Integrand size = 22, antiderivative size = 67 \[ \int (c+d x) \left (c^2-d^2 x^2\right )^{2/5} \, dx=\frac {5 \left (c^2-d^2 x^2\right )^{12/5} \operatorname {Hypergeometric2F1}\left (-\frac {2}{5},\frac {12}{5},\frac {17}{5},\frac {c+d x}{2 c}\right )}{6\ 2^{3/5} c^2 d \left (\frac {c-d x}{c}\right )^{12/5}} \] Output:
5/12*(-d^2*x^2+c^2)^(12/5)*hypergeom([-2/5, 12/5],[17/5],1/2*(d*x+c)/c)*2^ (2/5)/c^2/d/((-d*x+c)/c)^(12/5)
Time = 7.77 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19 \[ \int (c+d x) \left (c^2-d^2 x^2\right )^{2/5} \, dx=-\frac {5 \left (c^2-d^2 x^2\right )^{7/5}}{14 d}+\frac {c x \left (c^2-d^2 x^2\right )^{2/5} \operatorname {Hypergeometric2F1}\left (-\frac {2}{5},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\left (1-\frac {d^2 x^2}{c^2}\right )^{2/5}} \] Input:
Integrate[(c + d*x)*(c^2 - d^2*x^2)^(2/5),x]
Output:
(-5*(c^2 - d^2*x^2)^(7/5))/(14*d) + (c*x*(c^2 - d^2*x^2)^(2/5)*Hypergeomet ric2F1[-2/5, 1/2, 3/2, (d^2*x^2)/c^2])/(1 - (d^2*x^2)/c^2)^(2/5)
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {455, 238, 237}
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.
\(\displaystyle \int (c+d x) \left (c^2-d^2 x^2\right )^{2/5} \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle c \int \left (c^2-d^2 x^2\right )^{2/5}dx-\frac {5 \left (c^2-d^2 x^2\right )^{7/5}}{14 d}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {c \left (c^2-d^2 x^2\right )^{2/5} \int \left (1-\frac {d^2 x^2}{c^2}\right )^{2/5}dx}{\left (1-\frac {d^2 x^2}{c^2}\right )^{2/5}}-\frac {5 \left (c^2-d^2 x^2\right )^{7/5}}{14 d}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {c x \left (c^2-d^2 x^2\right )^{2/5} \operatorname {Hypergeometric2F1}\left (-\frac {2}{5},\frac {1}{2},\frac {3}{2},\frac {d^2 x^2}{c^2}\right )}{\left (1-\frac {d^2 x^2}{c^2}\right )^{2/5}}-\frac {5 \left (c^2-d^2 x^2\right )^{7/5}}{14 d}\) |
Input:
Int[(c + d*x)*(c^2 - d^2*x^2)^(2/5),x]
Output:
(-5*(c^2 - d^2*x^2)^(7/5))/(14*d) + (c*x*(c^2 - d^2*x^2)^(2/5)*Hypergeomet ric2F1[-2/5, 1/2, 3/2, (d^2*x^2)/c^2])/(1 - (d^2*x^2)/c^2)^(2/5)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[2*p ] && GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) ^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(1 + b*(x^2/a))^p, x], x] / ; FreeQ[{a, b, p}, x] && !IntegerQ[2*p] && !GtQ[a, 0]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
\[\int \left (d x +c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {2}{5}}d x\]
Input:
int((d*x+c)*(-d^2*x^2+c^2)^(2/5),x)
Output:
int((d*x+c)*(-d^2*x^2+c^2)^(2/5),x)
\[ \int (c+d x) \left (c^2-d^2 x^2\right )^{2/5} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{5}} {\left (d x + c\right )} \,d x } \] Input:
integrate((d*x+c)*(-d^2*x^2+c^2)^(2/5),x, algorithm="fricas")
Output:
integral((-d^2*x^2 + c^2)^(2/5)*(d*x + c), x)
Time = 1.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int (c+d x) \left (c^2-d^2 x^2\right )^{2/5} \, dx=c^{\frac {9}{5}} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{5}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {d^{2} x^{2} e^{2 i \pi }}{c^{2}}} \right )} + d \left (\begin {cases} \frac {x^{2} \left (c^{2}\right )^{\frac {2}{5}}}{2} & \text {for}\: d^{2} = 0 \\- \frac {5 \left (c^{2} - d^{2} x^{2}\right )^{\frac {7}{5}}}{14 d^{2}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((d*x+c)*(-d**2*x**2+c**2)**(2/5),x)
Output:
c**(9/5)*x*hyper((-2/5, 1/2), (3/2,), d**2*x**2*exp_polar(2*I*pi)/c**2) + d*Piecewise((x**2*(c**2)**(2/5)/2, Eq(d**2, 0)), (-5*(c**2 - d**2*x**2)**( 7/5)/(14*d**2), True))
\[ \int (c+d x) \left (c^2-d^2 x^2\right )^{2/5} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{5}} {\left (d x + c\right )} \,d x } \] Input:
integrate((d*x+c)*(-d^2*x^2+c^2)^(2/5),x, algorithm="maxima")
Output:
integrate((-d^2*x^2 + c^2)^(2/5)*(d*x + c), x)
\[ \int (c+d x) \left (c^2-d^2 x^2\right )^{2/5} \, dx=\int { {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {2}{5}} {\left (d x + c\right )} \,d x } \] Input:
integrate((d*x+c)*(-d^2*x^2+c^2)^(2/5),x, algorithm="giac")
Output:
integrate((-d^2*x^2 + c^2)^(2/5)*(d*x + c), x)
Time = 6.72 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int (c+d x) \left (c^2-d^2 x^2\right )^{2/5} \, dx=\frac {c\,x\,{\left (c^2-d^2\,x^2\right )}^{2/5}\,{{}}_2{\mathrm {F}}_1\left (-\frac {2}{5},\frac {1}{2};\ \frac {3}{2};\ \frac {d^2\,x^2}{c^2}\right )}{{\left (1-\frac {d^2\,x^2}{c^2}\right )}^{2/5}}-\frac {5\,{\left (c^2-d^2\,x^2\right )}^{7/5}}{14\,d} \] Input:
int((c^2 - d^2*x^2)^(2/5)*(c + d*x),x)
Output:
(c*x*(c^2 - d^2*x^2)^(2/5)*hypergeom([-2/5, 1/2], 3/2, (d^2*x^2)/c^2))/(1 - (d^2*x^2)/c^2)^(2/5) - (5*(c^2 - d^2*x^2)^(7/5))/(14*d)
\[ \int (c+d x) \left (c^2-d^2 x^2\right )^{2/5} \, dx=\frac {56 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{5}} \left (\int \frac {1}{\left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{5}}}d x \right ) c^{3} d -45 c^{4}+70 c^{3} d x +90 c^{2} d^{2} x^{2}-70 c \,d^{3} x^{3}-45 d^{4} x^{4}}{126 \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{5}} d} \] Input:
int((d*x+c)*(-d^2*x^2+c^2)^(2/5),x)
Output:
(56*(c**2 - d**2*x**2)**(3/5)*int((c**2 - d**2*x**2)**(2/5)/(c**2 - d**2*x **2),x)*c**3*d - 45*c**4 + 70*c**3*d*x + 90*c**2*d**2*x**2 - 70*c*d**3*x** 3 - 45*d**4*x**4)/(126*(c**2 - d**2*x**2)**(3/5)*d)