3.702 \(\int \frac{(c+d x^2)^{5/2}}{x^4 (a+b x^2)} \, dx\)

Optimal. Leaf size=130 \[ \frac{c \sqrt{c+d x^2} (b c-2 a d)}{a^2 x}+\frac{(b c-a d)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2} b}-\frac{c \left (c+d x^2\right )^{3/2}}{3 a x^3}+\frac{d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.178362, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {474, 580, 523, 217, 206, 377, 205} \[ \frac{c \sqrt{c+d x^2} (b c-2 a d)}{a^2 x}+\frac{(b c-a d)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2} b}-\frac{c \left (c+d x^2\right )^{3/2}}{3 a x^3}+\frac{d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^(5/2)/(x^4*(a + b*x^2)),x]

[Out]

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

Rule 474

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

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

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

Mathematica [A]  time = 0.096201, size = 125, normalized size = 0.96 \[ \frac{c \sqrt{c+d x^2} \left (3 b c x^2-a \left (c+7 d x^2\right )\right )}{3 a^2 x^3}+\frac{(b c-a d)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2} b}+\frac{d^{5/2} \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^(5/2)/(x^4*(a + b*x^2)),x]

[Out]

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

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Maple [B]  time = 0.013, size = 3346, normalized size = 25.7 \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)^(5/2)/x^4/(b*x^2+a),x)

[Out]

1/2*b^2/a^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^3+4/3/a*d^2/c^2*x*(d*x^2+c)^(5/2)-4/3/a*d/c^2/x*(d*x^2+c)^(7/2)-15/8*b/a^2*d*c*x*(d*x^2+c)^(
1/2)-b/a^2*d/c*x*(d*x^2+c)^(5/2)+b/a/(-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)*d*c-1/2/b*a/(-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)))*d^3-1/2/(-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)*d^2+1/2/b*d^(5/2)*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))+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)*d^2+1/2/b*d^(5/2)*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))-b/a/(-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)*d*c+1/2/b*a/(-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)))*d^3-1/4/a*d^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-5/4/a*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))*c-1/10*b^2/a^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)^(5/2)-1/4/a*d^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-5/4/a*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))*c-1/3/a/c/x^3*(d*x^2
+c)^(7/2)+5/2/a*d^2*x*(d*x^2+c)^(1/2)+5/2/a*d^(3/2)*c*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+1/10*b^2/a^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)^(5/2)-3/2/(-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)))*d^2*c+1/8*b/a^2*d
*((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)*x+15/16*b/a^2*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^2+1/6*b/a/(-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)*d-1/6*b^2/a^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)^(3/2)*c-1/2*b^2/a^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^2+3/2/(-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)))*d^2*c+b/a^2/c/x*(d*x^2+c)^(7/2)-5/4*b/a^2*d*x*(d*x^2+c)^(3/2)-15/8
*b/a^2*d^(1/2)*c^2*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+5/3/a*d^2/c*x*(d*x^2+c)^(3/2)+1/8*b/a^2*d*((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)*x+15/16*b/a^2*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^2-1/6*b/a/(-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)*d+1/6*b^2/a^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)^(3/2)*c+1/2*b^2/a^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^2+7/16*b/a^2*d*c*((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+7/16*b/a^2*d*c*((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
/2*b/a/(-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)))*d*c^2+3/2*b/a/(-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)))*d*c^2-1/2*b^2/a^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^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )}^{\frac{5}{2}}}{{\left (b x^{2} + a\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^4/(b*x^2+a),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^(5/2)/((b*x^2 + a)*x^4), x)

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Fricas [A]  time = 3.50601, size = 1932, normalized size = 14.86 \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)^(5/2)/x^4/(b*x^2+a),x, algorithm="fricas")

[Out]

[1/12*(6*a^2*d^(5/2)*x^3*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x
^3*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)) - 4*(a
*b*c^2 - (3*b^2*c^2 - 7*a*b*c*d)*x^2)*sqrt(d*x^2 + c))/(a^2*b*x^3), -1/12*(12*a^2*sqrt(-d)*d^2*x^3*arctan(sqrt
(-d)*x/sqrt(d*x^2 + c)) - 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^3*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)) + 4*(a*b*c^2 - (3*b^2*c^2 - 7*a*b*c*d)*x^2)*sqrt(d*x^2
+ c))/(a^2*b*x^3), 1/6*(3*a^2*d^(5/2)*x^3*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 3*(b^2*c^2 - 2*a*b
*c*d + a^2*d^2)*x^3*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)) - 2*(a*b*c^2 - (3*b^2*c^2 - 7*a*b*c*d)*x^2)*sqrt(d*x^2 + c))/(a^
2*b*x^3), -1/6*(6*a^2*sqrt(-d)*d^2*x^3*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*
x^3*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)) + 2*(a*b*c^2 - (3*b^2*c^2 - 7*a*b*c*d)*x^2)*sqrt(d*x^2 + c))/(a^2*b*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x^{2}\right )^{\frac{5}{2}}}{x^{4} \left (a + b x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(5/2)/x**4/(b*x**2+a),x)

[Out]

Integral((c + d*x**2)**(5/2)/(x**4*(a + b*x**2)), x)

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Giac [B]  time = 1.2001, size = 410, normalized size = 3.15 \begin{align*} -\frac{d^{\frac{5}{2}} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{2 \, b} - \frac{{\left (b^{3} c^{3} \sqrt{d} - 3 \, a b^{2} c^{2} d^{\frac{3}{2}} + 3 \, a^{2} b c d^{\frac{5}{2}} - a^{3} d^{\frac{7}{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}} a^{2} b} - \frac{2 \,{\left (3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b c^{3} \sqrt{d} - 9 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a c^{2} d^{\frac{3}{2}} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c^{4} \sqrt{d} + 12 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a c^{3} d^{\frac{3}{2}} + 3 \, b c^{5} \sqrt{d} - 7 \, a c^{4} d^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^4/(b*x^2+a),x, algorithm="giac")

[Out]

-1/2*d^(5/2)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/b - (b^3*c^3*sqrt(d) - 3*a*b^2*c^2*d^(3/2) + 3*a^2*b*c*d^(5/
2) - a^3*d^(7/2))*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)*a^2*b) - 2/3*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c^3*sqrt(d) - 9*(sqrt(d)*x - sqrt(d*x^2 +
c))^4*a*c^2*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^4*sqrt(d) + 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*
c^3*d^(3/2) + 3*b*c^5*sqrt(d) - 7*a*c^4*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*a^2)