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

Optimal. Leaf size=217 \[ \frac{x \sqrt{c+d x^2} \left (8 a^2 d^2-18 a b c d+11 b^2 c^2\right )}{16 b^3}+\frac{\left (40 a^2 b c d^2-16 a^3 d^3-30 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 b^4 \sqrt{d}}+\frac{d x^3 \sqrt{c+d x^2} (3 b c-2 a d)}{8 b^2}-\frac{\sqrt{a} (b c-a d)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^4}+\frac{d x^3 \left (c+d x^2\right )^{3/2}}{6 b} \]

[Out]

((11*b^2*c^2 - 18*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*b^3) + (d*(3*b*c - 2*a*d)*x^3*Sqrt[c + d*x^2])/(
8*b^2) + (d*x^3*(c + d*x^2)^(3/2))/(6*b) - (Sqrt[a]*(b*c - a*d)^(5/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt
[c + d*x^2])])/b^4 + ((5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*ArcTanh[(Sqrt[d]*x)/Sqrt[c +
d*x^2]])/(16*b^4*Sqrt[d])

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Rubi [A]  time = 0.392116, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {477, 581, 582, 523, 217, 206, 377, 205} \[ \frac{x \sqrt{c+d x^2} \left (8 a^2 d^2-18 a b c d+11 b^2 c^2\right )}{16 b^3}+\frac{\left (40 a^2 b c d^2-16 a^3 d^3-30 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 b^4 \sqrt{d}}+\frac{d x^3 \sqrt{c+d x^2} (3 b c-2 a d)}{8 b^2}-\frac{\sqrt{a} (b c-a d)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^4}+\frac{d x^3 \left (c+d x^2\right )^{3/2}}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*b^2*c^2 - 18*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*b^3) + (d*(3*b*c - 2*a*d)*x^3*Sqrt[c + d*x^2])/(
8*b^2) + (d*x^3*(c + d*x^2)^(3/2))/(6*b) - (Sqrt[a]*(b*c - a*d)^(5/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt
[c + d*x^2])])/b^4 + ((5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*ArcTanh[(Sqrt[d]*x)/Sqrt[c +
d*x^2]])/(16*b^4*Sqrt[d])

Rule 477

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

Rule 581

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

Rule 582

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

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{x^2 \left (c+d x^2\right )^{5/2}}{a+b x^2} \, dx &=\frac{d x^3 \left (c+d x^2\right )^{3/2}}{6 b}+\frac{\int \frac{x^2 \sqrt{c+d x^2} \left (3 c (2 b c-a d)+3 d (3 b c-2 a d) x^2\right )}{a+b x^2} \, dx}{6 b}\\ &=\frac{d (3 b c-2 a d) x^3 \sqrt{c+d x^2}}{8 b^2}+\frac{d x^3 \left (c+d x^2\right )^{3/2}}{6 b}+\frac{\int \frac{x^2 \left (3 c \left (8 b^2 c^2-13 a b c d+6 a^2 d^2\right )+3 d \left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x^2\right )}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{24 b^2}\\ &=\frac{\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 b^3}+\frac{d (3 b c-2 a d) x^3 \sqrt{c+d x^2}}{8 b^2}+\frac{d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac{\int \frac{3 a c d \left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right )-3 d \left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) x^2}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{48 b^3 d}\\ &=\frac{\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 b^3}+\frac{d (3 b c-2 a d) x^3 \sqrt{c+d x^2}}{8 b^2}+\frac{d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac{\left (a (b c-a d)^3\right ) \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{b^4}+\frac{\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{16 b^4}\\ &=\frac{\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 b^3}+\frac{d (3 b c-2 a d) x^3 \sqrt{c+d x^2}}{8 b^2}+\frac{d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac{\left (a (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{b^4}+\frac{\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{16 b^4}\\ &=\frac{\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 b^3}+\frac{d (3 b c-2 a d) x^3 \sqrt{c+d x^2}}{8 b^2}+\frac{d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac{\sqrt{a} (b c-a d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^4}+\frac{\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 b^4 \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 0.144749, size = 187, normalized size = 0.86 \[ \frac{b x \sqrt{c+d x^2} \left (24 a^2 d^2-6 a b d \left (9 c+2 d x^2\right )+b^2 \left (33 c^2+26 c d x^2+8 d^2 x^4\right )\right )+\frac{3 \left (40 a^2 b c d^2-16 a^3 d^3-30 a b^2 c^2 d+5 b^3 c^3\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{\sqrt{d}}-48 \sqrt{a} (b c-a d)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{48 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*Sqrt[c + d*x^2]*(24*a^2*d^2 - 6*a*b*d*(9*c + 2*d*x^2) + b^2*(33*c^2 + 26*c*d*x^2 + 8*d^2*x^4)) - 48*Sqrt[
a]*(b*c - a*d)^(5/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])] + (3*(5*b^3*c^3 - 30*a*b^2*c^2*d +
40*a^2*b*c*d^2 - 16*a^3*d^3)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/Sqrt[d])/(48*b^4)

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Maple [B]  time = 0.012, size = 3235, normalized size = 14.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*a^3/(-a*b)^(1/2)/b^3/(-(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-3/2*a^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)))*d*c^2-3/2*a^3/(-a*b)^(1/2)/b^3/(-(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+3/2*a^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)))*d*c^2+1/6/b*x*(d*x^2+c)^(5/2
)-1/2*a^4/(-a*b)^(1/2)/b^4/(-(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/(-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)))*c^3-1/2*a/(-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)))*c^3-7/16*a/b^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-a^2/(-a*b)^(1/2)/b^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-7/16*a/b^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+a^2/(-a*b)^(1/2)/b^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*a^4/(-a*b)^(1/2)/b^4/(-(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^3-1/8*a/b^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*a/b^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^2-1/10*a/(-
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)^(5/2)-1/2*a^3/b^4*
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))+1/10*a/(-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)^(5/2)-1/2*a^3/b^4*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))+5/24/b*c*x*(d*x^2+c)^(3/2)
+5/16/b*c^2*x*(d*x^2+c)^(1/2)+5/16/b*c^3/d^(1/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))-15/16*a/b^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*a^2/(-a*b)^(1/2)/b^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*a/(-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)^(3/2)*c+1/4*a^2/b^3*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^2/b^3*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+1/4*a^2/b^3*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^2/b^3*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
+1/2*a^3/(-a*b)^(1/2)/b^3*((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*a/(-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)
*c^2-1/2*a/(-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)
*c^2-1/2*a^3/(-a*b)^(1/2)/b^3*((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^2-1/8*a/b^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+1/
6*a/(-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)^(3/2)*c-1/6*
a^2/(-a*b)^(1/2)/b^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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 16.7969, size = 2535, normalized size = 11.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*
sqrt(d)*x - c) - 24*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*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)) - 2*(8*b^3*d^3*x^5 + 2*(13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c
^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d), -1/48*(3*(5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^
2*b*c*d^2 - 16*a^3*d^3)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 12*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*s
qrt(-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)) - (8*b^3*d^3*x^5
+ 2*(13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b
^4*d), -1/96*(48*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*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)) + 3*(5*b^3*c^3 - 30*a*b
^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(8*b^3*d^3
*x^5 + 2*(13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c)
)/(b^4*d), -1/48*(24*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*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)) + 3*(5*b^3*c^3 - 30
*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (8*b^3*d^3*x^5 + 2*(
13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14543, size = 373, normalized size = 1.72 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (\frac{4 \, d^{2} x^{2}}{b} + \frac{13 \, b^{9} c d^{5} - 6 \, a b^{8} d^{6}}{b^{10} d^{4}}\right )} x^{2} + \frac{3 \,{\left (11 \, b^{9} c^{2} d^{4} - 18 \, a b^{8} c d^{5} + 8 \, a^{2} b^{7} d^{6}\right )}}{b^{10} d^{4}}\right )} \sqrt{d x^{2} + c} x + \frac{{\left (a b^{3} c^{3} \sqrt{d} - 3 \, a^{2} b^{2} c^{2} d^{\frac{3}{2}} + 3 \, a^{3} b c d^{\frac{5}{2}} - a^{4} 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}} b^{4}} - \frac{{\left (5 \, b^{3} c^{3} - 30 \, a b^{2} c^{2} d + 40 \, a^{2} b c d^{2} - 16 \, a^{3} d^{3}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{32 \, b^{4} \sqrt{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*d^2*x^2/b + (13*b^9*c*d^5 - 6*a*b^8*d^6)/(b^10*d^4))*x^2 + 3*(11*b^9*c^2*d^4 - 18*a*b^8*c*d^5 + 8*a
^2*b^7*d^6)/(b^10*d^4))*sqrt(d*x^2 + c)*x + (a*b^3*c^3*sqrt(d) - 3*a^2*b^2*c^2*d^(3/2) + 3*a^3*b*c*d^(5/2) - a
^4*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)*b^4) - 1/32*(5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*log((sqrt(d)*x - sqrt(d*x^
2 + c))^2)/(b^4*sqrt(d))