3.689 \(\int \frac{(c+d x^2)^{3/2}}{a+b x^2} \, dx\)

Optimal. Leaf size=113 \[ \frac{(b c-a d)^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} b^2}+\frac{\sqrt{d} (3 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 b^2}+\frac{d x \sqrt{c+d x^2}}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0955573, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {416, 523, 217, 206, 377, 205} \[ \frac{(b c-a d)^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} b^2}+\frac{\sqrt{d} (3 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 b^2}+\frac{d x \sqrt{c+d x^2}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^(3/2)/(a + b*x^2),x]

[Out]

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

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, 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])

Rule 377

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

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

Mathematica [A]  time = 0.162279, size = 110, normalized size = 0.97 \[ \frac{\sqrt{d} (3 b c-2 a d) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )+\frac{2 (b c-a d)^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a}}+b d x \sqrt{c+d x^2}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*Sqrt[c + d*x^2] + (2*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/Sqrt[a] +
Sqrt[d]*(3*b*c - 2*a*d)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(2*b^2)

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Maple [B]  time = 0.009, size = 1875, normalized size = 16.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+1/4*d/b
*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x+3/4/b*d^(1/2)*ln((-d*(
-a*b)^(1/2)/b+(x+1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2)
)-(a*d-b*c)/b)^(1/2))*c+1/2/(-a*b)^(1/2)/b*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(
a*d-b*c)/b)^(1/2)*a*d-1/2/(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-
b*c)/b)^(1/2)*c-1/2/b^2*d^(3/2)*ln((-d*(-a*b)^(1/2)/b+(x+1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*
d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))*a+1/2/(-a*b)^(1/2)/b^2/(-(a*d-b*c)/b)^(1/2)*ln((
-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-
a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*a^2*d^2-1/(-a*b)^(1/2)/b/(-(a*d-b*
c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(
1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*a*d*c+1/2/(-a*b)^(
1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((
x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*c^2+
1/6/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+1/4*d/b*
((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x+3/4/b*d^(1/2)*ln((d*(-a
*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-
(a*d-b*c)/b)^(1/2))*c-1/2/(-a*b)^(1/2)/b*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*
d-b*c)/b)^(1/2)*a*d+1/2/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*
c)/b)^(1/2)*c-1/2/b^2*d^(3/2)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-a*b)^(1/2))^2*d+2
*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))*a-1/2/(-a*b)^(1/2)/b^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*
(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b
)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))*a^2*d^2+1/(-a*b)^(1/2)/b/(-(a*d-b*c)/
b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2
))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))*a*d*c-1/2/(-a*b)^(1/2
)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1
/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))*c^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((d*x^2+c)^(3/2)/(b*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.76252, size = 1582, normalized size = 14. \begin{align*} \left [\frac{2 \, \sqrt{d x^{2} + c} b d x -{\left (3 \, b c - 2 \, a d\right )} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) -{\left (b c - a d\right )} \sqrt{-\frac{b c - a d}{a}} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \,{\left (a^{2} c x -{\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, b^{2}}, \frac{2 \, \sqrt{d x^{2} + c} b d x - 2 \,{\left (3 \, b c - 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) -{\left (b c - a d\right )} \sqrt{-\frac{b c - a d}{a}} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \,{\left (a^{2} c x -{\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, b^{2}}, \frac{2 \, \sqrt{d x^{2} + c} b d x + 2 \,{\left (b c - a d\right )} \sqrt{\frac{b c - a d}{a}} \arctan \left (\frac{{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b c - a d}{a}}}{2 \,{\left ({\left (b c d - a d^{2}\right )} x^{3} +{\left (b c^{2} - a c d\right )} x\right )}}\right ) -{\left (3 \, b c - 2 \, a d\right )} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right )}{4 \, b^{2}}, \frac{\sqrt{d x^{2} + c} b d x -{\left (3 \, b c - 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) +{\left (b c - a d\right )} \sqrt{\frac{b c - a d}{a}} \arctan \left (\frac{{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b c - a d}{a}}}{2 \,{\left ({\left (b c d - a d^{2}\right )} x^{3} +{\left (b c^{2} - a c d\right )} x\right )}}\right )}{2 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(d*x^2 + c)*b*d*x - (3*b*c - 2*a*d)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - (b*c
- a*d)*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*
x^2 + 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/
b^2, 1/4*(2*sqrt(d*x^2 + c)*b*d*x - 2*(3*b*c - 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (b*c - a*d
)*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 +
4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/b^2, 1
/4*(2*sqrt(d*x^2 + c)*b*d*x + 2*(b*c - a*d)*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^
2 + c)*sqrt((b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)) - (3*b*c - 2*a*d)*sqrt(d)*log(-2*d*x^2 +
2*sqrt(d*x^2 + c)*sqrt(d)*x - c))/b^2, 1/2*(sqrt(d*x^2 + c)*b*d*x - (3*b*c - 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*
x/sqrt(d*x^2 + c)) + (b*c - a*d)*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt
((b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)))/b^2]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x^{2}\right )^{\frac{3}{2}}}{a + b x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x**2)**(3/2)/(a + b*x**2), x)

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Giac [A]  time = 1.13906, size = 205, normalized size = 1.81 \begin{align*} \frac{\sqrt{d x^{2} + c} d x}{2 \, b} - \frac{{\left (3 \, b c \sqrt{d} - 2 \, a d^{\frac{3}{2}}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{4 \, b^{2}} - \frac{{\left (b^{2} c^{2} \sqrt{d} - 2 \, a b c d^{\frac{3}{2}} + a^{2} d^{\frac{5}{2}}\right )} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{\sqrt{a b c d - a^{2} d^{2}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(d*x^2 + c)*d*x/b - 1/4*(3*b*c*sqrt(d) - 2*a*d^(3/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/b^2 - (b^2*
c^2*sqrt(d) - 2*a*b*c*d^(3/2) + a^2*d^(5/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt
(a*b*c*d - a^2*d^2))/(sqrt(a*b*c*d - a^2*d^2)*b^2)