3.465 \(\int \frac{1}{(d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}})^{5/2}} \, dx\)

Optimal. Leaf size=199 \[ -\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 d^3 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}-\frac{2 a f^2}{d^3 e \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}-\frac{\frac{a f^2}{d^2}+1}{3 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}+\frac{5 a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 d^{7/2} e} \]

[Out]

-(1 + (a*f^2)/d^2)/(3*e*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(3/2)) - (2*a*f^2)/(d^3*e*Sqrt[d + e*x + f*Sqrt[
a + (e^2*x^2)/f^2]]) - (a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(2*d^3*e*(e*x + f*Sqrt[a + (e^2*x^2)/
f^2])) + (5*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/(2*d^(7/2)*e)

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Rubi [A]  time = 0.176685, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2117, 897, 1259, 1261, 206} \[ -\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 d^3 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}-\frac{2 a f^2}{d^3 e \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}-\frac{\frac{a f^2}{d^2}+1}{3 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}+\frac{5 a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 d^{7/2} e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(-5/2),x]

[Out]

-(1 + (a*f^2)/d^2)/(3*e*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(3/2)) - (2*a*f^2)/(d^3*e*Sqrt[d + e*x + f*Sqrt[
a + (e^2*x^2)/f^2]]) - (a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(2*d^3*e*(e*x + f*Sqrt[a + (e^2*x^2)/
f^2])) + (5*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/(2*d^(7/2)*e)

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]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], 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] && IntegersQ[n,
p] && FractionQ[m]

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{d^2+a f^2-2 d x+x^2}{(d-x)^2 x^{5/2}} \, dx,x,d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}{2 e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{d^2+a f^2-2 d x^2+x^4}{x^4 \left (d-x^2\right )^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{e}\\ &=-\frac{a f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 d^3 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 d^2 \left (d^2+a f^2\right )-2 d \left (d^2-a f^2\right ) x^2+a f^2 x^4}{x^4 \left (d-x^2\right )} \, dx,x,\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{2 d^3 e}\\ &=-\frac{a f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 d^3 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}+\frac{\operatorname{Subst}\left (\int \left (\frac{2 \left (d^3+a d f^2\right )}{x^4}+\frac{4 a f^2}{x^2}+\frac{5 a f^2}{d-x^2}\right ) \, dx,x,\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{2 d^3 e}\\ &=-\frac{d^2+a f^2}{3 d^2 e \left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )^{3/2}}-\frac{2 a f^2}{d^3 e \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}-\frac{a f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 d^3 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}+\frac{\left (5 a f^2\right ) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{2 d^3 e}\\ &=-\frac{d^2+a f^2}{3 d^2 e \left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )^{3/2}}-\frac{2 a f^2}{d^3 e \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}-\frac{a f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 d^3 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}+\frac{5 a f^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{\sqrt{d}}\right )}{2 d^{7/2} e}\\ \end{align*}

Mathematica [A]  time = 0.613331, size = 186, normalized size = 0.93 \[ -\frac{\frac{2 d \left (a f^2+d^2\right )}{3 \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}+\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x}+\frac{4 a f^2}{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}-\frac{5 a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{\sqrt{d}}}{2 d^3 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(-5/2),x]

[Out]

-((2*d*(d^2 + a*f^2))/(3*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(3/2)) + (4*a*f^2)/Sqrt[d + e*x + f*Sqrt[a + (e
^2*x^2)/f^2]] + (a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) - (5*a*f^2
*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/Sqrt[d])/(2*d^3*e)

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Maple [F]  time = 0.009, size = 0, normalized size = 0. \begin{align*} \int \left ( d+ex+f\sqrt{a+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ) ^{-{\frac{5}{2}}}\, dx \end{align*}

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))^(5/2),x)

[Out]

int(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (e x + \sqrt{\frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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))^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + sqrt(e^2*x^2/f^2 + a)*f + d)^(-5/2), x)

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Fricas [B]  time = 1.55992, size = 1683, normalized size = 8.46 \begin{align*} \text{result too large to display} \end{align*}

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))^(5/2),x, algorithm="fricas")

[Out]

[1/12*(15*(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)*sqrt(d
)*log(a*f^2 - 2*d*e*x + 2*d*f*sqrt((e^2*x^2 + a*f^2)/f^2) - 2*(sqrt(d)*e*x - sqrt(d)*f*sqrt((e^2*x^2 + a*f^2)/
f^2))*sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)) + 2*(12*d^3*e^3*x^3 + 10*a^2*d^2*f^4 - 16*a*d^4*f^2 - 2*d
^6 - 8*(5*a*d^2*e^2*f^2 - d^4*e^2)*x^2 + (15*a^2*d*e*f^4 - 46*a*d^3*e*f^2 - d^5*e)*x - (15*a^2*d*f^5 + 12*d^3*
e^2*f*x^2 - 22*a*d^3*f^3 - d^5*f - 8*(5*a*d^2*e*f^3 - d^4*e*f)*x)*sqrt((e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sq
rt((e^2*x^2 + a*f^2)/f^2) + d))/(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), -1/6*(15*(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)*sqrt(-d)*arctan(sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)*sqrt(-d)/d) - (12*d^3*e^3*x^3 + 10*a^2*d^2*f
^4 - 16*a*d^4*f^2 - 2*d^6 - 8*(5*a*d^2*e^2*f^2 - d^4*e^2)*x^2 + (15*a^2*d*e*f^4 - 46*a*d^3*e*f^2 - d^5*e)*x -
(15*a^2*d*f^5 + 12*d^3*e^2*f*x^2 - 22*a*d^3*f^3 - d^5*f - 8*(5*a*d^2*e*f^3 - d^4*e*f)*x)*sqrt((e^2*x^2 + a*f^2
)/f^2))*sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d))/(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x + f \sqrt{a + \frac{e^{2} x^{2}}{f^{2}}}\right )^{\frac{5}{2}}}\, dx \end{align*}

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))**(5/2),x)

[Out]

Integral((d + e*x + f*sqrt(a + e**2*x**2/f**2))**(-5/2), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))^(5/2),x, algorithm="giac")

[Out]

Timed out