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

Optimal. Leaf size=104 \[ -\frac {2 \left (a e^2+c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} \sqrt {e f-d g}}-\frac {2 c \sqrt {f+g x} (d g+e f)}{e^2 g^2}+\frac {2 c (f+g x)^{3/2}}{3 e g^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 104, normalized size of antiderivative = 1.00, 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, 1153, 208} \begin {gather*} -\frac {2 \left (a e^2+c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} \sqrt {e f-d g}}-\frac {2 c \sqrt {f+g x} (d g+e f)}{e^2 g^2}+\frac {2 c (f+g x)^{3/2}}{3 e g^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 1153

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 92, normalized size = 0.88 \begin {gather*} \frac {2 c \sqrt {f+g x} (-3 d g-2 e f+e g x)}{3 e^2 g^2}-\frac {2 \left (a e^2+c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} \sqrt {e f-d g}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 105, normalized size = 1.01 \begin {gather*} \frac {2 c \sqrt {f+g x} (-3 d g+e (f+g x)-3 e f)}{3 e^2 g^2}-\frac {2 \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x} \sqrt {d g-e f}}{e f-d g}\right )}{e^{5/2} \sqrt {d g-e f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 297, normalized size = 2.86 \begin {gather*} \left [\frac {3 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {e^{2} f - d e g} g^{2} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {e^{2} f - d e g} \sqrt {g x + f}}{e x + d}\right ) - 2 \, {\left (2 \, c e^{3} f^{2} + c d e^{2} f g - 3 \, c d^{2} e g^{2} - {\left (c e^{3} f g - c d e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}}{3 \, {\left (e^{4} f g^{2} - d e^{3} g^{3}\right )}}, \frac {2 \, {\left (3 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {-e^{2} f + d e g} g^{2} \arctan \left (\frac {\sqrt {-e^{2} f + d e g} \sqrt {g x + f}}{e g x + e f}\right ) - {\left (2 \, c e^{3} f^{2} + c d e^{2} f g - 3 \, c d^{2} e g^{2} - {\left (c e^{3} f g - c d e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}\right )}}{3 \, {\left (e^{4} f g^{2} - d e^{3} g^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 107, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (c d^{2} + a e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right ) e^{\left (-2\right )}}{\sqrt {d g e - f e^{2}}} - \frac {2 \, {\left (3 \, \sqrt {g x + f} c d g^{5} e - {\left (g x + f\right )}^{\frac {3}{2}} c g^{4} e^{2} + 3 \, \sqrt {g x + f} c f g^{4} e^{2}\right )} e^{\left (-3\right )}}{3 \, g^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(c*d^2 + a*e^2)*arctan(sqrt(g*x + f)*e/sqrt(d*g*e - f*e^2))*e^(-2)/sqrt(d*g*e - f*e^2) - 2/3*(3*sqrt(g*x + f
)*c*d*g^5*e - (g*x + f)^(3/2)*c*g^4*e^2 + 3*sqrt(g*x + f)*c*f*g^4*e^2)*e^(-3)/g^6

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maple [A]  time = 0.02, size = 132, normalized size = 1.27 \begin {gather*} \frac {2 a \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\sqrt {\left (d g -e f \right ) e}}+\frac {2 c \,d^{2} \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\sqrt {\left (d g -e f \right ) e}\, e^{2}}-\frac {2 \sqrt {g x +f}\, c d}{e^{2} g}-\frac {2 \sqrt {g x +f}\, c f}{e \,g^{2}}+\frac {2 \left (g x +f \right )^{\frac {3}{2}} c}{3 e \,g^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d*g-e*f>0)', see `assume?` for
more details)Is d*g-e*f positive or negative?

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mupad [B]  time = 0.11, size = 107, normalized size = 1.03 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (c\,d^2+a\,e^2\right )}{e^{5/2}\,\sqrt {d\,g-e\,f}}-\sqrt {f+g\,x}\,\left (\frac {2\,c\,\left (d\,g^3-e\,f\,g^2\right )}{e^2\,g^4}+\frac {4\,c\,f}{e\,g^2}\right )+\frac {2\,c\,{\left (f+g\,x\right )}^{3/2}}{3\,e\,g^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atan((e^(1/2)*(f + g*x)^(1/2))/(d*g - e*f)^(1/2))*(a*e^2 + c*d^2))/(e^(5/2)*(d*g - e*f)^(1/2)) - (f + g*x)^
(1/2)*((2*c*(d*g^3 - e*f*g^2))/(e^2*g^4) + (4*c*f)/(e*g^2)) + (2*c*(f + g*x)^(3/2))/(3*e*g^2)

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sympy [A]  time = 48.66, size = 100, normalized size = 0.96 \begin {gather*} \frac {2 c \left (f + g x\right )^{\frac {3}{2}}}{3 e g^{2}} - \frac {2 c \sqrt {f + g x} \left (d g + e f\right )}{e^{2} g^{2}} - \frac {2 \left (a e^{2} + c d^{2}\right ) \operatorname {atan}{\left (\frac {1}{\sqrt {\frac {e}{d g - e f}} \sqrt {f + g x}} \right )}}{e^{2} \sqrt {\frac {e}{d g - e f}} \left (d g - e f\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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