3.5.61 \(\int (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}})^{3/2} \, dx\) [461]

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

[Out]

-3/2*a*f^2*arctanh((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/d^(1/2))*d^(1/2)/e+1/5*(d+e*x+f*(a+e^2*x^2/f^2)^(1/2)
)^(5/2)/e+a*f^2*(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/e-1/2*a*d*f^2*(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/e/(e
*x+f*(a+e^2*x^2/f^2)^(1/2))

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Rubi [A]
time = 0.11, antiderivative size = 183, normalized size of antiderivative = 1.00, 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 = {2142, 911, 1271, 1824, 212} \begin {gather*} \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} \end {gather*}

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 212

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

Rule 911

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 1271

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/(d + e*x^2))*(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))], 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 1824

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 2142

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 \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {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 ) \text {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*}

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Mathematica [A]
time = 1.28, size = 170, normalized size = 0.93 \begin {gather*} \frac {\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \left (2 (d+2 e x)^2 \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )+a f^2 \left (-d+16 e x+12 f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )\right )}{e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}-15 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 )}{10 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]*(2*(d + 2*e*x)^2*(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) + a*f^2*(-d + 1
6*e*x + 12*f*Sqrt[a + (e^2*x^2)/f^2])))/(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) - 15*a*Sqrt[d]*f^2*ArcTanh[Sqrt[d +
e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/(10*e)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (d +e x +f \sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [A]
time = 0.41, size = 335, normalized size = 1.83 \begin {gather*} \left [\frac {1}{20} \, {\left (15 \, a \sqrt {d} f^{2} \log \left (a f^{2} - 2 \, d x e + 2 \, d f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + 2 \, {\left (\sqrt {d} x e - \sqrt {d} f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + d}\right ) + 2 \, {\left (12 \, a f^{2} + 4 \, x^{2} e^{2} + 9 \, d x e + 2 \, d^{2} + {\left (4 \, f x e - d f\right )} \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + d}\right )} e^{\left (-1\right )}, \frac {1}{10} \, {\left (15 \, a \sqrt {-d} f^{2} \arctan \left (\frac {\sqrt {x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + d} \sqrt {-d}}{d}\right ) + {\left (12 \, a f^{2} + 4 \, x^{2} e^{2} + 9 \, d x e + 2 \, d^{2} + {\left (4 \, f x e - d f\right )} \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + d}\right )} e^{\left (-1\right )}\right ] \end {gather*}

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

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

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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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