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

Optimal. Leaf size=183 \[ \frac{\left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{5 e}+\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{e}-\frac{a d f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}-\frac{3 a \sqrt{d} f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.149687, antiderivative size = 183, 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, 1257, 1810, 206} \[ \frac{\left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{5 e}+\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{e}-\frac{a d f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}-\frac{3 a \sqrt{d} f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/e - (a*d*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(2*e
*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])) + (d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(5/2)/(5*e) - (3*a*Sqrt[d]*f^2*Arc
Tanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/(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 1257

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[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4
)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(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] && IGtQ[m/2, 0]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -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 \left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^{3/2} \left (d^2+a f^2-2 d x+x^2\right )}{(d-x)^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{x^4 \left (d^2+a f^2-2 d x^2+x^4\right )}{\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 d f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}-\frac{\operatorname{Subst}\left (\int \frac{a d f^2+2 a f^2 x^2-2 d x^4+2 x^6}{d-x^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{2 e}\\ &=-\frac{a d f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}-\frac{\operatorname{Subst}\left (\int \left (-2 a f^2-2 x^4+\frac{3 a d f^2}{d-x^2}\right ) \, dx,x,\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}\right )}{2 e}\\ &=\frac{a f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{e}-\frac{a d f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}+\frac{\left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )^{5/2}}{5 e}-\frac{\left (3 a d 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 e}\\ &=\frac{a f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{e}-\frac{a d f^2 \sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{2 e \left (e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )}+\frac{\left (d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}\right )^{5/2}}{5 e}-\frac{3 a \sqrt{d} f^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}}{\sqrt{d}}\right )}{2 e}\\ \end{align*}

Mathematica [A]  time = 0.233944, size = 175, normalized size = 0.96 \[ \frac{\frac{2}{5} \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )^{5/2}+2 a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}-\frac{a d f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x}-3 a \sqrt{d} f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.011, size = 0, normalized size = 0. \begin{align*} \int \left ( d+ex+f\sqrt{a+{\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((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x)

[Out]

int((d+e*x+f*(a+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{\left (e x + \sqrt{\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((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.35343, size = 752, normalized size = 4.11 \begin{align*} \left [\frac{15 \, a \sqrt{d} f^{2} \log \left (a f^{2} - 2 \, d e x + 2 \, d f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + 2 \,{\left (\sqrt{d} e x - \sqrt{d} f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d}\right ) + 2 \,{\left (4 \, e^{2} x^{2} + 12 \, a f^{2} + 9 \, d e x + 2 \, d^{2} +{\left (4 \, e f x - d f\right )} \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{20 \, e}, \frac{15 \, a \sqrt{-d} f^{2} \arctan \left (\frac{\sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d} \sqrt{-d}}{d}\right ) +{\left (4 \, e^{2} x^{2} + 12 \, a f^{2} + 9 \, d e x + 2 \, d^{2} +{\left (4 \, e f x - d f\right )} \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{10 \, e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/20*(15*a*sqrt(d)*f^2*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*s
qrt((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*(4*e^2*x^2 + 12*a*f^2 + 9*d*e*x
+ 2*d^2 + (4*e*f*x - 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))/e, 1/10
*(15*a*sqrt(-d)*f^2*arctan(sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)*sqrt(-d)/d) + (4*e^2*x^2 + 12*a*f^2 +
9*d*e*x + 2*d^2 + (4*e*f*x - 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))
/e]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x + f \sqrt{a + \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((d+e*x+f*(a+e**2*x**2/f**2)**(1/2))**(3/2),x)

[Out]

Integral((d + e*x + f*sqrt(a + 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{\left (e x + \sqrt{\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((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="giac")

[Out]

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