296, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {474, 473, 60, 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.

\(\displaystyle \int (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \, dx\)

\(\Big \downarrow \) 474

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

\(\Big \downarrow \) 473

\(\displaystyle \frac {(c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{5/3} \int \left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}dx}{\left (\frac {d x}{c}+1\right )^{7/3} \left (c^2-c d x\right )^{5/3}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{5/3} \left (\frac {4}{9} c^2 \int \frac {\left (\frac {d x}{c}+1\right )^{4/3}}{\sqrt [3]{c^2-c d x}}dx+\frac {c \left (c^2-c d x\right )^{2/3} \left (\frac {d x}{c}+1\right )^{7/3}}{3 d}\right )}{\left (\frac {d x}{c}+1\right )^{7/3} \left (c^2-c d x\right )^{5/3}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{5/3} \left (\frac {4}{9} c^2 \left (\frac {4}{3} \int \frac {\sqrt [3]{\frac {d x}{c}+1}}{\sqrt [3]{c^2-c d x}}dx-\frac {\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}{2 c d}\right )+\frac {c \left (c^2-c d x\right )^{2/3} \left (\frac {d x}{c}+1\right )^{7/3}}{3 d}\right )}{\left (\frac {d x}{c}+1\right )^{7/3} \left (c^2-c d x\right )^{5/3}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{5/3} \left (\frac {4}{9} c^2 \left (\frac {4}{3} \left (\frac {2}{3} \int \frac {1}{\left (\frac {d x}{c}+1\right )^{2/3} \sqrt [3]{c^2-c d x}}dx-\frac {\sqrt [3]{\frac {d x}{c}+1} \left (c^2-c d x\right )^{2/3}}{c d}\right )-\frac {\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}{2 c d}\right )+\frac {c \left (c^2-c d x\right )^{2/3} \left (\frac {d x}{c}+1\right )^{7/3}}{3 d}\right )}{\left (\frac {d x}{c}+1\right )^{7/3} \left (c^2-c d x\right )^{5/3}}\)

\(\Big \downarrow \) 72

\(\displaystyle \frac {(c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{5/3} \left (\frac {4}{9} c^2 \left (\frac {4}{3} \left (\frac {2}{3} \left (\frac {\sqrt {3} \sqrt [3]{c} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c^2-c d x}}{\sqrt {3} c^{2/3} \sqrt [3]{\frac {d x}{c}+1}}\right )}{d}+\frac {3 \sqrt [3]{c} \log \left (\frac {\sqrt [3]{c^2-c d x}}{c^{2/3} \sqrt [3]{\frac {d x}{c}+1}}+1\right )}{2 d}+\frac {\sqrt [3]{c} \log \left (\frac {d x}{c}+1\right )}{2 d}\right )-\frac {\sqrt [3]{\frac {d x}{c}+1} \left (c^2-c d x\right )^{2/3}}{c d}\right )-\frac {\left (\frac {d x}{c}+1\right )^{4/3} \left (c^2-c d x\right )^{2/3}}{2 c d}\right )+\frac {c \left (c^2-c d x\right )^{2/3} \left (\frac {d x}{c}+1\right )^{7/3}}{3 d}\right )}{\left (\frac {d x}{c}+1\right )^{7/3} \left (c^2-c d x\right )^{5/3}}\)

Input:

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

Output:

((c + d*x)^(2/3)*(c^2 - d^2*x^2)^(5/3)*((c*(1 + (d*x)/c)^(7/3)*(c^2 - c*d* 
x)^(2/3))/(3*d) + (4*c^2*(-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))/(Sqrt[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))/9) 
)/((1 + (d*x)/c)^(7/3)*(c^2 - c*d*x)^(5/3))
 

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.94 \[ \int (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \, dx=-\frac {32 \, \sqrt {3} {\left (c^{3} d x + c^{4}\right )} \arctan \left (\frac {2 \, \sqrt {3} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}} + \sqrt {3} {\left (d^{2} x^{2} - c^{2}\right )}}{3 \, {\left (d^{2} x^{2} - c^{2}\right )}}\right ) - 3 \, {\left (9 \, d^{2} x^{2} + 12 \, c d x - 13 \, c^{2}\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}} + 16 \, {\left (c^{3} d x + c^{4}\right )} \log \left (\frac {d^{2} x^{2} - c^{2} - {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {4}{3}} + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{d^{2} x^{2} - c^{2}}\right ) - 32 \, {\left (c^{3} d x + c^{4}\right )} \log \left (-\frac {d^{2} x^{2} - c^{2} - {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{d^{2} x^{2} - c^{2}}\right )}{81 \, {\left (d^{2} x + c d\right )}} \] Input:

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

Output:

-1/81*(32*sqrt(3)*(c^3*d*x + c^4)*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*(9*d^ 
2*x^2 + 12*c*d*x - 13*c^2)*(-d^2*x^2 + c^2)^(2/3)*(d*x + c)^(2/3) + 16*(c^ 
3*d*x + c^4)*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)) - 32*(c^3*d*x + 
c^4)*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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 301, normalized size of antiderivative = 0.99 \[ \int (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \, dx=-\frac {144 \, \sqrt {3} c^{3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) - {\left (16 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {1}{3}} - 1\right )}\right ) + \frac {3 \, {\left (2 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {8}{3}} - 35 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{3}} - {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {2}{3}}\right )} {\left (d x + c\right )}^{3}}{c^{3}} + 8 \, \log \left ({\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {2}{3}} - {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {1}{3}} + 1\right ) - 16 \, \log \left ({\left | {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {1}{3}} + 1 \right |}\right )\right )} c^{3} + 72 \, c^{3} \log \left ({\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {2}{3}} - {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {1}{3}} + 1\right ) - 144 \, c^{3} \log \left ({\left | {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {1}{3}} + 1 \right |}\right ) + \frac {54 \, {\left (2 \, c^{3} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{3}} - c^{3} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {2}{3}}\right )} {\left (d x + c\right )}^{2}}{c^{2}}}{324 \, d} \] Input:

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

Output:

-1/324*(144*sqrt(3)*c^3*arctan(1/3*sqrt(3)*(2*(2*c/(d*x + c) - 1)^(1/3) - 
1)) - (16*sqrt(3)*arctan(1/3*sqrt(3)*(2*(2*c/(d*x + c) - 1)^(1/3) - 1)) + 
3*(2*(2*c/(d*x + c) - 1)^(8/3) - 35*(2*c/(d*x + c) - 1)^(5/3) - (2*c/(d*x 
+ c) - 1)^(2/3))*(d*x + c)^3/c^3 + 8*log((2*c/(d*x + c) - 1)^(2/3) - (2*c/ 
(d*x + c) - 1)^(1/3) + 1) - 16*log(abs((2*c/(d*x + c) - 1)^(1/3) + 1)))*c^ 
3 + 72*c^3*log((2*c/(d*x + c) - 1)^(2/3) - (2*c/(d*x + c) - 1)^(1/3) + 1) 
- 144*c^3*log(abs((2*c/(d*x + c) - 1)^(1/3) + 1)) + 54*(2*c^3*(2*c/(d*x + 
c) - 1)^(5/3) - c^3*(2*c/(d*x + c) - 1)^(2/3))*(d*x + c)^2/c^2)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^{2/3} \left (c^2-d^2 x^2\right )^{2/3} \, dx=\frac {9 \left (d x +c \right )^{\frac {2}{3}} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}} c +3 \left (d x +c \right )^{\frac {2}{3}} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}} d x +16 \left (\int \frac {\left (d x +c \right )^{\frac {2}{3}} x}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {1}{3}}}d x \right ) c \,d^{2}}{9 d} \] Input:

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

Output:

(9*(c + d*x)**(2/3)*(c**2 - d**2*x**2)**(2/3)*c + 3*(c + d*x)**(2/3)*(c**2 
 - d**2*x**2)**(2/3)*d*x + 16*int(((c + d*x)**(2/3)*(c**2 - d**2*x**2)**(2 
/3)*x)/(c**2 - d**2*x**2),x)*c*d**2)/(9*d)