3.1913 \(\int \frac{\sqrt{a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=124 \[ \frac{\sqrt{c} \sqrt{d} \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{e^{3/2}}-\frac{2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e (d+e x)} \]

[Out]

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

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Rubi [A]  time = 0.054264, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.081, Rules used = {662, 621, 206} \[ \frac{\sqrt{c} \sqrt{d} \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{e^{3/2}}-\frac{2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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

Mathematica [A]  time = 0.692493, size = 164, normalized size = 1.32 \[ \frac{2 \sqrt{(d+e x) (a e+c d x)} \left (\frac{\sqrt{c} \sqrt{d} \sqrt{c d} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )}{\sqrt{c d^2-a e^2} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}}}-\frac{\sqrt{e}}{d+e x}\right )}{e^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-(Sqrt[e]/(d + e*x)) + (Sqrt[c]*Sqrt[d]*Sqrt[c*d]*ArcSinh[(Sqrt[c]*Sqrt[d]*S
qrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/(Sqrt[c*d^2 - a*e^2]*Sqrt[a*e + c*d*x]*Sqrt[(c*d*(
d + e*x))/(c*d^2 - a*e^2)])))/e^(3/2)

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Maple [B]  time = 0.048, size = 317, normalized size = 2.6 \begin{align*} -2\,{\frac{1}{{e}^{2} \left ( a{e}^{2}-c{d}^{2} \right ) } \left ( cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) \right ) ^{3/2} \left ({\frac{d}{e}}+x \right ) ^{-2}}+2\,{\frac{cd}{e \left ( a{e}^{2}-c{d}^{2} \right ) }\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) }}+{\frac{acde}{a{e}^{2}-c{d}^{2}}\ln \left ({ \left ({\frac{a{e}^{2}}{2}}-{\frac{c{d}^{2}}{2}}+ \left ({\frac{d}{e}}+x \right ) cde \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) } \right ){\frac{1}{\sqrt{dec}}}}-{\frac{{c}^{2}{d}^{3}}{e \left ( a{e}^{2}-c{d}^{2} \right ) }\ln \left ({ \left ({\frac{a{e}^{2}}{2}}-{\frac{c{d}^{2}}{2}}+ \left ({\frac{d}{e}}+x \right ) cde \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{cde \left ({\frac{d}{e}}+x \right ) ^{2}+ \left ( a{e}^{2}-c{d}^{2} \right ) \left ({\frac{d}{e}}+x \right ) } \right ){\frac{1}{\sqrt{dec}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/e^2/(a*e^2-c*d^2)/(d/e+x)^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)+2/e*d*c/(a*e^2-c*d^2)*(c*d*e*(d/e
+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)+e*d*c/(a*e^2-c*d^2)*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c
*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)*a-1/e*d^3*c^2/(a*e^2-c*d^2)*ln((1/2*a*e^2-1/2*c*d^2
+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.18697, size = 697, normalized size = 5.62 \begin{align*} \left [\frac{{\left (e x + d\right )} \sqrt{\frac{c d}{e}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 4 \,{\left (2 \, c d e^{2} x + c d^{2} e + a e^{3}\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{\frac{c d}{e}} + 8 \,{\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) - 4 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{2 \,{\left (e^{2} x + d e\right )}}, -\frac{{\left (e x + d\right )} \sqrt{-\frac{c d}{e}} \arctan \left (\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt{-\frac{c d}{e}}}{2 \,{\left (c^{2} d^{2} e x^{2} + a c d^{2} e +{\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )}}\right ) + 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{e^{2} x + d e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((e*x + d)*sqrt(c*d/e)*log(8*c^2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 + 4*(2*c*d*e^2*x + c*d^2
*e + a*e^3)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(c*d/e) + 8*(c^2*d^3*e + a*c*d*e^3)*x) - 4*sqrt(c*
d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^2*x + d*e), -((e*x + d)*sqrt(-c*d/e)*arctan(1/2*sqrt(c*d*e*x^2 + a*d*
e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-c*d/e)/(c^2*d^2*e*x^2 + a*c*d^2*e + (c^2*d^3 + a*c*d*
e^2)*x)) + 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^2*x + d*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((d + e*x)*(a*e + c*d*x))/(d + e*x)**2, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError