3.111 $$\int \frac{d+e x+f x^2}{(a+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=61 $\frac{f \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{c^{3/2}}-\frac{a e-x (c d-a f)}{a c \sqrt{a+c x^2}}$

[Out]

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

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Rubi [A]  time = 0.0357731, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {1814, 12, 217, 206} $\frac{f \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{c^{3/2}}-\frac{a e-x (c d-a f)}{a c \sqrt{a+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0625536, size = 74, normalized size = 1.21 $\frac{a^{3/2} f \sqrt{\frac{c x^2}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )+\sqrt{c} (c d x-a (e+f x))}{a c^{3/2} \sqrt{a+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[c]*(c*d*x - a*(e + f*x)) + a^(3/2)*f*Sqrt[1 + (c*x^2)/a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/(a*c^(3/2)*Sqrt[a
+ c*x^2])

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Maple [A]  time = 0.052, size = 69, normalized size = 1.1 \begin{align*} -{\frac{fx}{c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{f\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{e}{c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{\frac{dx}{a}{\frac{1}{\sqrt{c{x}^{2}+a}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7177, size = 396, normalized size = 6.49 \begin{align*} \left [\frac{{\left (a c f x^{2} + a^{2} f\right )} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) - 2 \,{\left (a c e -{\left (c^{2} d - a c f\right )} x\right )} \sqrt{c x^{2} + a}}{2 \,{\left (a c^{3} x^{2} + a^{2} c^{2}\right )}}, -\frac{{\left (a c f x^{2} + a^{2} f\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) +{\left (a c e -{\left (c^{2} d - a c f\right )} x\right )} \sqrt{c x^{2} + a}}{a c^{3} x^{2} + a^{2} c^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 5.60967, size = 87, normalized size = 1.43 \begin{align*} e \left (\begin{cases} - \frac{1}{c \sqrt{a + c x^{2}}} & \text{for}\: c \neq 0 \\\frac{x^{2}}{2 a^{\frac{3}{2}}} & \text{otherwise} \end{cases}\right ) + f \left (\frac{\operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{c^{\frac{3}{2}}} - \frac{x}{\sqrt{a} c \sqrt{1 + \frac{c x^{2}}{a}}}\right ) + \frac{d x}{a^{\frac{3}{2}} \sqrt{1 + \frac{c x^{2}}{a}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*Piecewise((-1/(c*sqrt(a + c*x**2)), Ne(c, 0)), (x**2/(2*a**(3/2)), True)) + f*(asinh(sqrt(c)*x/sqrt(a))/c**(
3/2) - x/(sqrt(a)*c*sqrt(1 + c*x**2/a))) + d*x/(a**(3/2)*sqrt(1 + c*x**2/a))

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Giac [A]  time = 1.18497, size = 85, normalized size = 1.39 \begin{align*} -\frac{\frac{e}{c} - \frac{{\left (c^{2} d - a c f\right )} x}{a c^{2}}}{\sqrt{c x^{2} + a}} - \frac{f \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{c^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(e/c - (c^2*d - a*c*f)*x/(a*c^2))/sqrt(c*x^2 + a) - f*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2)