Integrand size = 19, antiderivative size = 145 \[ \int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx=-\frac {e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right ) (d+e x)^2}-\frac {3 c d e \sqrt {a+c x^2}}{2 \left (c d^2+a e^2\right )^2 (d+e x)}-\frac {c \left (2 c d^2-a e^2\right ) \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{2 \left (c d^2+a e^2\right )^{5/2}} \]
-1/2*c*(-a*e^2+2*c*d^2)*arctanh((-c*d*x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a) ^(1/2))/(a*e^2+c*d^2)^(5/2)-1/2*e*(c*x^2+a)^(1/2)/(a*e^2+c*d^2)/(e*x+d)^2- 3/2*c*d*e*(c*x^2+a)^(1/2)/(a*e^2+c*d^2)^2/(e*x+d)
Time = 0.97 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx=-\frac {e \sqrt {a+c x^2} \left (a e^2+c d (4 d+3 e x)\right )}{2 \left (c d^2+a e^2\right )^2 (d+e x)^2}-\frac {c \left (2 c d^2-a e^2\right ) \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{\left (-c d^2-a e^2\right )^{5/2}} \]
-1/2*(e*Sqrt[a + c*x^2]*(a*e^2 + c*d*(4*d + 3*e*x)))/((c*d^2 + a*e^2)^2*(d + e*x)^2) - (c*(2*c*d^2 - a*e^2)*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + c *x^2])/Sqrt[-(c*d^2) - a*e^2]])/(-(c*d^2) - a*e^2)^(5/2)
Time = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {498, 25, 679, 488, 219}
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.
\(\displaystyle \int \frac {1}{\sqrt {a+c x^2} (d+e x)^3} \, dx\) |
\(\Big \downarrow \) 498 |
\(\displaystyle -\frac {c \int -\frac {2 d-e x}{(d+e x)^2 \sqrt {c x^2+a}}dx}{2 \left (a e^2+c d^2\right )}-\frac {e \sqrt {a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {c \int \frac {2 d-e x}{(d+e x)^2 \sqrt {c x^2+a}}dx}{2 \left (a e^2+c d^2\right )}-\frac {e \sqrt {a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 679 |
\(\displaystyle \frac {c \left (\frac {\left (2 c d^2-a e^2\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+a}}dx}{a e^2+c d^2}-\frac {3 d e \sqrt {a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}\right )}{2 \left (a e^2+c d^2\right )}-\frac {e \sqrt {a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {c \left (-\frac {\left (2 c d^2-a e^2\right ) \int \frac {1}{c d^2+a e^2-\frac {(a e-c d x)^2}{c x^2+a}}d\frac {a e-c d x}{\sqrt {c x^2+a}}}{a e^2+c d^2}-\frac {3 d e \sqrt {a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}\right )}{2 \left (a e^2+c d^2\right )}-\frac {e \sqrt {a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {c \left (-\frac {\left (2 c d^2-a e^2\right ) \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac {3 d e \sqrt {a+c x^2}}{(d+e x) \left (a e^2+c d^2\right )}\right )}{2 \left (a e^2+c d^2\right )}-\frac {e \sqrt {a+c x^2}}{2 (d+e x)^2 \left (a e^2+c d^2\right )}\) |
-1/2*(e*Sqrt[a + c*x^2])/((c*d^2 + a*e^2)*(d + e*x)^2) + (c*((-3*d*e*Sqrt[ a + c*x^2])/((c*d^2 + a*e^2)*(d + e*x)) - ((2*c*d^2 - a*e^2)*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(c*d^2 + a*e^2)^(3/2)))/( 2*(c*d^2 + a*e^2))
3.6.67.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
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)/((n + 1)*(b*c^2 + a*d^2))), x] + S imp[b/((n + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p*(c*(n + 1) - d*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[n , -1] && ((LtQ[n, -1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]) || (SumSimp lerQ[n, 1] && IntegerQ[p]) || ILtQ[Simplify[n + 2*p + 3], 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 )/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
Leaf count of result is larger than twice the leaf count of optimal. \(443\) vs. \(2(129)=258\).
Time = 2.02 (sec) , antiderivative size = 444, normalized size of antiderivative = 3.06
method | result | size |
default | \(\frac {-\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{2 \left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2}}+\frac {3 c d e \left (-\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{\left (e^{2} a +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}-\frac {c d e \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}\right )}{2 \left (e^{2} a +c \,d^{2}\right )}+\frac {c \,e^{2} \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{2 \left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}}{e^{3}}\) | \(444\) |
1/e^3*(-1/2/(a*e^2+c*d^2)*e^2/(x+d/e)^2*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^ 2+c*d^2)/e^2)^(1/2)+3/2*c*d*e/(a*e^2+c*d^2)*(-1/(a*e^2+c*d^2)*e^2/(x+d/e)* (c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)-c*d*e/(a*e^2+c*d^2)/ ((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(x+d/e)+2*((a*e^ 2+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2-2*c*d/e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2)) /(x+d/e)))+1/2*c/(a*e^2+c*d^2)*e^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+ c*d^2)/e^2-2*c*d/e*(x+d/e)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2-2*c*d/ e*(x+d/e)+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e)))
Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (130) = 260\).
Time = 0.43 (sec) , antiderivative size = 713, normalized size of antiderivative = 4.92 \[ \int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx=\left [-\frac {{\left (2 \, c^{2} d^{4} - a c d^{2} e^{2} + {\left (2 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x^{2} + 2 \, {\left (2 \, c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt {c d^{2} + a e^{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (4 \, c^{2} d^{4} e + 5 \, a c d^{2} e^{3} + a^{2} e^{5} + 3 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{4 \, {\left (c^{3} d^{8} + 3 \, a c^{2} d^{6} e^{2} + 3 \, a^{2} c d^{4} e^{4} + a^{3} d^{2} e^{6} + {\left (c^{3} d^{6} e^{2} + 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} + a^{3} e^{8}\right )} x^{2} + 2 \, {\left (c^{3} d^{7} e + 3 \, a c^{2} d^{5} e^{3} + 3 \, a^{2} c d^{3} e^{5} + a^{3} d e^{7}\right )} x\right )}}, -\frac {{\left (2 \, c^{2} d^{4} - a c d^{2} e^{2} + {\left (2 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} x^{2} + 2 \, {\left (2 \, c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) + {\left (4 \, c^{2} d^{4} e + 5 \, a c d^{2} e^{3} + a^{2} e^{5} + 3 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (c^{3} d^{8} + 3 \, a c^{2} d^{6} e^{2} + 3 \, a^{2} c d^{4} e^{4} + a^{3} d^{2} e^{6} + {\left (c^{3} d^{6} e^{2} + 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} + a^{3} e^{8}\right )} x^{2} + 2 \, {\left (c^{3} d^{7} e + 3 \, a c^{2} d^{5} e^{3} + 3 \, a^{2} c d^{3} e^{5} + a^{3} d e^{7}\right )} x\right )}}\right ] \]
[-1/4*((2*c^2*d^4 - a*c*d^2*e^2 + (2*c^2*d^2*e^2 - a*c*e^4)*x^2 + 2*(2*c^2 *d^3*e - a*c*d*e^3)*x)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2* a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)* sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(4*c^2*d^4*e + 5*a*c*d^2*e ^3 + a^2*e^5 + 3*(c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^8 + 3*a*c^2*d^6*e^2 + 3*a^2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c^2*d ^4*e^4 + 3*a^2*c*d^2*e^6 + a^3*e^8)*x^2 + 2*(c^3*d^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 + a^3*d*e^7)*x), -1/2*((2*c^2*d^4 - a*c*d^2*e^2 + (2*c^2* d^2*e^2 - a*c*e^4)*x^2 + 2*(2*c^2*d^3*e - a*c*d*e^3)*x)*sqrt(-c*d^2 - a*e^ 2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^ 2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) + (4*c^2*d^4*e + 5*a*c*d^2*e^3 + a^2*e^5 + 3*(c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^3*d^8 + 3*a*c^2*d^6* e^2 + 3*a^2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c^2*d^4*e^4 + 3*a ^2*c*d^2*e^6 + a^3*e^8)*x^2 + 2*(c^3*d^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3 *e^5 + a^3*d*e^7)*x)]
\[ \int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx=\int \frac {1}{\sqrt {a + c x^{2}} \left (d + e x\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (130) = 260\).
Time = 0.27 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.42 \[ \int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx=-c {\left (\frac {{\left (2 \, c d^{2} - a e^{2}\right )} \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right )}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c d^{2} - a e^{2}}} + \frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} c d^{2} e - {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} a e^{3} + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} c^{\frac {3}{2}} d^{3} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} a \sqrt {c} d e^{2} - 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a c d^{2} e - {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} a^{2} e^{3} + 3 \, a^{2} \sqrt {c} d e^{2}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} e + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} \sqrt {c} d - a e\right )}^{2}}\right )} \]
-c*((2*c*d^2 - a*e^2)*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d) /sqrt(-c*d^2 - a*e^2))/((c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c*d^2 - a*e^2)) + (2*(sqrt(c)*x - sqrt(c*x^2 + a))^3*c*d^2*e - (sqrt(c)*x - sqrt(c *x^2 + a))^3*a*e^3 + 6*(sqrt(c)*x - sqrt(c*x^2 + a))^2*c^(3/2)*d^3 - 3*(sq rt(c)*x - sqrt(c*x^2 + a))^2*a*sqrt(c)*d*e^2 - 10*(sqrt(c)*x - sqrt(c*x^2 + a))*a*c*d^2*e - (sqrt(c)*x - sqrt(c*x^2 + a))*a^2*e^3 + 3*a^2*sqrt(c)*d* e^2)/((c^2*d^4 + 2*a*c*d^2*e^2 + a^2*e^4)*((sqrt(c)*x - sqrt(c*x^2 + a))^2 *e + 2*(sqrt(c)*x - sqrt(c*x^2 + a))*sqrt(c)*d - a*e)^2))
Timed out. \[ \int \frac {1}{(d+e x)^3 \sqrt {a+c x^2}} \, dx=\int \frac {1}{\sqrt {c\,x^2+a}\,{\left (d+e\,x\right )}^3} \,d x \]