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

Optimal. Leaf size=269 \[ -\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\left (2 d e-b f^2\right )^2 \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac{3 f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{\sqrt{2} \sqrt{e} \left (2 d e-b f^2\right )^{5/2}}-\frac{4 \left (a e f^2-b d f^2+d^2 e\right )}{\left (2 d e-b f^2\right )^2 \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}} \]

[Out]

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

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Rubi [A]  time = 0.367633, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2116, 897, 1259, 453, 208} \[ -\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\left (2 d e-b f^2\right )^2 \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac{3 f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{\sqrt{2} \sqrt{e} \left (2 d e-b f^2\right )^{5/2}}-\frac{4 \left (a e f^2-b d f^2+d^2 e\right )}{\left (2 d e-b f^2\right )^2 \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{\left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{3/2}} \, 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^{3/2} \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 )\\ &=4 \operatorname{Subst}\left (\int \frac{d^2 e-(b d-a e) f^2+\left (-2 d e+b f^2\right ) x^2+e x^4}{x^2 \left (-2 d e+b f^2+2 e x^2\right )^2} \, dx,x,\sqrt{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 ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{\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{\operatorname{Subst}\left (\int \frac{8 e^2 \left (2 d e-b f^2\right ) \left (d^2 e-b d f^2+a e f^2\right )-2 e^2 \left (8 d^2 e^2-8 b d e f^2-4 a e^2 f^2+3 b^2 f^4\right ) x^2}{x^2 \left (-2 d e+b f^2+2 e x^2\right )} \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{2 e^2 \left (2 d e-b f^2\right )^2}\\ &=-\frac{4 \left (d^2 e-b d f^2+a e f^2\right )}{\left (2 d e-b f^2\right )^2 \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{\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{\left (3 f^2 \left (4 a e^2-b^2 f^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-2 d e+b f^2+2 e x^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{\left (2 d e-b f^2\right )^2}\\ &=-\frac{4 \left (d^2 e-b d f^2+a e f^2\right )}{\left (2 d e-b f^2\right )^2 \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{\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{3 f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{\sqrt{2 d e-b f^2}}\right )}{\sqrt{2} \sqrt{e} \left (2 d e-b f^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.481623, size = 257, normalized size = 0.96 \[ \frac{-\frac{2 e^2 \left (4 a e^2 f^2-b^2 f^4\right ) \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{2 e \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+e x\right )+b f^2}+\frac{3 e^{3/2} f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{\sqrt{d e-\frac{b f^2}{2}}}-\frac{8 e^2 \left (a e f^2-b d f^2+d^2 e\right )}{\sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}}{2 e^2 \left (b f^2-2 d e\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*e^2*(d^2*e - b*d*f^2 + a*e*f^2))/Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]] - (2*e^2*(4*a*e^2*f^2 -
b^2*f^4)*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])
) + (3*e^(3/2)*f^2*(4*a*e^2 - b^2*f^2)*ArcTanh[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]
])/Sqrt[2*d*e - b*f^2]])/Sqrt[d*e - (b*f^2)/2])/(2*e^2*(-2*d*e + b*f^2)^2)

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Maple [F]  time = 0.01, size = 0, normalized size = 0. \begin{align*} \int \left ( d+ex+f\sqrt{a+bx+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

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))^(3/2),x)

[Out]

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

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

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.45926, size = 2992, normalized size = 11.12 \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+b*x+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="fricas")

[Out]

[1/4*(3*(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)*sqrt(-2*b*e*f^2 + 4*d*e^2)*log(-b^2*f^4 + 4*(b*d*e - a*e^2)*f^2 - 4*(b*e^2*f^2 - 2*d*e^3)*x -
2*(2*sqrt(-2*b*e*f^2 + 4*d*e^2)*e*f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(-2*b*e*f^2 + 4*d*e^2)*(b*f^2
+ 2*e^2*x))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d) + 4*(b*e*f^3 - 2*d*e^2*f)*sqrt((b*f^2*x +
e^2*x^2 + a*f^2)/f^2)) + 4*(a*b^2*e*f^6 - 8*d^4*e^3 - (5*b^2*d^2*e - 2*a*b*d*e^2)*f^4 + 2*(7*b*d^3*e^2 - 4*a*d
^2*e^3)*f^2 + 2*(b^2*e^3*f^4 - 4*b*d*e^4*f^2 + 4*d^2*e^5)*x^2 + (b^3*e*f^6 - 4*d^3*e^4 - 2*(4*b^2*d*e^2 - 3*a*
b*e^3)*f^4 + 2*(7*b*d^2*e^3 - 6*a*d*e^4)*f^2)*x + 2*(2*d^3*e^3*f + (2*b^2*d*e - 3*a*b*e^2)*f^5 - (5*b*d^2*e^2
- 6*a*d*e^3)*f^3 - (b^2*e^2*f^5 - 4*b*d*e^3*f^3 + 4*d^2*e^4*f)*x)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(
e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/(a*b^3*e*f^8 + 8*d^5*e^4 - (b^3*d^2*e + 6*a*b^2*d*e^2)*f^6
+ 6*(b^2*d^3*e^2 + 2*a*b*d^2*e^3)*f^4 - 4*(3*b*d^4*e^3 + 2*a*d^3*e^4)*f^2 + (b^4*e*f^8 - 8*b^3*d*e^2*f^6 + 24
*b^2*d^2*e^3*f^4 - 32*b*d^3*e^4*f^2 + 16*d^4*e^5)*x), -1/2*(3*(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)*sqrt(2*b*e*f^2 - 4*d*e^2)*arctan(1/2*sqr
t(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d)*(sqrt(2*b*e*f^2 - 4*d*e^2)*f*sqrt((b*f^2*x + e^2*x^2 + a*
f^2)/f^2) - sqrt(2*b*e*f^2 - 4*d*e^2)*(e*x + d))/(a*e*f^2 - d^2*e + (b*e*f^2 - 2*d*e^2)*x)) - 2*(a*b^2*e*f^6 -
8*d^4*e^3 - (5*b^2*d^2*e - 2*a*b*d*e^2)*f^4 + 2*(7*b*d^3*e^2 - 4*a*d^2*e^3)*f^2 + 2*(b^2*e^3*f^4 - 4*b*d*e^4*
f^2 + 4*d^2*e^5)*x^2 + (b^3*e*f^6 - 4*d^3*e^4 - 2*(4*b^2*d*e^2 - 3*a*b*e^3)*f^4 + 2*(7*b*d^2*e^3 - 6*a*d*e^4)*
f^2)*x + 2*(2*d^3*e^3*f + (2*b^2*d*e - 3*a*b*e^2)*f^5 - (5*b*d^2*e^2 - 6*a*d*e^3)*f^3 - (b^2*e^2*f^5 - 4*b*d*e
^3*f^3 + 4*d^2*e^4*f)*x)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/
f^2) + d))/(a*b^3*e*f^8 + 8*d^5*e^4 - (b^3*d^2*e + 6*a*b^2*d*e^2)*f^6 + 6*(b^2*d^3*e^2 + 2*a*b*d^2*e^3)*f^4 -
4*(3*b*d^4*e^3 + 2*a*d^3*e^4)*f^2 + (b^4*e*f^8 - 8*b^3*d*e^2*f^6 + 24*b^2*d^2*e^3*f^4 - 32*b*d^3*e^4*f^2 + 16*
d^4*e^5)*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 + b x + \frac{e^{2} x^{2}}{f^{2}}}\right )^{\frac{3}{2}}}\, dx \end{align*}

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))**(3/2),x)

[Out]

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

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

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))^(3/2),x, algorithm="giac")

[Out]

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