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

Optimal. Leaf size=122 \[ -\frac {\sqrt {f+g x} \left (a+\frac {c d^2}{e^2}\right )}{(d+e x) (e f-d g)}+\frac {\left (a e^2 g+c d (4 e f-3 d g)\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} (e f-d g)^{3/2}}+\frac {2 c \sqrt {f+g x}}{e^2 g} \]

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Rubi [A]  time = 0.20, antiderivative size = 122, normalized size of antiderivative = 1.00, 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, 388, 208} \begin {gather*} -\frac {\sqrt {f+g x} \left (a+\frac {c d^2}{e^2}\right )}{(d+e x) (e f-d g)}+\frac {\left (a e^2 g+c d (4 e f-3 d g)\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{5/2} (e f-d g)^{3/2}}+\frac {2 c \sqrt {f+g x}}{e^2 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 388

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 171, normalized size = 1.40 \begin {gather*} \frac {\frac {\left (a e^2+c d^2\right ) \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 {4 c d \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e f-d g}}+\frac {2 c \sqrt {e} \sqrt {f+g x}}{g}}{e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.51, size = 179, normalized size = 1.47 \begin {gather*} \frac {\sqrt {f+g x} \left (a e^2 g^2+3 c d^2 g^2+2 c d e g (f+g x)-4 c d e f g+2 c e^2 f^2-2 c e^2 f (f+g x)\right )}{e^2 g (e f-d g) (-d g-e (f+g x)+e f)}+\frac {\left (-a e^2 g+3 c d^2 g-4 c d e f\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x} \sqrt {d g-e f}}{e f-d g}\right )}{e^{5/2} (d g-e f)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.41, size = 539, normalized size = 4.42 \begin {gather*} \left [-\frac {{\left (4 \, c d^{2} e f g - {\left (3 \, c d^{3} - a d e^{2}\right )} g^{2} + {\left (4 \, c d e^{2} f g - {\left (3 \, c d^{2} e - a e^{3}\right )} g^{2}\right )} x\right )} \sqrt {e^{2} f - d e g} \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 d e^{3} f^{2} - {\left (5 \, c d^{2} e^{2} + a e^{4}\right )} f g + {\left (3 \, c d^{3} e + a d e^{3}\right )} g^{2} + 2 \, {\left (c e^{4} f^{2} - 2 \, c d e^{3} f g + c d^{2} e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}}{2 \, {\left (d e^{5} f^{2} g - 2 \, d^{2} e^{4} f g^{2} + d^{3} e^{3} g^{3} + {\left (e^{6} f^{2} g - 2 \, d e^{5} f g^{2} + d^{2} e^{4} g^{3}\right )} x\right )}}, -\frac {{\left (4 \, c d^{2} e f g - {\left (3 \, c d^{3} - a d e^{2}\right )} g^{2} + {\left (4 \, c d e^{2} f g - {\left (3 \, c d^{2} e - a e^{3}\right )} g^{2}\right )} x\right )} \sqrt {-e^{2} f + d e g} \arctan \left (\frac {\sqrt {-e^{2} f + d e g} \sqrt {g x + f}}{e g x + e f}\right ) - {\left (2 \, c d e^{3} f^{2} - {\left (5 \, c d^{2} e^{2} + a e^{4}\right )} f g + {\left (3 \, c d^{3} e + a d e^{3}\right )} g^{2} + 2 \, {\left (c e^{4} f^{2} - 2 \, c d e^{3} f g + c d^{2} e^{2} g^{2}\right )} x\right )} \sqrt {g x + f}}{d e^{5} f^{2} g - 2 \, d^{2} e^{4} f g^{2} + d^{3} e^{3} g^{3} + {\left (e^{6} f^{2} g - 2 \, d e^{5} f g^{2} + d^{2} e^{4} g^{3}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((4*c*d^2*e*f*g - (3*c*d^3 - a*d*e^2)*g^2 + (4*c*d*e^2*f*g - (3*c*d^2*e - a*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*c*d*e^3*f^2 - (5*c*d^2*
e^2 + a*e^4)*f*g + (3*c*d^3*e + a*d*e^3)*g^2 + 2*(c*e^4*f^2 - 2*c*d*e^3*f*g + c*d^2*e^2*g^2)*x)*sqrt(g*x + f))
/(d*e^5*f^2*g - 2*d^2*e^4*f*g^2 + d^3*e^3*g^3 + (e^6*f^2*g - 2*d*e^5*f*g^2 + d^2*e^4*g^3)*x), -((4*c*d^2*e*f*g
- (3*c*d^3 - a*d*e^2)*g^2 + (4*c*d*e^2*f*g - (3*c*d^2*e - a*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*c*d*e^3*f^2 - (5*c*d^2*e^2 + a*e^4)*f*g + (3*c*d^3*e + a*d*e^3)
*g^2 + 2*(c*e^4*f^2 - 2*c*d*e^3*f*g + c*d^2*e^2*g^2)*x)*sqrt(g*x + f))/(d*e^5*f^2*g - 2*d^2*e^4*f*g^2 + d^3*e^
3*g^3 + (e^6*f^2*g - 2*d*e^5*f*g^2 + d^2*e^4*g^3)*x)]

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giac [A]  time = 0.17, size = 148, normalized size = 1.21 \begin {gather*} \frac {2 \, \sqrt {g x + f} c e^{\left (-2\right )}}{g} - \frac {{\left (3 \, c d^{2} g - 4 \, c d f e - a g e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right )}{{\left (d g e^{2} - f e^{3}\right )} \sqrt {d g e - f e^{2}}} + \frac {\sqrt {g x + f} c d^{2} g + \sqrt {g x + f} a g e^{2}}{{\left (d g e^{2} - f e^{3}\right )} {\left (d g + {\left (g x + f\right )} e - f e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 237, normalized size = 1.94 \begin {gather*} \frac {a g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}}-\frac {3 c \,d^{2} g \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}\, e^{2}}+\frac {4 c d f \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right ) \sqrt {\left (d g -e f \right ) e}\, e}+\frac {\sqrt {g x +f}\, a g}{\left (d g -e f \right ) \left (e g x +d g \right )}+\frac {\sqrt {g x +f}\, c \,d^{2} g}{\left (d g -e f \right ) \left (e g x +d g \right ) e^{2}}+\frac {2 \sqrt {g x +f}\, c}{e^{2} g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2/(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 = 2.68, size = 128, normalized size = 1.05 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (-3\,c\,g\,d^2+4\,c\,f\,d\,e+a\,g\,e^2\right )}{e^{5/2}\,{\left (d\,g-e\,f\right )}^{3/2}}+\frac {\sqrt {f+g\,x}\,\left (c\,g\,d^2+a\,g\,e^2\right )}{\left (d\,g-e\,f\right )\,\left (e^3\,\left (f+g\,x\right )-e^3\,f+d\,e^2\,g\right )}+\frac {2\,c\,\sqrt {f+g\,x}}{e^2\,g} \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)^2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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