3.60 \(\int \frac{(a+b x^2)^{3/2}}{(c+d x^2)^4} \, dx\)

Optimal. Leaf size=199 \[ \frac{a^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{16 c^{7/2} (b c-a d)^{3/2}}+\frac{x \left (a+b x^2\right )^{3/2} (6 b c-5 a d)}{24 c^2 \left (c+d x^2\right )^2 (b c-a d)}+\frac{a x \sqrt{a+b x^2} (6 b c-5 a d)}{16 c^3 \left (c+d x^2\right ) (b c-a d)}-\frac{d x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3 (b c-a d)} \]

[Out]

-(d*x*(a + b*x^2)^(5/2))/(6*c*(b*c - a*d)*(c + d*x^2)^3) + ((6*b*c - 5*a*d)*x*(a + b*x^2)^(3/2))/(24*c^2*(b*c
- a*d)*(c + d*x^2)^2) + (a*(6*b*c - 5*a*d)*x*Sqrt[a + b*x^2])/(16*c^3*(b*c - a*d)*(c + d*x^2)) + (a^2*(6*b*c -
5*a*d)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(16*c^(7/2)*(b*c - a*d)^(3/2))

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Rubi [A]  time = 0.11543, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {382, 378, 377, 208} \[ \frac{a^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{16 c^{7/2} (b c-a d)^{3/2}}+\frac{x \left (a+b x^2\right )^{3/2} (6 b c-5 a d)}{24 c^2 \left (c+d x^2\right )^2 (b c-a d)}+\frac{a x \sqrt{a+b x^2} (6 b c-5 a d)}{16 c^3 \left (c+d x^2\right ) (b c-a d)}-\frac{d x \left (a+b x^2\right )^{5/2}}{6 c \left (c+d x^2\right )^3 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/(c + d*x^2)^4,x]

[Out]

-(d*x*(a + b*x^2)^(5/2))/(6*c*(b*c - a*d)*(c + d*x^2)^3) + ((6*b*c - 5*a*d)*x*(a + b*x^2)^(3/2))/(24*c^2*(b*c
- a*d)*(c + d*x^2)^2) + (a*(6*b*c - 5*a*d)*x*Sqrt[a + b*x^2])/(16*c^3*(b*c - a*d)*(c + d*x^2)) + (a^2*(6*b*c -
5*a*d)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(16*c^(7/2)*(b*c - a*d)^(3/2))

Rule 382

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

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 )^4} \, dx &=-\frac{d x \left (a+b x^2\right )^{5/2}}{6 c (b c-a d) \left (c+d x^2\right )^3}+\frac{(6 b c-5 a d) \int \frac{\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx}{6 c (b c-a d)}\\ &=-\frac{d x \left (a+b x^2\right )^{5/2}}{6 c (b c-a d) \left (c+d x^2\right )^3}+\frac{(6 b c-5 a d) x \left (a+b x^2\right )^{3/2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac{(a (6 b c-5 a d)) \int \frac{\sqrt{a+b x^2}}{\left (c+d x^2\right )^2} \, dx}{8 c^2 (b c-a d)}\\ &=-\frac{d x \left (a+b x^2\right )^{5/2}}{6 c (b c-a d) \left (c+d x^2\right )^3}+\frac{(6 b c-5 a d) x \left (a+b x^2\right )^{3/2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac{a (6 b c-5 a d) x \sqrt{a+b x^2}}{16 c^3 (b c-a d) \left (c+d x^2\right )}+\frac{\left (a^2 (6 b c-5 a d)\right ) \int \frac{1}{\sqrt{a+b x^2} \left (c+d x^2\right )} \, dx}{16 c^3 (b c-a d)}\\ &=-\frac{d x \left (a+b x^2\right )^{5/2}}{6 c (b c-a d) \left (c+d x^2\right )^3}+\frac{(6 b c-5 a d) x \left (a+b x^2\right )^{3/2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac{a (6 b c-5 a d) x \sqrt{a+b x^2}}{16 c^3 (b c-a d) \left (c+d x^2\right )}+\frac{\left (a^2 (6 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{c-(b c-a d) x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{16 c^3 (b c-a d)}\\ &=-\frac{d x \left (a+b x^2\right )^{5/2}}{6 c (b c-a d) \left (c+d x^2\right )^3}+\frac{(6 b c-5 a d) x \left (a+b x^2\right )^{3/2}}{24 c^2 (b c-a d) \left (c+d x^2\right )^2}+\frac{a (6 b c-5 a d) x \sqrt{a+b x^2}}{16 c^3 (b c-a d) \left (c+d x^2\right )}+\frac{a^2 (6 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{c} \sqrt{a+b x^2}}\right )}{16 c^{7/2} (b c-a d)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.775098, size = 247, normalized size = 1.24 \[ \frac{a x \left (\frac{b x^2}{a}+1\right ) \left (c \left (a^2 b \left (11 c^2 d x^2-30 c^3+32 c d^2 x^4+15 d^3 x^6\right )+a^3 d \left (33 c^2+40 c d x^2+15 d^2 x^4\right )-2 a b^2 c x^2 \left (21 c^2+13 c d x^2+4 d^2 x^4\right )-4 b^3 c^2 x^4 \left (3 c+d x^2\right )\right )+\frac{3 a^2 \left (c+d x^2\right )^3 (5 a d-6 b c) \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )}{\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}}\right )}{48 c^4 \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/(c + d*x^2)^4,x]

[Out]

(a*x*(1 + (b*x^2)/a)*(c*(-4*b^3*c^2*x^4*(3*c + d*x^2) - 2*a*b^2*c*x^2*(21*c^2 + 13*c*d*x^2 + 4*d^2*x^4) + a^3*
d*(33*c^2 + 40*c*d*x^2 + 15*d^2*x^4) + a^2*b*(-30*c^3 + 11*c^2*d*x^2 + 32*c*d^2*x^4 + 15*d^3*x^6)) + (3*a^2*(-
6*b*c + 5*a*d)*(c + d*x^2)^3*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/Sqrt[((b*c - a*d)*x^2)/(c*(a +
b*x^2))]))/(48*c^4*(-(b*c) + a*d)*(a + b*x^2)^(3/2)*(c + d*x^2)^3)

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Maple [B]  time = 0.03, size = 13766, normalized size = 69.2 \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)^4,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 )}^{4}}\,{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)^4,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/2)/(d*x^2 + c)^4, x)

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Fricas [B]  time = 3.95139, size = 1994, normalized size = 10.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(d*x^2+c)^4,x, algorithm="fricas")

[Out]

[1/192*(3*(6*a^2*b*c^4 - 5*a^3*c^3*d + (6*a^2*b*c*d^3 - 5*a^3*d^4)*x^6 + 3*(6*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*x^4
+ 3*(6*a^2*b*c^3*d - 5*a^3*c^2*d^2)*x^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*((4*b^3*c^4*d + 4*a*b^2*c^3*d^2 - 23*a^2*b*c^2*d^3 + 15*a^3*c*d^4)*x^5 + 2*(6*b^3
*c^5 + 5*a*b^2*c^4*d - 31*a^2*b*c^3*d^2 + 20*a^3*c^2*d^3)*x^3 + 3*(10*a*b^2*c^5 - 21*a^2*b*c^4*d + 11*a^3*c^3*
d^2)*x)*sqrt(b*x^2 + a))/(b^2*c^9 - 2*a*b*c^8*d + a^2*c^7*d^2 + (b^2*c^6*d^3 - 2*a*b*c^5*d^4 + a^2*c^4*d^5)*x^
6 + 3*(b^2*c^7*d^2 - 2*a*b*c^6*d^3 + a^2*c^5*d^4)*x^4 + 3*(b^2*c^8*d - 2*a*b*c^7*d^2 + a^2*c^6*d^3)*x^2), -1/9
6*(3*(6*a^2*b*c^4 - 5*a^3*c^3*d + (6*a^2*b*c*d^3 - 5*a^3*d^4)*x^6 + 3*(6*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*x^4 + 3*
(6*a^2*b*c^3*d - 5*a^3*c^2*d^2)*x^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*((4*b^3*c^4*d + 4*a*b^2*c^3*d^2 -
23*a^2*b*c^2*d^3 + 15*a^3*c*d^4)*x^5 + 2*(6*b^3*c^5 + 5*a*b^2*c^4*d - 31*a^2*b*c^3*d^2 + 20*a^3*c^2*d^3)*x^3
+ 3*(10*a*b^2*c^5 - 21*a^2*b*c^4*d + 11*a^3*c^3*d^2)*x)*sqrt(b*x^2 + a))/(b^2*c^9 - 2*a*b*c^8*d + a^2*c^7*d^2
+ (b^2*c^6*d^3 - 2*a*b*c^5*d^4 + a^2*c^4*d^5)*x^6 + 3*(b^2*c^7*d^2 - 2*a*b*c^6*d^3 + a^2*c^5*d^4)*x^4 + 3*(b^2
*c^8*d - 2*a*b*c^7*d^2 + a^2*c^6*d^3)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/(d*x**2+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 20.881, size = 1241, normalized size = 6.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(d*x^2+c)^4,x, algorithm="giac")

[Out]

-1/16*(6*a^2*b^(3/2)*c - 5*a^3*sqrt(b)*d)*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))/((b*c^4 - a*c^3*d)*sqrt(-b^2*c^2 + a*b*c*d)) - 1/24*(18*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^
2*b^(3/2)*c*d^4 - 15*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^3*sqrt(b)*d^5 - 96*(sqrt(b)*x - sqrt(b*x^2 + a))^8*b^(
9/2)*c^4*d + 96*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a*b^(7/2)*c^3*d^2 + 180*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^2*b^
(5/2)*c^2*d^3 - 240*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^3*b^(3/2)*c*d^4 + 75*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^4
*sqrt(b)*d^5 - 128*(sqrt(b)*x - sqrt(b*x^2 + a))^6*b^(11/2)*c^5 - 64*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*b^(9/2)
*c^4*d + 720*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^2*b^(7/2)*c^3*d^2 - 968*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^3*b^(
5/2)*c^2*d^3 + 620*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^4*b^(3/2)*c*d^4 - 150*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^5
*sqrt(b)*d^5 - 96*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^(9/2)*c^4*d - 288*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^3*
b^(7/2)*c^3*d^2 + 864*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^4*b^(5/2)*c^2*d^3 - 600*(sqrt(b)*x - sqrt(b*x^2 + a))^
4*a^5*b^(3/2)*c*d^4 + 150*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^6*sqrt(b)*d^5 - 48*(sqrt(b)*x - sqrt(b*x^2 + a))^2
*a^4*b^(7/2)*c^3*d^2 - 72*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^5*b^(5/2)*c^2*d^3 + 210*(sqrt(b)*x - sqrt(b*x^2 +
a))^2*a^6*b^(3/2)*c*d^4 - 75*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^7*sqrt(b)*d^5 - 4*a^6*b^(5/2)*c^2*d^3 - 8*a^7*b
^(3/2)*c*d^4 + 15*a^8*sqrt(b)*d^5)/((b*c^4*d^2 - a*c^3*d^3)*((sqrt(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)^3)