Integrand size = 29, antiderivative size = 178 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\frac {c \sqrt {c d^2-c e^2 x^2}}{4 e (d+e x)^{5/2}}-\frac {c \sqrt {c d^2-c e^2 x^2}}{16 d e (d+e x)^{3/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}-\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{16 \sqrt {2} d^{3/2} e} \]
-1/3*(-c*e^2*x^2+c*d^2)^(3/2)/e/(e*x+d)^(9/2)-1/32*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))/d^(3/2)/e*2^( 1/2)+1/4*c*(-c*e^2*x^2+c*d^2)^(1/2)/e/(e*x+d)^(5/2)-1/16*c*(-c*e^2*x^2+c*d ^2)^(1/2)/d/e/(e*x+d)^(3/2)
Time = 0.79 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.85 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=-\frac {\left (c \left (d^2-e^2 x^2\right )\right )^{3/2} \left (2 \sqrt {d} \sqrt {d^2-e^2 x^2} \left (7 d^2-22 d e x+3 e^2 x^2\right )+3 \sqrt {2} (d+e x)^{7/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )\right )}{96 d^{3/2} e (d+e x)^{7/2} \left (d^2-e^2 x^2\right )^{3/2}} \]
-1/96*((c*(d^2 - e^2*x^2))^(3/2)*(2*Sqrt[d]*Sqrt[d^2 - e^2*x^2]*(7*d^2 - 2 2*d*e*x + 3*e^2*x^2) + 3*Sqrt[2]*(d + e*x)^(7/2)*ArcTanh[(Sqrt[2]*Sqrt[d]* Sqrt[d + e*x])/Sqrt[d^2 - e^2*x^2]]))/(d^(3/2)*e*(d + e*x)^(7/2)*(d^2 - e^ 2*x^2)^(3/2))
Time = 0.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {465, 465, 470, 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.
| \(\displaystyle \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx\) |
| \(\Big \downarrow \) 465 |
| \(\displaystyle -\frac {1}{2} c \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{7/2}}dx-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}\) |
| \(\Big \downarrow \) 465 |
| \(\displaystyle -\frac {1}{2} c \left (-\frac {1}{4} c \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}dx-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}\) |
| \(\Big \downarrow \) 470 |
| \(\displaystyle -\frac {1}{2} c \left (-\frac {1}{4} c \left (\frac {\int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}dx}{4 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\right )-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}\) |
| \(\Big \downarrow \) 471 |
| \(\displaystyle -\frac {1}{2} c \left (-\frac {1}{4} c \left (\frac {e \int \frac {1}{\frac {e^2 \left (c d^2-c e^2 x^2\right )}{d+e x}-2 c d e^2}d\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}}{2 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\right )-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {1}{2} c \left (-\frac {1}{4} c \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{2 \sqrt {2} \sqrt {c} d^{3/2} e}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\right )-\frac {\sqrt {c d^2-c e^2 x^2}}{2 e (d+e x)^{5/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{9/2}}\) |
-1/3*(c*d^2 - c*e^2*x^2)^(3/2)/(e*(d + e*x)^(9/2)) - (c*(-1/2*Sqrt[c*d^2 - c*e^2*x^2]/(e*(d + e*x)^(5/2)) - (c*(-1/2*Sqrt[c*d^2 - c*e^2*x^2]/(c*d*e* (d + e*x)^(3/2)) - ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d ]*Sqrt[d + e*x])]/(2*Sqrt[2]*Sqrt[c]*d^(3/2)*e)))/4))/2
3.9.77.3.1 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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 2*p + 2)/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 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 = 2.36 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.35
| method | result | size |
| default | \(-\frac {\sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, c \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,e^{3} x^{3}+9 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c d \,e^{2} x^{2}+9 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{2} e x +3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{3}+6 e^{2} x^{2} \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}-44 d e x \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}+14 \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\, d^{2}\right )}{96 \left (e x +d \right )^{\frac {7}{2}} \sqrt {c \left (-e x +d \right )}\, e d \sqrt {c d}}\) | \(241\) |
-1/96*(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^3*x^3+9*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2 ^(1/2)/(c*d)^(1/2))*c*d*e^2*x^2+9*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2 ^(1/2)/(c*d)^(1/2))*c*d^2*e*x+3*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^( 1/2)/(c*d)^(1/2))*c*d^3+6*e^2*x^2*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)-44*d*e*x* (c*(-e*x+d))^(1/2)*(c*d)^(1/2)+14*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)*d^2)/(e*x +d)^(7/2)/(c*(-e*x+d))^(1/2)/e/d/(c*d)^(1/2)
Time = 0.45 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.49 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\left [\frac {3 \, \sqrt {\frac {1}{2}} {\left (c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + 6 \, c d^{2} e^{2} x^{2} + 4 \, c d^{3} e x + c d^{4}\right )} \sqrt {\frac {c}{d}} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 4 \, \sqrt {\frac {1}{2}} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {\frac {c}{d}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, {\left (3 \, c e^{2} x^{2} - 22 \, c d e x + 7 \, c d^{2}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{96 \, {\left (d e^{5} x^{4} + 4 \, d^{2} e^{4} x^{3} + 6 \, d^{3} e^{3} x^{2} + 4 \, d^{4} e^{2} x + d^{5} e\right )}}, -\frac {3 \, \sqrt {\frac {1}{2}} {\left (c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + 6 \, c d^{2} e^{2} x^{2} + 4 \, c d^{3} e x + c d^{4}\right )} \sqrt {-\frac {c}{d}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {-\frac {c}{d}}}{c e^{2} x^{2} - c d^{2}}\right ) + {\left (3 \, c e^{2} x^{2} - 22 \, c d e x + 7 \, c d^{2}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{48 \, {\left (d e^{5} x^{4} + 4 \, d^{2} e^{4} x^{3} + 6 \, d^{3} e^{3} x^{2} + 4 \, d^{4} e^{2} x + d^{5} e\right )}}\right ] \]
[1/96*(3*sqrt(1/2)*(c*e^4*x^4 + 4*c*d*e^3*x^3 + 6*c*d^2*e^2*x^2 + 4*c*d^3* e*x + c*d^4)*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*(3*c*e^2*x^2 - 22*c*d*e*x + 7*c*d^2)*sqrt(-c*e^2*x^2 + c*d^2)*sq rt(e*x + d))/(d*e^5*x^4 + 4*d^2*e^4*x^3 + 6*d^3*e^3*x^2 + 4*d^4*e^2*x + d^ 5*e), -1/48*(3*sqrt(1/2)*(c*e^4*x^4 + 4*c*d*e^3*x^3 + 6*c*d^2*e^2*x^2 + 4* c*d^3*e*x + c*d^4)*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)) + (3*c*e^2*x^2 - 22*c*d*e* x + 7*c*d^2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d))/(d*e^5*x^4 + 4*d^2*e^ 4*x^3 + 6*d^3*e^3*x^2 + 4*d^4*e^2*x + d^5*e)]
\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\int \frac {\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {11}{2}}}\, dx \]
\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\int { \frac {{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}}}{{\left (e x + d\right )}^{\frac {11}{2}}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\frac {\frac {3 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d} d} + \frac {2 \, {\left (12 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{4} d^{2} - 16 \, {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c^{3} d - 3 \, {\left ({\left (e x + d\right )} c - 2 \, c d\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{2}\right )}}{{\left (e x + d\right )}^{3} c^{3} d}}{96 \, e} \]
1/96*(3*sqrt(2)*c^2*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c* d))/(sqrt(-c*d)*d) + 2*(12*sqrt(-(e*x + d)*c + 2*c*d)*c^4*d^2 - 16*(-(e*x + d)*c + 2*c*d)^(3/2)*c^3*d - 3*((e*x + d)*c - 2*c*d)^2*sqrt(-(e*x + d)*c + 2*c*d)*c^2)/((e*x + d)^3*c^3*d))/e
Timed out. \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx=\int \frac {{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^{11/2}} \,d x \]