Integrand size = 29, antiderivative size = 152 \[ \int \frac {(d+e x)^{11/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 (d+e x)^{9/2}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}-\frac {4 (d+e x)^{5/2}}{c^2 e \sqrt {c d^2-c e^2 x^2}}-\frac {64 d \sqrt {c d^2-c e^2 x^2}}{3 c^3 e \sqrt {d+e x}}-\frac {16 \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}{3 c^3 e} \] Output:
2/3*(e*x+d)^(9/2)/c/e/(-c*e^2*x^2+c*d^2)^(3/2)-4*(e*x+d)^(5/2)/c^2/e/(-c*e ^2*x^2+c*d^2)^(1/2)-64/3*d*(-c*e^2*x^2+c*d^2)^(1/2)/c^3/e/(e*x+d)^(1/2)-16 /3*(e*x+d)^(1/2)*(-c*e^2*x^2+c*d^2)^(1/2)/c^3/e
Time = 0.40 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.49 \[ \int \frac {(d+e x)^{11/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {d+e x} \left (45 d^3-69 d^2 e x+15 d e^2 x^2+e^3 x^3\right )}{3 c^2 e (-d+e x) \sqrt {c \left (d^2-e^2 x^2\right )}} \] Input:
Integrate[(d + e*x)^(11/2)/(c*d^2 - c*e^2*x^2)^(5/2),x]
Output:
(2*Sqrt[d + e*x]*(45*d^3 - 69*d^2*e*x + 15*d*e^2*x^2 + e^3*x^3))/(3*c^2*e* (-d + e*x)*Sqrt[c*(d^2 - e^2*x^2)])
Time = 0.42 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {459, 459, 459, 458}
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.
\(\displaystyle \int \frac {(d+e x)^{11/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 459 |
\(\displaystyle 4 d \int \frac {(d+e x)^{9/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}}dx-\frac {2 (d+e x)^{9/2}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle 4 d \left (8 d \int \frac {(d+e x)^{7/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}}dx-\frac {2 (d+e x)^{7/2}}{c e \left (c d^2-c e^2 x^2\right )^{3/2}}\right )-\frac {2 (d+e x)^{9/2}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 459 |
\(\displaystyle 4 d \left (8 d \left (\frac {2 (d+e x)^{5/2}}{c e \left (c d^2-c e^2 x^2\right )^{3/2}}-4 d \int \frac {(d+e x)^{5/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}}dx\right )-\frac {2 (d+e x)^{7/2}}{c e \left (c d^2-c e^2 x^2\right )^{3/2}}\right )-\frac {2 (d+e x)^{9/2}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle 4 d \left (8 d \left (\frac {2 (d+e x)^{5/2}}{c e \left (c d^2-c e^2 x^2\right )^{3/2}}-\frac {8 d (d+e x)^{3/2}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\right )-\frac {2 (d+e x)^{7/2}}{c e \left (c d^2-c e^2 x^2\right )^{3/2}}\right )-\frac {2 (d+e x)^{9/2}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\) |
Input:
Int[(d + e*x)^(11/2)/(c*d^2 - c*e^2*x^2)^(5/2),x]
Output:
(-2*(d + e*x)^(9/2))/(3*c*e*(c*d^2 - c*e^2*x^2)^(3/2)) + 4*d*((-2*(d + e*x )^(7/2))/(c*e*(c*d^2 - c*e^2*x^2)^(3/2)) + 8*d*((-8*d*(d + e*x)^(3/2))/(3* c*e*(c*d^2 - c*e^2*x^2)^(3/2)) + (2*(d + e*x)^(5/2))/(c*e*(c*d^2 - c*e^2*x ^2)^(3/2))))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c , d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* (Simplify[n + p]/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[Simplif y[n + p], 0]
Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.43
method | result | size |
gosper | \(-\frac {2 \left (-e x +d \right ) \left (e^{3} x^{3}+15 d \,e^{2} x^{2}-69 d^{2} e x +45 d^{3}\right ) \left (e x +d \right )^{\frac {5}{2}}}{3 e \left (-c \,e^{2} x^{2}+c \,d^{2}\right )^{\frac {5}{2}}}\) | \(65\) |
orering | \(-\frac {2 \left (-e x +d \right ) \left (e^{3} x^{3}+15 d \,e^{2} x^{2}-69 d^{2} e x +45 d^{3}\right ) \left (e x +d \right )^{\frac {5}{2}}}{3 e \left (-c \,e^{2} x^{2}+c \,d^{2}\right )^{\frac {5}{2}}}\) | \(65\) |
default | \(-\frac {2 \sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, \left (e^{3} x^{3}+15 d \,e^{2} x^{2}-69 d^{2} e x +45 d^{3}\right )}{3 \sqrt {e x +d}\, c^{3} \left (-e x +d \right )^{2} e}\) | \(69\) |
risch | \(-\frac {2 \left (e x +17 d \right ) \left (-e x +d \right ) \sqrt {-\frac {c \left (e^{2} x^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}}{3 e \sqrt {-c \left (e x -d \right )}\, \sqrt {-c \left (e^{2} x^{2}-d^{2}\right )}\, c^{2}}-\frac {8 d^{2} \left (-9 e x +7 d \right ) \sqrt {-\frac {c \left (e^{2} x^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}}{3 e \sqrt {c \left (-e x +d \right )}\, \left (-e x +d \right ) \sqrt {-c \left (e^{2} x^{2}-d^{2}\right )}\, c^{2}}\) | \(172\) |
Input:
int((e*x+d)^(11/2)/(-c*e^2*x^2+c*d^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*(-e*x+d)*(e^3*x^3+15*d*e^2*x^2-69*d^2*e*x+45*d^3)*(e*x+d)^(5/2)/e/(-c *e^2*x^2+c*d^2)^(5/2)
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.66 \[ \int \frac {(d+e x)^{11/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (e^{3} x^{3} + 15 \, d e^{2} x^{2} - 69 \, d^{2} e x + 45 \, d^{3}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{3 \, {\left (c^{3} e^{4} x^{3} - c^{3} d e^{3} x^{2} - c^{3} d^{2} e^{2} x + c^{3} d^{3} e\right )}} \] Input:
integrate((e*x+d)^(11/2)/(-c*e^2*x^2+c*d^2)^(5/2),x, algorithm="fricas")
Output:
-2/3*(e^3*x^3 + 15*d*e^2*x^2 - 69*d^2*e*x + 45*d^3)*sqrt(-c*e^2*x^2 + c*d^ 2)*sqrt(e*x + d)/(c^3*e^4*x^3 - c^3*d*e^3*x^2 - c^3*d^2*e^2*x + c^3*d^3*e)
\[ \int \frac {(d+e x)^{11/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {11}{2}}}{\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((e*x+d)**(11/2)/(-c*e**2*x**2+c*d**2)**(5/2),x)
Output:
Integral((d + e*x)**(11/2)/(-c*(-d + e*x)*(d + e*x))**(5/2), x)
Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.38 \[ \int \frac {(d+e x)^{11/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (e^{3} x^{3} + 15 \, d e^{2} x^{2} - 69 \, d^{2} e x + 45 \, d^{3}\right )}}{3 \, {\left (c^{\frac {5}{2}} e^{2} x - c^{\frac {5}{2}} d e\right )} \sqrt {-e x + d}} \] Input:
integrate((e*x+d)^(11/2)/(-c*e^2*x^2+c*d^2)^(5/2),x, algorithm="maxima")
Output:
2/3*(e^3*x^3 + 15*d*e^2*x^2 - 69*d^2*e*x + 45*d^3)/((c^(5/2)*e^2*x - c^(5/ 2)*d*e)*sqrt(-e*x + d))
Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.72 \[ \int \frac {(d+e x)^{11/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (\frac {4 \, {\left (2 \, c d^{3} + 9 \, {\left ({\left (e x + d\right )} c - 2 \, c d\right )} d^{2}\right )}}{{\left ({\left (e x + d\right )} c - 2 \, c d\right )} \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{2}} + \frac {18 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{9} d - {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c^{8}}{c^{12}}\right )}}{3 \, e} \] Input:
integrate((e*x+d)^(11/2)/(-c*e^2*x^2+c*d^2)^(5/2),x, algorithm="giac")
Output:
-2/3*(4*(2*c*d^3 + 9*((e*x + d)*c - 2*c*d)*d^2)/(((e*x + d)*c - 2*c*d)*sqr t(-(e*x + d)*c + 2*c*d)*c^2) + (18*sqrt(-(e*x + d)*c + 2*c*d)*c^9*d - (-(e *x + d)*c + 2*c*d)^(3/2)*c^8)/c^12)/e
Time = 6.60 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.82 \[ \int \frac {(d+e x)^{11/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c\,d^2-c\,e^2\,x^2}\,\left (\frac {30\,d^3\,\sqrt {d+e\,x}}{c^3\,e^4}+\frac {2\,x^3\,\sqrt {d+e\,x}}{3\,c^3\,e}+\frac {10\,d\,x^2\,\sqrt {d+e\,x}}{c^3\,e^2}-\frac {46\,d^2\,x\,\sqrt {d+e\,x}}{c^3\,e^3}\right )}{x^3+\frac {d^3}{e^3}-\frac {d\,x^2}{e}-\frac {d^2\,x}{e^2}} \] Input:
int((d + e*x)^(11/2)/(c*d^2 - c*e^2*x^2)^(5/2),x)
Output:
-((c*d^2 - c*e^2*x^2)^(1/2)*((30*d^3*(d + e*x)^(1/2))/(c^3*e^4) + (2*x^3*( d + e*x)^(1/2))/(3*c^3*e) + (10*d*x^2*(d + e*x)^(1/2))/(c^3*e^2) - (46*d^2 *x*(d + e*x)^(1/2))/(c^3*e^3)))/(x^3 + d^3/e^3 - (d*x^2)/e - (d^2*x)/e^2)
Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.38 \[ \int \frac {(d+e x)^{11/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {c}\, \left (-e^{3} x^{3}-15 d \,e^{2} x^{2}+69 d^{2} e x -45 d^{3}\right )}{3 \sqrt {-e x +d}\, c^{3} e \left (-e x +d \right )} \] Input:
int((e*x+d)^(11/2)/(-c*e^2*x^2+c*d^2)^(5/2),x)
Output:
(2*sqrt(c)*( - 45*d**3 + 69*d**2*e*x - 15*d*e**2*x**2 - e**3*x**3))/(3*sqr t(d - e*x)*c**3*e*(d - e*x))