3.477 \(\int \frac {1}{(d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}})^2} \, dx\)

Optimal. Leaf size=266 \[ \frac {2 f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )}{\left (2 d e-b f^2\right )^3}-\frac {f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^2 \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac {2 f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}{\left (2 d e-b f^2\right )^3}-\frac {2 \left (a e f^2-b d f^2+d^2 e\right )}{\left (2 d e-b f^2\right )^2 \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )} \]

[Out]

2*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)^3-2*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)^3-2*(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))-f^2*(-b^2*f^2+4*a*e^2)/(-b*f^2+2*d*e)^2/(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/
f^2)^(1/2)))

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Rubi [A]  time = 0.23, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2116, 893} \[ \frac {2 f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )}{\left (2 d e-b f^2\right )^3}-\frac {f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^2 \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac {2 f^2 \left (4 a e^2-b^2 f^2\right ) \log \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}{\left (2 d e-b f^2\right )^3}-\frac {2 \left (a e f^2-b d f^2+d^2 e\right )}{\left (2 d e-b f^2\right )^2 \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(-2),x]

[Out]

(-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])) - (f^2*(4*a*e
^2 - b^2*f^2))/((2*d*e - b*f^2)^2*(b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) + (2*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)^3 - (2*f^2*(4*a*e^2 - b^2*f^2)*Log
[b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2])])/(2*d*e - b*f^2)^3

Rule 893

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}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2116

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol]
:> Dist[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]

Rubi steps

\begin {align*} \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^2} \, dx &=2 \operatorname {Subst}\left (\int \frac {d^2 e-(b d-a e) f^2-\left (2 d e-b f^2\right ) x+e x^2}{x^2 \left (-2 d e+b f^2+2 e x\right )^2} \, dx,x,d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {d^2 e-b d f^2+a e f^2}{\left (2 d e-b f^2\right )^2 x^2}+\frac {4 a e^2 f^2-b^2 f^4}{\left (2 d e-b f^2\right )^3 x}+\frac {4 a e^3 f^2-b^2 e f^4}{\left (2 d e-b f^2\right )^2 \left (2 d e-b f^2-2 e x\right )^2}+\frac {2 \left (4 a e^3 f^2-b^2 e f^4\right )}{\left (2 d e-b f^2\right )^3 \left (2 d e-b f^2-2 e x\right )}\right ) \, dx,x,d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )\\ &=-\frac {2 \left (d^2 e-b d f^2+a e f^2\right )}{\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 )}-\frac {f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^2 \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 {2 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 )^3}-\frac {2 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 )^3}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 237, normalized size = 0.89 \[ -\frac {\frac {f^2 \left (b^2 f^2-4 a e^2\right ) \left (b f^2-2 d e\right )}{2 e \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+e x\right )+b f^2}+2 f^2 \left (b^2 f^2-4 a e^2\right ) \log \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x\right )-2 f^2 \left (b^2 f^2-4 a e^2\right ) \log \left (-2 e \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+e x\right )-b f^2\right )+\frac {2 \left (2 d e-b f^2\right ) \left (a e f^2-b d f^2+d^2 e\right )}{f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x}}{\left (2 d e-b f^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(-2),x]

[Out]

-(((2*(2*d*e - b*f^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)]) + (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)])) + 2*f^2*(-4*a*e^2 + b^2*f
^2)*Log[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]] - 2*f^2*(-4*a*e^2 + b^2*f^2)*Log[-(b*f^2) - 2*e*(e*x + f*Sq
rt[a + x*(b + (e^2*x)/f^2)])])/(2*d*e - b*f^2)^3)

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fricas [B]  time = 2.56, size = 826, normalized size = 3.11 \[ -\frac {a b^{2} f^{6} + {\left (3 \, b^{2} d^{2} - 14 \, a b d e + 8 \, a^{2} e^{2}\right )} f^{4} - 2 \, {\left (b d^{3} e - 4 \, a d^{2} e^{2}\right )} f^{2} - 4 \, {\left (b^{2} e^{2} f^{4} - 4 \, b d e^{3} f^{2} + 4 \, d^{2} e^{4}\right )} x^{2} + {\left (b^{3} f^{6} - 8 \, b^{2} d e f^{4} + 20 \, b d^{2} e^{2} f^{2} - 16 \, d^{3} e^{3}\right )} x - 2 \, {\left (a b^{2} f^{6} + 4 \, a d^{2} e^{2} f^{2} - {\left (b^{2} d^{2} + 4 \, a^{2} e^{2}\right )} f^{4} + {\left (b^{3} f^{6} + 8 \, a d e^{3} f^{2} - 2 \, {\left (b^{2} d e + 2 \, a b e^{2}\right )} f^{4}\right )} x\right )} \log \left (-4 \, a d e^{2} f^{2} - {\left (b^{2} d - 4 \, a b e\right )} f^{4} + 4 \, {\left (b e^{3} f^{2} - 2 \, d e^{4}\right )} x^{2} + {\left (3 \, b^{2} e f^{4} - 4 \, {\left (2 \, b d e^{2} - a e^{3}\right )} f^{2}\right )} x - {\left (b^{2} f^{5} - 4 \, {\left (b d e - a e^{2}\right )} f^{3} + 4 \, {\left (b e^{2} f^{3} - 2 \, d e^{3} f\right )} x\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) - 2 \, {\left (a b^{2} f^{6} + 4 \, a d^{2} e^{2} f^{2} - {\left (b^{2} d^{2} + 4 \, a^{2} e^{2}\right )} f^{4} + {\left (b^{3} f^{6} + 8 \, a d e^{3} f^{2} - 2 \, {\left (b^{2} d e + 2 \, a b e^{2}\right )} f^{4}\right )} x\right )} \log \left (a f^{2} - d^{2} + {\left (b f^{2} - 2 \, d e\right )} x\right ) + 2 \, {\left (a b^{2} f^{6} + 4 \, a d^{2} e^{2} f^{2} - {\left (b^{2} d^{2} + 4 \, a^{2} e^{2}\right )} f^{4} + {\left (b^{3} f^{6} + 8 \, a d e^{3} f^{2} - 2 \, {\left (b^{2} d e + 2 \, a b e^{2}\right )} f^{4}\right )} x\right )} \log \left (-e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - d\right ) - 4 \, {\left ({\left (b^{2} d - 2 \, a b e\right )} f^{5} - 2 \, {\left (b d^{2} e - 2 \, a d e^{2}\right )} f^{3} - {\left (b^{2} e f^{5} - 4 \, b d e^{2} f^{3} + 4 \, d^{2} e^{3} f\right )} x\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}}{2 \, {\left (a b^{3} f^{8} + 8 \, d^{5} e^{3} - {\left (b^{3} d^{2} + 6 \, a b^{2} d e\right )} f^{6} + 6 \, {\left (b^{2} d^{3} e + 2 \, a b d^{2} e^{2}\right )} f^{4} - 4 \, {\left (3 \, b d^{4} e^{2} + 2 \, a d^{3} e^{3}\right )} f^{2} + {\left (b^{4} f^{8} - 8 \, b^{3} d e f^{6} + 24 \, b^{2} d^{2} e^{2} f^{4} - 32 \, b d^{3} e^{3} f^{2} + 16 \, d^{4} e^{4}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^2,x, algorithm="fricas")

[Out]

-1/2*(a*b^2*f^6 + (3*b^2*d^2 - 14*a*b*d*e + 8*a^2*e^2)*f^4 - 2*(b*d^3*e - 4*a*d^2*e^2)*f^2 - 4*(b^2*e^2*f^4 -
4*b*d*e^3*f^2 + 4*d^2*e^4)*x^2 + (b^3*f^6 - 8*b^2*d*e*f^4 + 20*b*d^2*e^2*f^2 - 16*d^3*e^3)*x - 2*(a*b^2*f^6 +
4*a*d^2*e^2*f^2 - (b^2*d^2 + 4*a^2*e^2)*f^4 + (b^3*f^6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^2)*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)) -
2*(a*b^2*f^6 + 4*a*d^2*e^2*f^2 - (b^2*d^2 + 4*a^2*e^2)*f^4 + (b^3*f^6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^
2)*f^4)*x)*log(a*f^2 - d^2 + (b*f^2 - 2*d*e)*x) + 2*(a*b^2*f^6 + 4*a*d^2*e^2*f^2 - (b^2*d^2 + 4*a^2*e^2)*f^4 +
(b^3*f^6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^2)*f^4)*x)*log(-e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2)
- d) - 4*((b^2*d - 2*a*b*e)*f^5 - 2*(b*d^2*e - 2*a*d*e^2)*f^3 - (b^2*e*f^5 - 4*b*d*e^2*f^3 + 4*d^2*e^3*f)*x)*
sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))/(a*b^3*f^8 + 8*d^5*e^3 - (b^3*d^2 + 6*a*b^2*d*e)*f^6 + 6*(b^2*d^3*e + 2
*a*b*d^2*e^2)*f^4 - 4*(3*b*d^4*e^2 + 2*a*d^3*e^3)*f^2 + (b^4*f^8 - 8*b^3*d*e*f^6 + 24*b^2*d^2*e^2*f^4 - 32*b*d
^3*e^3*f^2 + 16*d^4*e^4)*x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.05, size = 58067, normalized size = 218.30 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d+(b*x+e^2/f^2*x^2+a)^(1/2)*f)^2,x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^2,x, algorithm="maxima")

[Out]

integrate((e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d)^(-2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^2,x)

[Out]

int(1/(d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x + f \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+b*x+e**2*x**2/f**2)**(1/2))**2,x)

[Out]

Integral((d + e*x + f*sqrt(a + b*x + e**2*x**2/f**2))**(-2), x)

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