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

Optimal. Leaf size=113 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{8 c^{5/2} \sqrt{b c-a d}}+\frac{3 a x \sqrt{a+b x^2}}{8 c^2 \left (c+d x^2\right )}+\frac{x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0574365, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {378, 377, 208} \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{8 c^{5/2} \sqrt{b c-a d}}+\frac{3 a x \sqrt{a+b x^2}}{8 c^2 \left (c+d x^2\right )}+\frac{x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 378

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

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

Mathematica [A]  time = 0.670222, size = 163, normalized size = 1.44 \[ \frac{x \sqrt{a+b x^2} \left (\frac{\sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \left (5 a c+3 a d x^2+2 b c x^2\right )}{\left (c+d x^2\right ) \sqrt{\frac{d x^2}{c}+1}}+\frac{3 a \sin ^{-1}\left (\frac{\sqrt{x^2 \left (\frac{d}{c}-\frac{b}{a}\right )}}{\sqrt{\frac{d x^2}{c}+1}}\right )}{\sqrt{\frac{x^2 (a d-b c)}{a c}}}\right )}{8 c^3 \sqrt{\frac{b x^2}{a}+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*((Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*(5*a*c + 2*b*c*x^2 + 3*a*d*x^2))/((c + d*x^2)*Sqrt[
1 + (d*x^2)/c]) + (3*a*ArcSin[Sqrt[(-(b/a) + d/c)*x^2]/Sqrt[1 + (d*x^2)/c]])/Sqrt[((-(b*c) + a*d)*x^2)/(a*c)])
)/(8*c^3*Sqrt[1 + (b*x^2)/a])

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Maple [B]  time = 0.022, size = 9059, normalized size = 80.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.51491, size = 1072, normalized size = 9.49 \begin{align*} \left [\frac{3 \,{\left (a^{2} d^{2} x^{4} + 2 \, a^{2} c d x^{2} + a^{2} c^{2}\right )} \sqrt{b c^{2} - a c d} \log \left (\frac{{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \,{\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} + 4 \,{\left ({\left (2 \, b c - a d\right )} x^{3} + a c x\right )} \sqrt{b c^{2} - a c d} \sqrt{b x^{2} + a}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right ) + 4 \,{\left ({\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3} + 5 \,{\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt{b x^{2} + a}}{32 \,{\left (b c^{6} - a c^{5} d +{\left (b c^{4} d^{2} - a c^{3} d^{3}\right )} x^{4} + 2 \,{\left (b c^{5} d - a c^{4} d^{2}\right )} x^{2}\right )}}, -\frac{3 \,{\left (a^{2} d^{2} x^{4} + 2 \, a^{2} c d x^{2} + a^{2} c^{2}\right )} \sqrt{-b c^{2} + a c d} \arctan \left (\frac{\sqrt{-b c^{2} + a c d}{\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt{b x^{2} + a}}{2 \,{\left ({\left (b^{2} c^{2} - a b c d\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - 2 \,{\left ({\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3} + 5 \,{\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt{b x^{2} + a}}{16 \,{\left (b c^{6} - a c^{5} d +{\left (b c^{4} d^{2} - a c^{3} d^{3}\right )} x^{4} + 2 \,{\left (b c^{5} d - a c^{4} d^{2}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(a^2*d^2*x^4 + 2*a^2*c*d*x^2 + a^2*c^2)*sqrt(b*c^2 - a*c*d)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d^2)*x^
4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 + 4*((2*b*c - a*d)*x^3 + a*c*x)*sqrt(b*c^2 - a*c*d)*sqrt(b*x^2 + a
))/(d^2*x^4 + 2*c*d*x^2 + c^2)) + 4*((2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^3 + 5*(a*b*c^3 - a^2*c^2*d)*x)*sq
rt(b*x^2 + a))/(b*c^6 - a*c^5*d + (b*c^4*d^2 - a*c^3*d^3)*x^4 + 2*(b*c^5*d - a*c^4*d^2)*x^2), -1/16*(3*(a^2*d^
2*x^4 + 2*a^2*c*d*x^2 + a^2*c^2)*sqrt(-b*c^2 + a*c*d)*arctan(1/2*sqrt(-b*c^2 + a*c*d)*((2*b*c - a*d)*x^2 + a*c
)*sqrt(b*x^2 + a)/((b^2*c^2 - a*b*c*d)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - 2*((2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2
)*x^3 + 5*(a*b*c^3 - a^2*c^2*d)*x)*sqrt(b*x^2 + a))/(b*c^6 - a*c^5*d + (b*c^4*d^2 - a*c^3*d^3)*x^4 + 2*(b*c^5*
d - a*c^4*d^2)*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.30446, size = 609, normalized size = 5.39 \begin{align*} -\frac{3 \, a^{2} \sqrt{b} \arctan \left (\frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt{-b^{2} c^{2} + a b c d}}\right )}{8 \, \sqrt{-b^{2} c^{2} + a b c d} c^{2}} + \frac{8 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{6} b^{\frac{5}{2}} c^{2} d - 3 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{6} a^{2} \sqrt{b} d^{3} + 16 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} b^{\frac{7}{2}} c^{3} + 8 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} a b^{\frac{5}{2}} c^{2} d - 18 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} a^{2} b^{\frac{3}{2}} c d^{2} + 9 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} a^{3} \sqrt{b} d^{3} + 8 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a^{2} b^{\frac{5}{2}} c^{2} d + 16 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a^{3} b^{\frac{3}{2}} c d^{2} - 9 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a^{4} \sqrt{b} d^{3} + 2 \, a^{4} b^{\frac{3}{2}} c d^{2} + 3 \, a^{5} \sqrt{b} d^{3}}{4 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} d + 4 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} b c - 2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a d + a^{2} d\right )}^{2} c^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*a^2*sqrt(b)*arctan(1/2*((sqrt(b)*x - sqrt(b*x^2 + a))^2*d + 2*b*c - a*d)/sqrt(-b^2*c^2 + a*b*c*d))/(sqrt(
-b^2*c^2 + a*b*c*d)*c^2) + 1/4*(8*(sqrt(b)*x - sqrt(b*x^2 + a))^6*b^(5/2)*c^2*d - 3*(sqrt(b)*x - sqrt(b*x^2 +
a))^6*a^2*sqrt(b)*d^3 + 16*(sqrt(b)*x - sqrt(b*x^2 + a))^4*b^(7/2)*c^3 + 8*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a*b
^(5/2)*c^2*d - 18*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^(3/2)*c*d^2 + 9*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^3*sq
rt(b)*d^3 + 8*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^2*b^(5/2)*c^2*d + 16*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3*b^(3/
2)*c*d^2 - 9*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^4*sqrt(b)*d^3 + 2*a^4*b^(3/2)*c*d^2 + 3*a^5*sqrt(b)*d^3)/(((sqr
t(b)*x - sqrt(b*x^2 + a))^4*d + 4*(sqrt(b)*x - sqrt(b*x^2 + a))^2*b*c - 2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a*d
+ a^2*d)^2*c^2*d^2)