Integrand size = 24, antiderivative size = 684 \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^3} \, dx=-\frac {3 \left (c^2-d^2 x^2\right )^{2/3}}{4 d (c+d x)^2}+\frac {3 \left (c^2-d^2 x^2\right )^{2/3}}{4 c d (c+d x)}-\frac {3 x}{4 c \left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}-\frac {3 \sqrt [4]{3} \sqrt {2+\sqrt {3}} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} E\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right )|-7+4 \sqrt {3}\right )}{8 \sqrt [3]{c} d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}}+\frac {3^{3/4} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right ) \sqrt {\frac {c^{4/3}+c^{2/3} \sqrt [3]{c^2-d^2 x^2}+\left (c^2-d^2 x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}{\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{2 \sqrt {2} \sqrt [3]{c} d^2 x \sqrt {-\frac {c^{2/3} \left (c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) c^{2/3}-\sqrt [3]{c^2-d^2 x^2}\right )^2}}} \] Output:
-3/4*(-d^2*x^2+c^2)^(2/3)/d/(d*x+c)^2+3/4*(-d^2*x^2+c^2)^(2/3)/c/d/(d*x+c) -3/4*x/c/((1-3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3))-3/8*3^(1/4)*(1/2*6^(1/ 2)+1/2*2^(1/2))*(c^(2/3)-(-d^2*x^2+c^2)^(1/3))*((c^(4/3)+c^(2/3)*(-d^2*x^2 +c^2)^(1/3)+(-d^2*x^2+c^2)^(2/3))/((1-3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3 ))^2)^(1/2)*EllipticE(((1+3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3))/((1-3^(1/ 2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3)),2*I-I*3^(1/2))/c^(1/3)/d^2/x/(-c^(2/3)*( c^(2/3)-(-d^2*x^2+c^2)^(1/3))/((1-3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3))^2 )^(1/2)+1/4*3^(3/4)*(c^(2/3)-(-d^2*x^2+c^2)^(1/3))*((c^(4/3)+c^(2/3)*(-d^2 *x^2+c^2)^(1/3)+(-d^2*x^2+c^2)^(2/3))/((1-3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^ (1/3))^2)^(1/2)*EllipticF(((1+3^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3))/((1-3 ^(1/2))*c^(2/3)-(-d^2*x^2+c^2)^(1/3)),2*I-I*3^(1/2))*2^(1/2)/c^(1/3)/d^2/x /(-c^(2/3)*(c^(2/3)-(-d^2*x^2+c^2)^(1/3))/((1-3^(1/2))*c^(2/3)-(-d^2*x^2+c ^2)^(1/3))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.60 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.12 \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^3} \, dx=-\frac {3 (c-d x) \sqrt [3]{1+\frac {d x}{c}} \left (c^2-d^2 x^2\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {5}{3},\frac {7}{3},\frac {8}{3},\frac {c-d x}{2 c}\right )}{20 \sqrt [3]{2} c^2 d (c+d x)} \] Input:
Integrate[(c^2 - d^2*x^2)^(2/3)/(c + d*x)^3,x]
Output:
(-3*(c - d*x)*(1 + (d*x)/c)^(1/3)*(c^2 - d^2*x^2)^(2/3)*Hypergeometric2F1[ 5/3, 7/3, 8/3, (c - d*x)/(2*c)])/(20*2^(1/3)*c^2*d*(c + d*x))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.10, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {506, 74}
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.
| \(\displaystyle \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^3} \, dx\) |
| \(\Big \downarrow \) 506 |
| \(\displaystyle -\frac {\left (\frac {1}{c+d x}\right )^{4/3} \left (c^2-d^2 x^2\right )^{2/3} \int \frac {\left (1-\frac {2 c}{c+d x}\right )^{2/3}}{\sqrt [3]{\frac {1}{c+d x}}}d\frac {1}{c+d x}}{d \left (1-\frac {2 c}{c+d x}\right )^{2/3}}\) |
| \(\Big \downarrow \) 74 |
| \(\displaystyle -\frac {3 \left (c^2-d^2 x^2\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {2}{3},\frac {5}{3},\frac {2 c}{c+d x}\right )}{2 d (c+d x)^2 \left (1-\frac {2 c}{c+d x}\right )^{2/3}}\) |
Input:
Int[(c^2 - d^2*x^2)^(2/3)/(c + d*x)^3,x]
Output:
(-3*(c^2 - d^2*x^2)^(2/3)*Hypergeometric2F1[-2/3, 2/3, 5/3, (2*c)/(c + d*x )])/(2*d*(c + d*x)^2*(1 - (2*c)/(c + d*x))^(2/3))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {q = Rt[-a/b, 2]}, Simp[(-(a + b*x^2)^p)*((1/(c + d*x))^(2*p)/(d*(1 - (c - d*q)/(c + d*x))^p*(1 - (c + d*q)/(c + d*x))^p)) Subst[Int[(1 - (c - d*q)* x)^p*((1 - (c + d*q)*x)^p/x^(n + 2*p + 2)), x], x, 1/(c + d*x)], x]] /; Fre eQ[{a, b, c, d, p}, x] && ILtQ[n, -1] && NegQ[a/b]
\[\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}}{\left (d x +c \right )^{3}}d x\]
Input:
int((-d^2*x^2+c^2)^(2/3)/(d*x+c)^3,x)
Output:
int((-d^2*x^2+c^2)^(2/3)/(d*x+c)^3,x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^3} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^(2/3)/(d*x+c)^3,x, algorithm="fricas")
Output:
integral((-d^2*x^2 + c^2)^(2/3)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^3} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {2}{3}}}{\left (c + d x\right )^{3}}\, dx \] Input:
integrate((-d**2*x**2+c**2)**(2/3)/(d*x+c)**3,x)
Output:
Integral((-(-c + d*x)*(c + d*x))**(2/3)/(c + d*x)**3, x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^3} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^(2/3)/(d*x+c)^3,x, algorithm="maxima")
Output:
integrate((-d^2*x^2 + c^2)^(2/3)/(d*x + c)^3, x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^3} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{3}}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^(2/3)/(d*x+c)^3,x, algorithm="giac")
Output:
integrate((-d^2*x^2 + c^2)^(2/3)/(d*x + c)^3, x)
Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^3} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{2/3}}{{\left (c+d\,x\right )}^3} \,d x \] Input:
int((c^2 - d^2*x^2)^(2/3)/(c + d*x)^3,x)
Output:
int((c^2 - d^2*x^2)^(2/3)/(c + d*x)^3, x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^{2/3}}{(c+d x)^3} \, dx=\frac {-3 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}}-4 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}} x}{-d^{4} x^{4}-2 c \,d^{3} x^{3}+2 c^{3} d x +c^{4}}d x \right ) c^{2} d^{2}-8 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}} x}{-d^{4} x^{4}-2 c \,d^{3} x^{3}+2 c^{3} d x +c^{4}}d x \right ) c \,d^{3} x -4 \left (\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{3}} x}{-d^{4} x^{4}-2 c \,d^{3} x^{3}+2 c^{3} d x +c^{4}}d x \right ) d^{4} x^{2}}{6 d \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:
int((-d^2*x^2+c^2)^(2/3)/(d*x+c)^3,x)
Output:
( - 3*(c**2 - d**2*x**2)**(2/3) - 4*int(((c**2 - d**2*x**2)**(2/3)*x)/(c** 4 + 2*c**3*d*x - 2*c*d**3*x**3 - d**4*x**4),x)*c**2*d**2 - 8*int(((c**2 - d**2*x**2)**(2/3)*x)/(c**4 + 2*c**3*d*x - 2*c*d**3*x**3 - d**4*x**4),x)*c* d**3*x - 4*int(((c**2 - d**2*x**2)**(2/3)*x)/(c**4 + 2*c**3*d*x - 2*c*d**3 *x**3 - d**4*x**4),x)*d**4*x**2)/(6*d*(c**2 + 2*c*d*x + d**2*x**2))