3.695 \(\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{3/2} (f+g x)^2} \, dx\)

Optimal. Leaf size=178 \[ -\frac{3 c d \sqrt{c d f-a e g} \tan ^{-1}\left (\frac{\sqrt{g} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d f-a e g}}\right )}{g^{5/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{g (d+e x)^{3/2} (f+g x)}+\frac{3 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g^2 \sqrt{d+e x}} \]

[Out]

(3*c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(g^2*Sqrt[d + e*x]) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x
^2)^(3/2)/(g*(d + e*x)^(3/2)*(f + g*x)) - (3*c*d*Sqrt[c*d*f - a*e*g]*ArcTan[(Sqrt[g]*Sqrt[a*d*e + (c*d^2 + a*e
^2)*x + c*d*e*x^2])/(Sqrt[c*d*f - a*e*g]*Sqrt[d + e*x])])/g^(5/2)

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Rubi [A]  time = 0.250813, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {862, 864, 874, 205} \[ -\frac{3 c d \sqrt{c d f-a e g} \tan ^{-1}\left (\frac{\sqrt{g} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d f-a e g}}\right )}{g^{5/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{g (d+e x)^{3/2} (f+g x)}+\frac{3 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{g^2 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(g^2*Sqrt[d + e*x]) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x
^2)^(3/2)/(g*(d + e*x)^(3/2)*(f + g*x)) - (3*c*d*Sqrt[c*d*f - a*e*g]*ArcTan[(Sqrt[g]*Sqrt[a*d*e + (c*d^2 + a*e
^2)*x + c*d*e*x^2])/(Sqrt[c*d*f - a*e*g]*Sqrt[d + e*x])])/g^(5/2)

Rule 862

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

Rule 864

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

Rule 874

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[
2*e^2, Subst[Int[1/(c*(e*f + d*g) - b*e*g + e^2*g*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; Fre
eQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 205

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

Rubi steps

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

Mathematica [C]  time = 0.0617112, size = 75, normalized size = 0.42 \[ \frac{2 c d ((d+e x) (a e+c d x))^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{g (a e+c d x)}{a e g-c d f}\right )}{5 (d+e x)^{5/2} (c d f-a e g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*d*((a*e + c*d*x)*(d + e*x))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, (g*(a*e + c*d*x))/(-(c*d*f) + a*e*g)])/(
5*(c*d*f - a*e*g)^2*(d + e*x)^(5/2))

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Maple [A]  time = 0.341, size = 306, normalized size = 1.7 \begin{align*}{\frac{1}{{g}^{2} \left ( gx+f \right ) }\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( -3\,{\it Artanh} \left ({\frac{\sqrt{cdx+ae}g}{\sqrt{ \left ( aeg-cdf \right ) g}}} \right ) xacde{g}^{2}+3\,{\it Artanh} \left ({\frac{\sqrt{cdx+ae}g}{\sqrt{ \left ( aeg-cdf \right ) g}}} \right ) x{c}^{2}{d}^{2}fg-3\,{\it Artanh} \left ({\frac{\sqrt{cdx+ae}g}{\sqrt{ \left ( aeg-cdf \right ) g}}} \right ) acdefg+3\,{\it Artanh} \left ({\frac{\sqrt{cdx+ae}g}{\sqrt{ \left ( aeg-cdf \right ) g}}} \right ){c}^{2}{d}^{2}{f}^{2}+2\,\sqrt{ \left ( aeg-cdf \right ) g}\sqrt{cdx+ae}xcdg-\sqrt{ \left ( aeg-cdf \right ) g}\sqrt{cdx+ae}aeg+3\,\sqrt{ \left ( aeg-cdf \right ) g}\sqrt{cdx+ae}cdf \right ){\frac{1}{\sqrt{ex+d}}}{\frac{1}{\sqrt{cdx+ae}}}{\frac{1}{\sqrt{ \left ( aeg-cdf \right ) g}}}} \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)^(3/2)/(e*x+d)^(3/2)/(g*x+f)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{3}{2}}}{{\left (e x + d\right )}^{\frac{3}{2}}{\left (g x + f\right )}^{2}}\,{d x} \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)^(3/2)/(e*x+d)^(3/2)/(g*x+f)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.95448, size = 975, normalized size = 5.48 \begin{align*} \left [\frac{3 \,{\left (c d e g x^{2} + c d^{2} f +{\left (c d e f + c d^{2} g\right )} x\right )} \sqrt{-\frac{c d f - a e g}{g}} \log \left (-\frac{c d e g x^{2} - c d^{2} f + 2 \, a d e g - 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d} g \sqrt{-\frac{c d f - a e g}{g}} -{\left (c d e f -{\left (c d^{2} + 2 \, a e^{2}\right )} g\right )} x}{e g x^{2} + d f +{\left (e f + d g\right )} x}\right ) + 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d g x + 3 \, c d f - a e g\right )} \sqrt{e x + d}}{2 \,{\left (e g^{3} x^{2} + d f g^{2} +{\left (e f g^{2} + d g^{3}\right )} x\right )}}, \frac{3 \,{\left (c d e g x^{2} + c d^{2} f +{\left (c d e f + c d^{2} g\right )} x\right )} \sqrt{\frac{c d f - a e g}{g}} \arctan \left (\frac{\sqrt{e x + d} \sqrt{\frac{c d f - a e g}{g}}}{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}\right ) + \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d g x + 3 \, c d f - a e g\right )} \sqrt{e x + d}}{e g^{3} x^{2} + d f g^{2} +{\left (e f g^{2} + d g^{3}\right )} x}\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)^(3/2)/(e*x+d)^(3/2)/(g*x+f)^2,x, algorithm="fricas")

[Out]

[1/2*(3*(c*d*e*g*x^2 + c*d^2*f + (c*d*e*f + c*d^2*g)*x)*sqrt(-(c*d*f - a*e*g)/g)*log(-(c*d*e*g*x^2 - c*d^2*f +
2*a*d*e*g - 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*g*sqrt(-(c*d*f - a*e*g)/g) - (c*d*e*f
- (c*d^2 + 2*a*e^2)*g)*x)/(e*g*x^2 + d*f + (e*f + d*g)*x)) + 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2
*c*d*g*x + 3*c*d*f - a*e*g)*sqrt(e*x + d))/(e*g^3*x^2 + d*f*g^2 + (e*f*g^2 + d*g^3)*x), (3*(c*d*e*g*x^2 + c*d^
2*f + (c*d*e*f + c*d^2*g)*x)*sqrt((c*d*f - a*e*g)/g)*arctan(sqrt(e*x + d)*sqrt((c*d*f - a*e*g)/g)/sqrt(c*d*e*x
^2 + a*d*e + (c*d^2 + a*e^2)*x)) + sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*g*x + 3*c*d*f - a*e*g)*s
qrt(e*x + d))/(e*g^3*x^2 + d*f*g^2 + (e*f*g^2 + d*g^3)*x)]

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Sympy [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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2)/(e*x+d)**(3/2)/(g*x+f)**2,x)

[Out]

Timed out

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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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(3/2)/(g*x+f)^2,x, algorithm="giac")

[Out]

Timed out