Integrand size = 29, antiderivative size = 139 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\frac {3 c \sqrt {c d^2-c e^2 x^2}}{4 e (d+e x)^{3/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}-\frac {3 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{4 \sqrt {2} \sqrt {d} e} \]
-1/2*(-c*e^2*x^2+c*d^2)^(3/2)/e/(e*x+d)^(7/2)-3/8*c^(3/2)*arctanh(1/2*(-c* e^2*x^2+c*d^2)^(1/2)*2^(1/2)/c^(1/2)/d^(1/2)/(e*x+d)^(1/2))/e*2^(1/2)/d^(1 /2)+3/4*c*(-c*e^2*x^2+c*d^2)^(1/2)/e/(e*x+d)^(3/2)
Time = 0.75 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.78 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\frac {c \sqrt {c \left (d^2-e^2 x^2\right )} \left (\frac {2 (d+5 e x)}{(d+e x)^{5/2}}-\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{\sqrt {d} \sqrt {d^2-e^2 x^2}}\right )}{8 e} \]
(c*Sqrt[c*(d^2 - e^2*x^2)]*((2*(d + 5*e*x))/(d + e*x)^(5/2) - (3*Sqrt[2]*A rcTanh[(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[d^2 - e^2*x^2]])/(Sqrt[d]*Sqrt [d^2 - e^2*x^2])))/(8*e)
Time = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {465, 465, 471, 221}
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.
\(\displaystyle \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx\) |
\(\Big \downarrow \) 465 |
\(\displaystyle -\frac {3}{4} c \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{5/2}}dx-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}\) |
\(\Big \downarrow \) 465 |
\(\displaystyle -\frac {3}{4} c \left (-\frac {1}{2} c \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}dx-\frac {\sqrt {c d^2-c e^2 x^2}}{e (d+e x)^{3/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle -\frac {3}{4} c \left (-c e \int \frac {1}{\frac {e^2 \left (c d^2-c e^2 x^2\right )}{d+e x}-2 c d e^2}d\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}-\frac {\sqrt {c d^2-c e^2 x^2}}{e (d+e x)^{3/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {3}{4} c \left (\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {2} \sqrt {d} e}-\frac {\sqrt {c d^2-c e^2 x^2}}{e (d+e x)^{3/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}\) |
-1/2*(c*d^2 - c*e^2*x^2)^(3/2)/(e*(d + e*x)^(7/2)) - (3*c*(-(Sqrt[c*d^2 - c*e^2*x^2]/(e*(d + e*x)^(3/2))) + (Sqrt[c]*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2] /(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(Sqrt[2]*Sqrt[d]*e)))/4
3.9.76.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + p + 1))), x] - Simp[b*(p/(d^2*(n + p + 1))) Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n + 2*p + 1, 0]) && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim p[2*d Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] ], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
Time = 2.35 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.27
method | result | size |
default | \(-\frac {\sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, c \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,e^{2} x^{2}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c d e x +3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{2}-10 e x \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}-2 \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\, d \right )}{8 \left (e x +d \right )^{\frac {5}{2}} \sqrt {c \left (-e x +d \right )}\, e \sqrt {c d}}\) | \(176\) |
-1/8*(c*(-e^2*x^2+d^2))^(1/2)*c*(3*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)* 2^(1/2)/(c*d)^(1/2))*c*e^2*x^2+6*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^ (1/2)/(c*d)^(1/2))*c*d*e*x+3*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2 )/(c*d)^(1/2))*c*d^2-10*e*x*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)-2*(c*(-e*x+d))^ (1/2)*(c*d)^(1/2)*d)/(e*x+d)^(5/2)/(c*(-e*x+d))^(1/2)/e/(c*d)^(1/2)
Time = 0.40 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.64 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\left [\frac {3 \, \sqrt {\frac {1}{2}} {\left (c e^{3} x^{3} + 3 \, c d e^{2} x^{2} + 3 \, c d^{2} e x + c d^{3}\right )} \sqrt {\frac {c}{d}} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 4 \, \sqrt {\frac {1}{2}} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {\frac {c}{d}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (5 \, c e x + c d\right )} \sqrt {e x + d}}{8 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}}, -\frac {3 \, \sqrt {\frac {1}{2}} {\left (c e^{3} x^{3} + 3 \, c d e^{2} x^{2} + 3 \, c d^{2} e x + c d^{3}\right )} \sqrt {-\frac {c}{d}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {-\frac {c}{d}}}{c e^{2} x^{2} - c d^{2}}\right ) - \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (5 \, c e x + c d\right )} \sqrt {e x + d}}{4 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}}\right ] \]
[1/8*(3*sqrt(1/2)*(c*e^3*x^3 + 3*c*d*e^2*x^2 + 3*c*d^2*e*x + c*d^3)*sqrt(c /d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 + 4*sqrt(1/2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*d*sqrt(c/d))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*sqrt(-c*e ^2*x^2 + c*d^2)*(5*c*e*x + c*d)*sqrt(e*x + d))/(e^4*x^3 + 3*d*e^3*x^2 + 3* d^2*e^2*x + d^3*e), -1/4*(3*sqrt(1/2)*(c*e^3*x^3 + 3*c*d*e^2*x^2 + 3*c*d^2 *e*x + c*d^3)*sqrt(-c/d)*arctan(2*sqrt(1/2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt( e*x + d)*d*sqrt(-c/d)/(c*e^2*x^2 - c*d^2)) - sqrt(-c*e^2*x^2 + c*d^2)*(5*c *e*x + c*d)*sqrt(e*x + d))/(e^4*x^3 + 3*d*e^3*x^2 + 3*d^2*e^2*x + d^3*e)]
\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\int \frac {\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {9}{2}}}\, dx \]
\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\int { \frac {{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}}}{{\left (e x + d\right )}^{\frac {9}{2}}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.73 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\frac {\frac {3 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d}} + \frac {2 \, {\left (6 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{3} d - 5 \, {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c^{2}\right )}}{{\left (e x + d\right )}^{2} c^{2}}}{8 \, e} \]
1/8*(3*sqrt(2)*c^2*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d ))/sqrt(-c*d) + 2*(6*sqrt(-(e*x + d)*c + 2*c*d)*c^3*d - 5*(-(e*x + d)*c + 2*c*d)^(3/2)*c^2)/((e*x + d)^2*c^2))/e
Timed out. \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\int \frac {{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^{9/2}} \,d x \]