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

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

[Out]

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

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Rubi [A]  time = 0.113364, 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, 205} \[ \frac{c^2 (5 b c-6 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{16 a^{7/2} (b c-a d)^{3/2}}+\frac{x \left (c+d x^2\right )^{3/2} (5 b c-6 a d)}{24 a^2 \left (a+b x^2\right )^2 (b c-a d)}+\frac{c x \sqrt{c+d x^2} (5 b c-6 a d)}{16 a^3 \left (a+b x^2\right ) (b c-a d)}+\frac{b x \left (c+d x^2\right )^{5/2}}{6 a \left (a+b x^2\right )^3 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 5.26762, size = 179, normalized size = 0.9 \[ \frac{\frac{3 c^2 (5 b c-6 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{(b c-a d)^{3/2}}-\frac{\sqrt{a} x \sqrt{c+d x^2} \left (a^2 b \left (33 c^2-22 c d x^2-4 d^2 x^4\right )-6 a^3 d \left (5 c+2 d x^2\right )+8 a b^2 c x^2 \left (5 c-d x^2\right )+15 b^3 c^2 x^4\right )}{\left (a+b x^2\right )^3 (a d-b c)}}{48 a^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((Sqrt[a]*x*Sqrt[c + d*x^2]*(15*b^3*c^2*x^4 + 8*a*b^2*c*x^2*(5*c - d*x^2) - 6*a^3*d*(5*c + 2*d*x^2) + a^2*b*
(33*c^2 - 22*c*d*x^2 - 4*d^2*x^4)))/((-(b*c) + a*d)*(a + b*x^2)^3)) + (3*c^2*(5*b*c - 6*a*d)*ArcTan[(Sqrt[b*c
- a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(b*c - a*d)^(3/2))/(48*a^(7/2))

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Maple [B]  time = 0.044, size = 13964, normalized size = 70.2 \begin{align*} \text{output 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)^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 (d x^{2} + c\right )}^{\frac{3}{2}}}{{\left (b x^{2} + a\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 4.26016, 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((d*x^2+c)^(3/2)/(b*x^2+a)^4,x, algorithm="fricas")

[Out]

[-1/192*(3*(5*a^3*b*c^3 - 6*a^4*c^2*d + (5*b^4*c^3 - 6*a*b^3*c^2*d)*x^6 + 3*(5*a*b^3*c^3 - 6*a^2*b^2*c^2*d)*x^
4 + 3*(5*a^2*b^2*c^3 - 6*a^3*b*c^2*d)*x^2)*sqrt(-a*b*c + a^2*d)*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*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(
b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*((15*a*b^4*c^3 - 23*a^2*b^3*c^2*d + 4*a^3*b^2*c*d^2 + 4*a^4*b*d^3)*x^5 + 2*(20
*a^2*b^3*c^3 - 31*a^3*b^2*c^2*d + 5*a^4*b*c*d^2 + 6*a^5*d^3)*x^3 + 3*(11*a^3*b^2*c^3 - 21*a^4*b*c^2*d + 10*a^5
*c*d^2)*x)*sqrt(d*x^2 + c))/(a^7*b^2*c^2 - 2*a^8*b*c*d + a^9*d^2 + (a^4*b^5*c^2 - 2*a^5*b^4*c*d + a^6*b^3*d^2)
*x^6 + 3*(a^5*b^4*c^2 - 2*a^6*b^3*c*d + a^7*b^2*d^2)*x^4 + 3*(a^6*b^3*c^2 - 2*a^7*b^2*c*d + a^8*b*d^2)*x^2), 1
/96*(3*(5*a^3*b*c^3 - 6*a^4*c^2*d + (5*b^4*c^3 - 6*a*b^3*c^2*d)*x^6 + 3*(5*a*b^3*c^3 - 6*a^2*b^2*c^2*d)*x^4 +
3*(5*a^2*b^2*c^3 - 6*a^3*b*c^2*d)*x^2)*sqrt(a*b*c - a^2*d)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 -
a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) + 2*((15*a*b^4*c^3 - 23*a^2*b^3*c^2*d
+ 4*a^3*b^2*c*d^2 + 4*a^4*b*d^3)*x^5 + 2*(20*a^2*b^3*c^3 - 31*a^3*b^2*c^2*d + 5*a^4*b*c*d^2 + 6*a^5*d^3)*x^3
+ 3*(11*a^3*b^2*c^3 - 21*a^4*b*c^2*d + 10*a^5*c*d^2)*x)*sqrt(d*x^2 + c))/(a^7*b^2*c^2 - 2*a^8*b*c*d + a^9*d^2
+ (a^4*b^5*c^2 - 2*a^5*b^4*c*d + a^6*b^3*d^2)*x^6 + 3*(a^5*b^4*c^2 - 2*a^6*b^3*c*d + a^7*b^2*d^2)*x^4 + 3*(a^6
*b^3*c^2 - 2*a^7*b^2*c*d + a^8*b*d^2)*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((d*x**2+c)**(3/2)/(b*x**2+a)**4,x)

[Out]

Timed out

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Giac [B]  time = 23.1522, 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((d*x^2+c)^(3/2)/(b*x^2+a)^4,x, algorithm="giac")

[Out]

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