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

Optimal. Leaf size=112 \[ -\frac{2 \left (a e^2+c d^2\right )}{e^2 \sqrt{d+e x} (e f-d g)}-\frac{2 \left (a g^2+c f^2\right ) \tan ^{-1}\left (\frac{\sqrt{g} \sqrt{d+e x}}{\sqrt{e f-d g}}\right )}{g^{3/2} (e f-d g)^{3/2}}+\frac{2 c \sqrt{d+e x}}{e^2 g} \]

[Out]

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

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Rubi [A]  time = 0.215486, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {898, 1261, 205} \[ -\frac{2 \left (a e^2+c d^2\right )}{e^2 \sqrt{d+e x} (e f-d g)}-\frac{2 \left (a g^2+c f^2\right ) \tan ^{-1}\left (\frac{\sqrt{g} \sqrt{d+e x}}{\sqrt{e f-d g}}\right )}{g^{3/2} (e f-d g)^{3/2}}+\frac{2 c \sqrt{d+e x}}{e^2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 898

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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

Mathematica [C]  time = 0.0936026, size = 91, normalized size = 0.81 \[ \frac{2 c (e f-d g) (2 d g+e (f+g x))-2 e^2 \left (a g^2+c f^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{g (d+e x)}{d g-e f}\right )}{e^2 g^2 \sqrt{d+e x} (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*(e*f - d*g)*(2*d*g + e*(f + g*x)) - 2*e^2*(c*f^2 + a*g^2)*Hypergeometric2F1[-1/2, 1, 1/2, (g*(d + e*x))/(
-(e*f) + d*g)])/(e^2*g^2*(e*f - d*g)*Sqrt[d + e*x])

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Maple [A]  time = 0.239, size = 114, normalized size = 1. \begin{align*} 2\,{\frac{1}{{e}^{2}} \left ({\frac{c\sqrt{ex+d}}{g}}-{\frac{{e}^{2} \left ( a{g}^{2}+c{f}^{2} \right ) }{ \left ( dg-ef \right ) g\sqrt{ \left ( dg-ef \right ) g}}{\it Artanh} \left ({\frac{\sqrt{ex+d}g}{\sqrt{ \left ( dg-ef \right ) g}}} \right ) }-{\frac{-a{e}^{2}-c{d}^{2}}{ \left ( dg-ef \right ) \sqrt{ex+d}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/e^2*(c/g*(e*x+d)^(1/2)-e^2*(a*g^2+c*f^2)/(d*g-e*f)/g/((d*g-e*f)*g)^(1/2)*arctanh((e*x+d)^(1/2)*g/((d*g-e*f)*
g)^(1/2))-(-a*e^2-c*d^2)/(d*g-e*f)/(e*x+d)^(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((c*x^2+a)/(e*x+d)^(3/2)/(g*x+f),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.89517, size = 1030, normalized size = 9.2 \begin{align*} \left [\frac{{\left (c d e^{2} f^{2} + a d e^{2} g^{2} +{\left (c e^{3} f^{2} + a e^{3} g^{2}\right )} x\right )} \sqrt{-e f g + d g^{2}} \log \left (\frac{e g x - e f + 2 \, d g - 2 \, \sqrt{-e f g + d g^{2}} \sqrt{e x + d}}{g x + f}\right ) + 2 \,{\left (c d e^{2} f^{2} g -{\left (3 \, c d^{2} e + a e^{3}\right )} f g^{2} +{\left (2 \, c d^{3} + a d e^{2}\right )} g^{3} +{\left (c e^{3} f^{2} g - 2 \, c d e^{2} f g^{2} + c d^{2} e g^{3}\right )} x\right )} \sqrt{e x + d}}{d e^{4} f^{2} g^{2} - 2 \, d^{2} e^{3} f g^{3} + d^{3} e^{2} g^{4} +{\left (e^{5} f^{2} g^{2} - 2 \, d e^{4} f g^{3} + d^{2} e^{3} g^{4}\right )} x}, \frac{2 \,{\left ({\left (c d e^{2} f^{2} + a d e^{2} g^{2} +{\left (c e^{3} f^{2} + a e^{3} g^{2}\right )} x\right )} \sqrt{e f g - d g^{2}} \arctan \left (\frac{\sqrt{e f g - d g^{2}} \sqrt{e x + d}}{e g x + d g}\right ) +{\left (c d e^{2} f^{2} g -{\left (3 \, c d^{2} e + a e^{3}\right )} f g^{2} +{\left (2 \, c d^{3} + a d e^{2}\right )} g^{3} +{\left (c e^{3} f^{2} g - 2 \, c d e^{2} f g^{2} + c d^{2} e g^{3}\right )} x\right )} \sqrt{e x + d}\right )}}{d e^{4} f^{2} g^{2} - 2 \, d^{2} e^{3} f g^{3} + d^{3} e^{2} g^{4} +{\left (e^{5} f^{2} g^{2} - 2 \, d e^{4} f g^{3} + d^{2} e^{3} g^{4}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((c*d*e^2*f^2 + a*d*e^2*g^2 + (c*e^3*f^2 + a*e^3*g^2)*x)*sqrt(-e*f*g + d*g^2)*log((e*g*x - e*f + 2*d*g - 2*sq
rt(-e*f*g + d*g^2)*sqrt(e*x + d))/(g*x + f)) + 2*(c*d*e^2*f^2*g - (3*c*d^2*e + a*e^3)*f*g^2 + (2*c*d^3 + a*d*e
^2)*g^3 + (c*e^3*f^2*g - 2*c*d*e^2*f*g^2 + c*d^2*e*g^3)*x)*sqrt(e*x + d))/(d*e^4*f^2*g^2 - 2*d^2*e^3*f*g^3 + d
^3*e^2*g^4 + (e^5*f^2*g^2 - 2*d*e^4*f*g^3 + d^2*e^3*g^4)*x), 2*((c*d*e^2*f^2 + a*d*e^2*g^2 + (c*e^3*f^2 + a*e^
3*g^2)*x)*sqrt(e*f*g - d*g^2)*arctan(sqrt(e*f*g - d*g^2)*sqrt(e*x + d)/(e*g*x + d*g)) + (c*d*e^2*f^2*g - (3*c*
d^2*e + a*e^3)*f*g^2 + (2*c*d^3 + a*d*e^2)*g^3 + (c*e^3*f^2*g - 2*c*d*e^2*f*g^2 + c*d^2*e*g^3)*x)*sqrt(e*x + d
))/(d*e^4*f^2*g^2 - 2*d^2*e^3*f*g^3 + d^3*e^2*g^4 + (e^5*f^2*g^2 - 2*d*e^4*f*g^3 + d^2*e^3*g^4)*x)]

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Sympy [A]  time = 26.087, size = 107, normalized size = 0.96 \begin{align*} \frac{2 c \sqrt{d + e x}}{e^{2} g} + \frac{2 \left (a g^{2} + c f^{2}\right ) \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- \frac{d g - e f}{g}}} \right )}}{g^{2} \sqrt{- \frac{d g - e f}{g}} \left (d g - e f\right )} + \frac{2 \left (a e^{2} + c d^{2}\right )}{e^{2} \sqrt{d + e x} \left (d g - e f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*sqrt(d + e*x)/(e**2*g) + 2*(a*g**2 + c*f**2)*atan(sqrt(d + e*x)/sqrt(-(d*g - e*f)/g))/(g**2*sqrt(-(d*g - e
*f)/g)*(d*g - e*f)) + 2*(a*e**2 + c*d**2)/(e**2*sqrt(d + e*x)*(d*g - e*f))

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Giac [A]  time = 1.14253, size = 157, normalized size = 1.4 \begin{align*} \frac{2 \, \sqrt{x e + d} c e^{\left (-2\right )}}{g} + \frac{2 \,{\left (c f^{2} + a g^{2}\right )} \arctan \left (\frac{\sqrt{x e + d} g}{\sqrt{-d g^{2} + f g e}}\right )}{{\left (d g^{2} - f g e\right )} \sqrt{-d g^{2} + f g e}} + \frac{2 \,{\left (c d^{2} + a e^{2}\right )}}{{\left (d g e^{2} - f e^{3}\right )} \sqrt{x e + d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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