3.3.35 \(\int \frac {1}{(d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}})^3} \, dx\)

Optimal. Leaf size=193 \[ -\frac {3 a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}{2 d^4 e}+\frac {3 a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )}{2 d^4 e}-\frac {a f^2}{2 d^3 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}-\frac {a f^2}{d^3 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )}-\frac {\frac {a f^2}{d^2}+1}{4 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^2} \]

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Rubi [A]  time = 0.13, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2117, 893} \begin {gather*} -\frac {a f^2}{2 d^3 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}-\frac {a f^2}{d^3 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )}-\frac {\frac {a f^2}{d^2}+1}{4 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^2}-\frac {3 a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}{2 d^4 e}+\frac {3 a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )}{2 d^4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(-3),x]

[Out]

-(a*f^2)/(2*d^3*e*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])) - (1 + (a*f^2)/d^2)/(4*e*(d + e*x + f*Sqrt[a + (e^2*x^2)/
f^2])^2) - (a*f^2)/(d^3*e*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])) - (3*a*f^2*Log[e*x + f*Sqrt[a + (e^2*x^2)/f^2
]])/(2*d^4*e) + (3*a*f^2*Log[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(2*d^4*e)

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 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, 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+\frac {e^2 x^2}{f^2}}\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {d^2+a f^2-2 d x+x^2}{(d-x)^2 x^3} \, dx,x,d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{2 e}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a f^2}{d^3 (d-x)^2}+\frac {3 a f^2}{d^4 (d-x)}+\frac {d^2+a f^2}{d^2 x^3}+\frac {2 a f^2}{d^3 x^2}+\frac {3 a f^2}{d^4 x}\right ) \, dx,x,d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{2 e}\\ &=-\frac {a f^2}{2 d^3 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}-\frac {1+\frac {a f^2}{d^2}}{4 e \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^2}-\frac {a f^2}{d^3 e \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}-\frac {3 a f^2 \log \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{2 d^4 e}+\frac {3 a f^2 \log \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}{2 d^4 e}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 180, normalized size = 0.93 \begin {gather*} \frac {-\frac {3 a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}{d^4}+\frac {3 a f^2 \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )}{d^4}+\frac {a f^2}{d^3 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}-\frac {2 a f^2}{d^3 \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )}-\frac {\frac {a f^2}{d^2}+1}{2 \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^2}}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(-3),x]

[Out]

((a*f^2)/(d^3*(-(e*x) - f*Sqrt[a + (e^2*x^2)/f^2])) - (1 + (a*f^2)/d^2)/(2*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2
])^2) - (2*a*f^2)/(d^3*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])) - (3*a*f^2*Log[e*x + f*Sqrt[a + (e^2*x^2)/f^2]])
/d^4 + (3*a*f^2*Log[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/d^4)/(2*e)

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IntegrateAlgebraic [B]  time = 2.54, size = 528, normalized size = 2.74 \begin {gather*} \frac {\sqrt {a+\frac {e^2 x^2}{f^2}} \left (-3 a^2 f^5+5 a d^2 f^3+9 a d e f^3 x-3 d^3 e f x-4 d^2 e^2 f x^2\right )}{2 d^3 e \left (-a f^2+d^2+2 d e x\right )^2}+\frac {3 a^4 f^8 x-9 a^3 d^2 f^6 x-9 a^3 d e f^6 x^2+3 a^2 d^4 f^4 x+15 a^2 d^3 e f^4 x^2+4 a^2 d^2 e^2 f^4 x^3+a d^6 f^2 x-3 a d^5 e f^2 x^2-8 a d^4 e^2 f^2 x^3+2 d^8 x+5 d^7 e x^2+4 d^6 e^2 x^3}{2 d^3 \left (d^2-a f^2\right )^2 \left (-a f^2+d^2+2 d e x\right )^2}+\frac {3 a f^3 \sqrt {\frac {e^2}{f^2}} \log \left (d \sqrt {a+\frac {e^2 x^2}{f^2}}+a f+d x \left (-\sqrt {\frac {e^2}{f^2}}\right )\right )}{4 d^4 e^2}-\frac {3 a \left (e f^2-f^3 \sqrt {\frac {e^2}{f^2}}\right ) \log \left (\sqrt {a+\frac {e^2 x^2}{f^2}}-x \sqrt {\frac {e^2}{f^2}}\right )}{4 d^4 e^2}+\frac {3 a \left (e f^2-f^3 \sqrt {\frac {e^2}{f^2}}\right ) \log \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d-f x \sqrt {\frac {e^2}{f^2}}\right )}{4 d^4 e^2}+\frac {3 a f^2 \log \left (d^4 e \sqrt {a+\frac {e^2 x^2}{f^2}}+a d^3 e f+d^4 (-e) x \sqrt {\frac {e^2}{f^2}}\right )}{4 d^4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(-3),x]

[Out]

(Sqrt[a + (e^2*x^2)/f^2]*(5*a*d^2*f^3 - 3*a^2*f^5 - 3*d^3*e*f*x + 9*a*d*e*f^3*x - 4*d^2*e^2*f*x^2))/(2*d^3*e*(
d^2 - a*f^2 + 2*d*e*x)^2) + (2*d^8*x + a*d^6*f^2*x + 3*a^2*d^4*f^4*x - 9*a^3*d^2*f^6*x + 3*a^4*f^8*x + 5*d^7*e
*x^2 - 3*a*d^5*e*f^2*x^2 + 15*a^2*d^3*e*f^4*x^2 - 9*a^3*d*e*f^6*x^2 + 4*d^6*e^2*x^3 - 8*a*d^4*e^2*f^2*x^3 + 4*
a^2*d^2*e^2*f^4*x^3)/(2*d^3*(d^2 - a*f^2)^2*(d^2 - a*f^2 + 2*d*e*x)^2) - (3*a*(e*f^2 - Sqrt[e^2/f^2]*f^3)*Log[
-(Sqrt[e^2/f^2]*x) + Sqrt[a + (e^2*x^2)/f^2]])/(4*d^4*e^2) + (3*a*Sqrt[e^2/f^2]*f^3*Log[a*f - d*Sqrt[e^2/f^2]*
x + d*Sqrt[a + (e^2*x^2)/f^2]])/(4*d^4*e^2) + (3*a*f^2*Log[a*d^3*e*f - d^4*e*Sqrt[e^2/f^2]*x + d^4*e*Sqrt[a +
(e^2*x^2)/f^2]])/(4*d^4*e) + (3*a*(e*f^2 - Sqrt[e^2/f^2]*f^3)*Log[d - Sqrt[e^2/f^2]*f*x + f*Sqrt[a + (e^2*x^2)
/f^2]])/(4*d^4*e^2)

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fricas [B]  time = 0.71, size = 536, normalized size = 2.78 \begin {gather*} \frac {5 \, a^{3} f^{6} + 8 \, d^{3} e^{3} x^{3} - 6 \, a^{2} d^{2} f^{4} - 3 \, a d^{4} f^{2} + 2 \, {\left (a d^{2} e^{2} f^{2} + 5 \, d^{4} e^{2}\right )} x^{2} - 2 \, {\left (7 \, a^{2} d e f^{4} + a d^{3} e f^{2} - 2 \, d^{5} e\right )} x + 3 \, {\left (a^{3} f^{6} + 4 \, a d^{2} e^{2} f^{2} x^{2} - 2 \, a^{2} d^{2} f^{4} + a d^{4} f^{2} - 4 \, {\left (a^{2} d e f^{4} - a d^{3} e f^{2}\right )} x\right )} \log \left (-a e f^{2} x + 2 \, d e^{2} x^{2} + a d f^{2} + {\left (a f^{3} - 2 \, d e f x\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 3 \, {\left (a^{3} f^{6} + 4 \, a d^{2} e^{2} f^{2} x^{2} - 2 \, a^{2} d^{2} f^{4} + a d^{4} f^{2} - 4 \, {\left (a^{2} d e f^{4} - a d^{3} e f^{2}\right )} x\right )} \log \left (-a f^{2} + 2 \, d e x + d^{2}\right ) - 3 \, {\left (a^{3} f^{6} + 4 \, a d^{2} e^{2} f^{2} x^{2} - 2 \, a^{2} d^{2} f^{4} + a d^{4} f^{2} - 4 \, {\left (a^{2} d e f^{4} - a d^{3} e f^{2}\right )} x\right )} \log \left (-e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} - d\right ) - 2 \, {\left (3 \, a^{2} d f^{5} + 4 \, d^{3} e^{2} f x^{2} - 5 \, a d^{3} f^{3} - 3 \, {\left (3 \, a d^{2} e f^{3} - d^{4} e f\right )} x\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}}{4 \, {\left (a^{2} d^{4} e f^{4} + 4 \, d^{6} e^{3} x^{2} - 2 \, a d^{6} e f^{2} + d^{8} e - 4 \, {\left (a d^{5} e^{2} f^{2} - d^{7} e^{2}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^3,x, algorithm="fricas")

[Out]

1/4*(5*a^3*f^6 + 8*d^3*e^3*x^3 - 6*a^2*d^2*f^4 - 3*a*d^4*f^2 + 2*(a*d^2*e^2*f^2 + 5*d^4*e^2)*x^2 - 2*(7*a^2*d*
e*f^4 + a*d^3*e*f^2 - 2*d^5*e)*x + 3*(a^3*f^6 + 4*a*d^2*e^2*f^2*x^2 - 2*a^2*d^2*f^4 + a*d^4*f^2 - 4*(a^2*d*e*f
^4 - a*d^3*e*f^2)*x)*log(-a*e*f^2*x + 2*d*e^2*x^2 + a*d*f^2 + (a*f^3 - 2*d*e*f*x)*sqrt((e^2*x^2 + a*f^2)/f^2))
+ 3*(a^3*f^6 + 4*a*d^2*e^2*f^2*x^2 - 2*a^2*d^2*f^4 + a*d^4*f^2 - 4*(a^2*d*e*f^4 - a*d^3*e*f^2)*x)*log(-a*f^2
+ 2*d*e*x + d^2) - 3*(a^3*f^6 + 4*a*d^2*e^2*f^2*x^2 - 2*a^2*d^2*f^4 + a*d^4*f^2 - 4*(a^2*d*e*f^4 - a*d^3*e*f^2
)*x)*log(-e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) - d) - 2*(3*a^2*d*f^5 + 4*d^3*e^2*f*x^2 - 5*a*d^3*f^3 - 3*(3*a*d
^2*e*f^3 - d^4*e*f)*x)*sqrt((e^2*x^2 + a*f^2)/f^2))/(a^2*d^4*e*f^4 + 4*d^6*e^3*x^2 - 2*a*d^6*e*f^2 + d^8*e - 4
*(a*d^5*e^2*f^2 - d^7*e^2)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^3,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.05, size = 9721, normalized size = 50.37 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d+(e^2/f^2*x^2+a)^(1/2)*f)^3,x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^3,x, algorithm="maxima")

[Out]

integrate((e*x + sqrt(e^2*x^2/f^2 + a)*f + d)^(-3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d + e*x + f*(a + (e^2*x^2)/f^2)^(1/2))^3,x)

[Out]

int(1/(d + e*x + f*(a + (e^2*x^2)/f^2)^(1/2))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e**2*x**2/f**2)**(1/2))**3,x)

[Out]

Integral((d + e*x + f*sqrt(a + e**2*x**2/f**2))**(-3), x)

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