Integrand size = 28, antiderivative size = 330 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3} \, dx=-\frac {d^2 e-b d f^2+a e f^2}{\left (2 d e-b f^2\right )^2 \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2}-\frac {2 f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^3 \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )}-\frac {2 e f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^3 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac {6 e f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )}{\left (2 d e-b f^2\right )^4}-\frac {6 e f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}{\left (2 d e-b f^2\right )^4} \]
6*e*f^2*(-b^2*f^2+4*a*e^2)*ln(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))/(-b*f^2+2 *d*e)^4-6*e*f^2*(-b^2*f^2+4*a*e^2)*ln(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/ f^2)^(1/2)))/(-b*f^2+2*d*e)^4+(-a*e*f^2+b*d*f^2-d^2*e)/(-b*f^2+2*d*e)^2/(d +e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^2-2*f^2*(-b^2*f^2+4*a*e^2)/(-b*f^2+2*d*e )^3/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))-2*e*f^2*(-b^2*f^2+4*a*e^2)/(-b*f^2 +2*d*e)^3/(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/f^2)^(1/2)))
Time = 10.51 (sec) , antiderivative size = 300, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3} \, dx=-\frac {\frac {\left (-2 d e+b f^2\right )^2 \left (d^2 e-b d f^2+a e f^2\right )}{\left (d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )^2}+\frac {2 f^2 \left (-2 d e+b f^2\right ) \left (-4 a e^2+b^2 f^2\right )}{d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}+\frac {2 e f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right )}{b f^2+2 e \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )}-6 e f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+6 e f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (-b f^2-2 e \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}{\left (-2 d e+b f^2\right )^4} \]
-((((-2*d*e + b*f^2)^2*(d^2*e - b*d*f^2 + a*e*f^2))/(d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])^2 + (2*f^2*(-2*d*e + b*f^2)*(-4*a*e^2 + b^2*f^2))/(d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + (2*e*f^2*(2*d*e - b*f^2)*(4*a* e^2 - b^2*f^2))/(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])) - 6* e*f^2*(4*a*e^2 - b^2*f^2)*Log[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]] + 6*e*f^2*(4*a*e^2 - b^2*f^2)*Log[-(b*f^2) - 2*e*(e*x + f*Sqrt[a + x*(b + ( e^2*x)/f^2)])])/(-2*d*e + b*f^2)^4)
Time = 0.50 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2541, 1195, 2009}
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.
\(\displaystyle \int \frac {1}{\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^3} \, dx\) |
\(\Big \downarrow \) 2541 |
\(\displaystyle 2 \int \frac {e d^2-b f^2 d+a e f^2+e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2-\left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}{\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^3 \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle 2 \int \left (\frac {e d^2-b f^2 d+a e f^2}{\left (2 d e-b f^2\right )^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^3}+\frac {3 \left (4 a e^3 f^2-b^2 e f^4\right )}{\left (2 d e-b f^2\right )^4 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}+\frac {6 \left (4 a e^4 f^2-b^2 e^2 f^4\right )}{\left (2 d e-b f^2\right )^4 \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}+\frac {4 a e^2 f^2-b^2 f^4}{\left (2 d e-b f^2\right )^3 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2}+\frac {2 \left (4 a e^4 f^2-b^2 e^2 f^4\right )}{\left (2 d e-b f^2\right )^3 \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}\right )d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (-\frac {f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^3 \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )}+\frac {e f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^3 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}+\frac {3 e f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )}{\left (2 d e-b f^2\right )^4}-\frac {3 e f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}{\left (2 d e-b f^2\right )^4}-\frac {a e f^2-b d f^2+d^2 e}{2 \left (2 d e-b f^2\right )^2 \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^2}\right )\) |
2*(-1/2*(d^2*e - b*d*f^2 + a*e*f^2)/((2*d*e - b*f^2)^2*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^2) - (f^2*(4*a*e^2 - b^2*f^2))/((2*d*e - b*f^2)^3 *(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])) + (e*f^2*(4*a*e^2 - b^2*f^2) )/((2*d*e - b*f^2)^3*(2*d*e - b*f^2 - 2*e*(d + e*x + f*Sqrt[a + b*x + (e^2 *x^2)/f^2]))) + (3*e*f^2*(4*a*e^2 - b^2*f^2)*Log[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/(2*d*e - b*f^2)^4 - (3*e*f^2*(4*a*e^2 - b^2*f^2)*Log[2* d*e - b*f^2 - 2*e*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])])/(2*d*e - b *f^2)^4)
3.5.78.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c _.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Simp[2 Subst[Int[(g + h*x^n)^p*((d ^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2*e*x )^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e , f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(29132\) vs. \(2(320)=640\).
Time = 0.29 (sec) , antiderivative size = 29133, normalized size of antiderivative = 88.28
Leaf count of result is larger than twice the leaf count of optimal. 1954 vs. \(2 (311) = 622\).
Time = 12.00 (sec) , antiderivative size = 1954, normalized size of antiderivative = 5.92 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3} \, dx=\text {Too large to display} \]
((3*a*b^3*d - 4*a^2*b^2*e)*f^8 - (b^3*d^3 + 4*a*b^2*d^2*e + 10*a^2*b*d*e^2 - 20*a^3*e^3)*f^6 - 4*(b^2*d^4*e - 8*a*b*d^3*e^2 + 6*a^2*d^2*e^3)*f^4 - 4 *(b^3*e^3*f^6 - 6*b^2*d*e^4*f^4 + 12*b*d^2*e^5*f^2 - 8*d^3*e^6)*x^3 + 2*(b *d^5*e^2 - 6*a*d^4*e^3)*f^2 - (b^4*e*f^8 - 2*a*b^2*e^3*f^6 - 40*d^4*e^5 - 2*(11*b^2*d^2*e^3 - 4*a*b*d*e^4)*f^4 + 8*(7*b*d^3*e^4 - a*d^2*e^5)*f^2)*x^ 2 + (16*d^5*e^4 + (3*b^4*d - 5*a*b^3*e)*f^8 - (7*b^3*d^2*e + 10*a*b^2*d*e^ 2 - 28*a^2*b*e^3)*f^6 + 2*(5*b^2*d^3*e^2 + 22*a*b*d^2*e^3 - 28*a^2*d*e^4)* f^4 - 8*(3*b*d^4*e^3 + a*d^3*e^4)*f^2)*x - 3*(a^2*b^2*e*f^8 - 4*a*d^4*e^3* f^2 - 2*(a*b^2*d^2*e + 2*a^3*e^3)*f^6 + (b^2*d^4*e + 8*a^2*d^2*e^3)*f^4 + (b^4*e*f^8 - 16*a*d^2*e^5*f^2 - 4*(b^3*d*e^2 + a*b^2*e^3)*f^6 + 4*(b^2*d^2 *e^3 + 4*a*b*d*e^4)*f^4)*x^2 + 2*(a*b^3*e*f^8 - 8*a*d^3*e^4*f^2 - (b^3*d^2 *e + 2*a*b^2*d*e^2 + 4*a^2*b*e^3)*f^6 + 2*(b^2*d^3*e^2 + 2*a*b*d^2*e^3 + 4 *a^2*d*e^4)*f^4)*x)*log(-4*a*d*e^2*f^2 - (b^2*d - 4*a*b*e)*f^4 + 4*(b*e^3* f^2 - 2*d*e^4)*x^2 + (3*b^2*e*f^4 - 4*(2*b*d*e^2 - a*e^3)*f^2)*x - (b^2*f^ 5 - 4*(b*d*e - a*e^2)*f^3 + 4*(b*e^2*f^3 - 2*d*e^3*f)*x)*sqrt((b*f^2*x + e ^2*x^2 + a*f^2)/f^2)) - 3*(a^2*b^2*e*f^8 - 4*a*d^4*e^3*f^2 - 2*(a*b^2*d^2* e + 2*a^3*e^3)*f^6 + (b^2*d^4*e + 8*a^2*d^2*e^3)*f^4 + (b^4*e*f^8 - 16*a*d ^2*e^5*f^2 - 4*(b^3*d*e^2 + a*b^2*e^3)*f^6 + 4*(b^2*d^2*e^3 + 4*a*b*d*e^4) *f^4)*x^2 + 2*(a*b^3*e*f^8 - 8*a*d^3*e^4*f^2 - (b^3*d^2*e + 2*a*b^2*d*e^2 + 4*a^2*b*e^3)*f^6 + 2*(b^2*d^3*e^2 + 2*a*b*d^2*e^3 + 4*a^2*d*e^4)*f^4)...
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3} \, dx=\int \frac {1}{\left (d + e x + f \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}}\right )^{3}}\, dx \]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3} \, dx=\int { \frac {1}{{\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^3} \, dx=\int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^3} \,d x \]