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

Optimal. Leaf size=178 \[ -\frac{\left (3 a e^2 g^2+c \left (3 d^2 g^2-8 d e f g+8 e^2 f^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{4 e^{5/2} (e f-d g)^{5/2}}-\frac{\sqrt{f+g x} \left (a+\frac{c d^2}{e^2}\right )}{2 (d+e x)^2 (e f-d g)}+\frac{\sqrt{f+g x} \left (3 a e^2 g+c d (8 e f-5 d g)\right )}{4 e^2 (d+e x) (e f-d g)^2} \]

[Out]

-((a + (c*d^2)/e^2)*Sqrt[f + g*x])/(2*(e*f - d*g)*(d + e*x)^2) + ((3*a*e^2*g + c*d*(8*e*f - 5*d*g))*Sqrt[f + g
*x])/(4*e^2*(e*f - d*g)^2*(d + e*x)) - ((3*a*e^2*g^2 + c*(8*e^2*f^2 - 8*d*e*f*g + 3*d^2*g^2))*ArcTanh[(Sqrt[e]
*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(4*e^(5/2)*(e*f - d*g)^(5/2))

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Rubi [A]  time = 0.30305, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {898, 1157, 385, 208} \[ -\frac{\left (3 a e^2 g^2+c \left (3 d^2 g^2-8 d e f g+8 e^2 f^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{4 e^{5/2} (e f-d g)^{5/2}}-\frac{\sqrt{f+g x} \left (a+\frac{c d^2}{e^2}\right )}{2 (d+e x)^2 (e f-d g)}+\frac{\sqrt{f+g x} \left (3 a e^2 g+c d (8 e f-5 d g)\right )}{4 e^2 (d+e x) (e f-d g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + (c*d^2)/e^2)*Sqrt[f + g*x])/(2*(e*f - d*g)*(d + e*x)^2) + ((3*a*e^2*g + c*d*(8*e*f - 5*d*g))*Sqrt[f + g
*x])/(4*e^2*(e*f - d*g)^2*(d + e*x)) - ((3*a*e^2*g^2 + c*(8*e^2*f^2 - 8*d*e*f*g + 3*d^2*g^2))*ArcTanh[(Sqrt[e]
*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(4*e^(5/2)*(e*f - d*g)^(5/2))

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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 208

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

Rubi steps

\begin{align*} \int \frac{a+c x^2}{(d+e x)^3 \sqrt{f+g x}} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{\frac{c f^2+a g^2}{g^2}-\frac{2 c f x^2}{g^2}+\frac{c x^4}{g^2}}{\left (\frac{-e f+d g}{g}+\frac{e x^2}{g}\right )^3} \, dx,x,\sqrt{f+g x}\right )}{g}\\ &=-\frac{\left (a+\frac{c d^2}{e^2}\right ) \sqrt{f+g x}}{2 (e f-d g) (d+e x)^2}+\frac{\operatorname{Subst}\left (\int \frac{-3 a+\frac{c d^2}{e^2}-\frac{4 c f^2}{g^2}+\frac{4 c (e f-d g) x^2}{e g^2}}{\left (\frac{-e f+d g}{g}+\frac{e x^2}{g}\right )^2} \, dx,x,\sqrt{f+g x}\right )}{2 (e f-d g)}\\ &=-\frac{\left (a+\frac{c d^2}{e^2}\right ) \sqrt{f+g x}}{2 (e f-d g) (d+e x)^2}+\frac{\left (3 a e^2 g+c d (8 e f-5 d g)\right ) \sqrt{f+g x}}{4 e^2 (e f-d g)^2 (d+e x)}+\frac{\left (3 a e^2 g^2+c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{-e f+d g}{g}+\frac{e x^2}{g}} \, dx,x,\sqrt{f+g x}\right )}{4 e^2 g (e f-d g)^2}\\ &=-\frac{\left (a+\frac{c d^2}{e^2}\right ) \sqrt{f+g x}}{2 (e f-d g) (d+e x)^2}+\frac{\left (3 a e^2 g+c d (8 e f-5 d g)\right ) \sqrt{f+g x}}{4 e^2 (e f-d g)^2 (d+e x)}-\frac{\left (3 a e^2 g^2+c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{4 e^{5/2} (e f-d g)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.885958, size = 207, normalized size = 1.16 \[ \frac{2 \left (\frac{\sqrt{e} g^2 \sqrt{f+g x} \left (a e^2+c d^2\right ) \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{e (f+g x)}{e f-d g}\right )}{(d g-e f)^3}-\frac{c d \left (\sqrt{e} \sqrt{f+g x} (d g-e f)+g (d+e x) \sqrt{d g-e f} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{d g-e f}}\right )\right )}{(d+e x) (e f-d g)^2}-\frac{c \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{\sqrt{e f-d g}}\right )}{e^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-((c*d*(Sqrt[e]*(-(e*f) + d*g)*Sqrt[f + g*x] + g*Sqrt[-(e*f) + d*g]*(d + e*x)*ArcTan[(Sqrt[e]*Sqrt[f + g*x
])/Sqrt[-(e*f) + d*g]]))/((e*f - d*g)^2*(d + e*x))) - (c*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/Sqr
t[e*f - d*g] + (Sqrt[e]*(c*d^2 + a*e^2)*g^2*Sqrt[f + g*x]*Hypergeometric2F1[1/2, 3, 3/2, (e*(f + g*x))/(e*f -
d*g)])/(-(e*f) + d*g)^3))/e^(5/2)

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Maple [B]  time = 0.219, size = 384, normalized size = 2.2 \begin{align*} 2\,{\frac{1}{ \left ( e \left ( gx+f \right ) +dg-ef \right ) ^{2}} \left ( 1/8\,{\frac{g \left ( 3\,a{e}^{2}g-5\,c{d}^{2}g+8\,cdef \right ) \left ( gx+f \right ) ^{3/2}}{e \left ({d}^{2}{g}^{2}-2\,defg+{e}^{2}{f}^{2} \right ) }}+1/8\,{\frac{ \left ( 5\,a{e}^{2}g-3\,c{d}^{2}g+8\,cdef \right ) g\sqrt{gx+f}}{{e}^{2} \left ( dg-ef \right ) }} \right ) }+{\frac{3\,a{g}^{2}}{4\,{d}^{2}{g}^{2}-8\,defg+4\,{e}^{2}{f}^{2}}\arctan \left ({e\sqrt{gx+f}{\frac{1}{\sqrt{ \left ( dg-ef \right ) e}}}} \right ){\frac{1}{\sqrt{ \left ( dg-ef \right ) e}}}}+{\frac{3\,c{d}^{2}{g}^{2}}{ \left ( 4\,{d}^{2}{g}^{2}-8\,defg+4\,{e}^{2}{f}^{2} \right ){e}^{2}}\arctan \left ({e\sqrt{gx+f}{\frac{1}{\sqrt{ \left ( dg-ef \right ) e}}}} \right ){\frac{1}{\sqrt{ \left ( dg-ef \right ) e}}}}-2\,{\frac{cdfg}{e \left ({d}^{2}{g}^{2}-2\,defg+{e}^{2}{f}^{2} \right ) \sqrt{ \left ( dg-ef \right ) e}}\arctan \left ({\frac{e\sqrt{gx+f}}{\sqrt{ \left ( dg-ef \right ) e}}} \right ) }+2\,{\frac{c{f}^{2}}{ \left ({d}^{2}{g}^{2}-2\,defg+{e}^{2}{f}^{2} \right ) \sqrt{ \left ( dg-ef \right ) e}}\arctan \left ({\frac{e\sqrt{gx+f}}{\sqrt{ \left ( dg-ef \right ) e}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1/8*g*(3*a*e^2*g-5*c*d^2*g+8*c*d*e*f)/e/(d^2*g^2-2*d*e*f*g+e^2*f^2)*(g*x+f)^(3/2)+1/8*(5*a*e^2*g-3*c*d^2*g+
8*c*d*e*f)/e^2*g/(d*g-e*f)*(g*x+f)^(1/2))/(e*(g*x+f)+d*g-e*f)^2+3/4/(d^2*g^2-2*d*e*f*g+e^2*f^2)/((d*g-e*f)*e)^
(1/2)*arctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2))*a*g^2+3/4/(d^2*g^2-2*d*e*f*g+e^2*f^2)/e^2/((d*g-e*f)*e)^(1/2
)*arctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2))*c*d^2*g^2-2/(d^2*g^2-2*d*e*f*g+e^2*f^2)/e/((d*g-e*f)*e)^(1/2)*ar
ctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2))*c*d*f*g+2/(d^2*g^2-2*d*e*f*g+e^2*f^2)/((d*g-e*f)*e)^(1/2)*arctan(e*(
g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2))*c*f^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/(g*x+f)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.00828, size = 1835, normalized size = 10.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((8*c*d^2*e^2*f^2 - 8*c*d^3*e*f*g + 3*(c*d^4 + a*d^2*e^2)*g^2 + (8*c*e^4*f^2 - 8*c*d*e^3*f*g + 3*(c*d^2*e
^2 + a*e^4)*g^2)*x^2 + 2*(8*c*d*e^3*f^2 - 8*c*d^2*e^2*f*g + 3*(c*d^3*e + a*d*e^3)*g^2)*x)*sqrt(e^2*f - d*e*g)*
log((e*g*x + 2*e*f - d*g - 2*sqrt(e^2*f - d*e*g)*sqrt(g*x + f))/(e*x + d)) + 2*(2*(3*c*d^2*e^3 - a*e^5)*f^2 -
(9*c*d^3*e^2 - 7*a*d*e^4)*f*g + (3*c*d^4*e - 5*a*d^2*e^3)*g^2 + (8*c*d*e^4*f^2 - (13*c*d^2*e^3 - 3*a*e^5)*f*g
+ (5*c*d^3*e^2 - 3*a*d*e^4)*g^2)*x)*sqrt(g*x + f))/(d^2*e^6*f^3 - 3*d^3*e^5*f^2*g + 3*d^4*e^4*f*g^2 - d^5*e^3*
g^3 + (e^8*f^3 - 3*d*e^7*f^2*g + 3*d^2*e^6*f*g^2 - d^3*e^5*g^3)*x^2 + 2*(d*e^7*f^3 - 3*d^2*e^6*f^2*g + 3*d^3*e
^5*f*g^2 - d^4*e^4*g^3)*x), 1/4*((8*c*d^2*e^2*f^2 - 8*c*d^3*e*f*g + 3*(c*d^4 + a*d^2*e^2)*g^2 + (8*c*e^4*f^2 -
8*c*d*e^3*f*g + 3*(c*d^2*e^2 + a*e^4)*g^2)*x^2 + 2*(8*c*d*e^3*f^2 - 8*c*d^2*e^2*f*g + 3*(c*d^3*e + a*d*e^3)*g
^2)*x)*sqrt(-e^2*f + d*e*g)*arctan(sqrt(-e^2*f + d*e*g)*sqrt(g*x + f)/(e*g*x + e*f)) + (2*(3*c*d^2*e^3 - a*e^5
)*f^2 - (9*c*d^3*e^2 - 7*a*d*e^4)*f*g + (3*c*d^4*e - 5*a*d^2*e^3)*g^2 + (8*c*d*e^4*f^2 - (13*c*d^2*e^3 - 3*a*e
^5)*f*g + (5*c*d^3*e^2 - 3*a*d*e^4)*g^2)*x)*sqrt(g*x + f))/(d^2*e^6*f^3 - 3*d^3*e^5*f^2*g + 3*d^4*e^4*f*g^2 -
d^5*e^3*g^3 + (e^8*f^3 - 3*d*e^7*f^2*g + 3*d^2*e^6*f*g^2 - d^3*e^5*g^3)*x^2 + 2*(d*e^7*f^3 - 3*d^2*e^6*f^2*g +
3*d^3*e^5*f*g^2 - d^4*e^4*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((c*x**2+a)/(e*x+d)**3/(g*x+f)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.14981, size = 375, normalized size = 2.11 \begin{align*} \frac{{\left (3 \, c d^{2} g^{2} - 8 \, c d f g e + 8 \, c f^{2} e^{2} + 3 \, a g^{2} e^{2}\right )} \arctan \left (\frac{\sqrt{g x + f} e}{\sqrt{d g e - f e^{2}}}\right )}{4 \,{\left (d^{2} g^{2} e^{2} - 2 \, d f g e^{3} + f^{2} e^{4}\right )} \sqrt{d g e - f e^{2}}} - \frac{3 \, \sqrt{g x + f} c d^{3} g^{3} + 5 \,{\left (g x + f\right )}^{\frac{3}{2}} c d^{2} g^{2} e - 11 \, \sqrt{g x + f} c d^{2} f g^{2} e - 8 \,{\left (g x + f\right )}^{\frac{3}{2}} c d f g e^{2} + 8 \, \sqrt{g x + f} c d f^{2} g e^{2} - 5 \, \sqrt{g x + f} a d g^{3} e^{2} - 3 \,{\left (g x + f\right )}^{\frac{3}{2}} a g^{2} e^{3} + 5 \, \sqrt{g x + f} a f g^{2} e^{3}}{4 \,{\left (d^{2} g^{2} e^{2} - 2 \, d f g e^{3} + f^{2} e^{4}\right )}{\left (d g +{\left (g x + f\right )} e - f e\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*c*d^2*g^2 - 8*c*d*f*g*e + 8*c*f^2*e^2 + 3*a*g^2*e^2)*arctan(sqrt(g*x + f)*e/sqrt(d*g*e - f*e^2))/((d^2*
g^2*e^2 - 2*d*f*g*e^3 + f^2*e^4)*sqrt(d*g*e - f*e^2)) - 1/4*(3*sqrt(g*x + f)*c*d^3*g^3 + 5*(g*x + f)^(3/2)*c*d
^2*g^2*e - 11*sqrt(g*x + f)*c*d^2*f*g^2*e - 8*(g*x + f)^(3/2)*c*d*f*g*e^2 + 8*sqrt(g*x + f)*c*d*f^2*g*e^2 - 5*
sqrt(g*x + f)*a*d*g^3*e^2 - 3*(g*x + f)^(3/2)*a*g^2*e^3 + 5*sqrt(g*x + f)*a*f*g^2*e^3)/((d^2*g^2*e^2 - 2*d*f*g
*e^3 + f^2*e^4)*(d*g + (g*x + f)*e - f*e)^2)