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

Optimal. Leaf size=210 \[ \frac{x \sqrt{c+d x^2} \left (8 a^2 d^2-10 a b c d+b^2 c^2\right )}{16 b^3 d}-\frac{(b c-2 a d) \left (-8 a^2 d^2+8 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 b^4 d^{3/2}}+\frac{a^{3/2} (b c-a d)^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^4}+\frac{x^3 \sqrt{c+d x^2} (7 b c-6 a d)}{24 b^2}+\frac{d x^5 \sqrt{c+d x^2}}{6 b} \]

[Out]

((b^2*c^2 - 10*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*b^3*d) + ((7*b*c - 6*a*d)*x^3*Sqrt[c + d*x^2])/(24*
b^2) + (d*x^5*Sqrt[c + d*x^2])/(6*b) + (a^(3/2)*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c +
d*x^2])])/b^4 - ((b*c - 2*a*d)*(b^2*c^2 + 8*a*b*c*d - 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*b^
4*d^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.402702, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {477, 582, 523, 217, 206, 377, 205} \[ \frac{x \sqrt{c+d x^2} \left (8 a^2 d^2-10 a b c d+b^2 c^2\right )}{16 b^3 d}-\frac{(b c-2 a d) \left (-8 a^2 d^2+8 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 b^4 d^{3/2}}+\frac{a^{3/2} (b c-a d)^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^4}+\frac{x^3 \sqrt{c+d x^2} (7 b c-6 a d)}{24 b^2}+\frac{d x^5 \sqrt{c+d x^2}}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*c^2 - 10*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*b^3*d) + ((7*b*c - 6*a*d)*x^3*Sqrt[c + d*x^2])/(24*
b^2) + (d*x^5*Sqrt[c + d*x^2])/(6*b) + (a^(3/2)*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c +
d*x^2])])/b^4 - ((b*c - 2*a*d)*(b^2*c^2 + 8*a*b*c*d - 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*b^
4*d^(3/2))

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 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^4 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx &=\frac{d x^5 \sqrt{c+d x^2}}{6 b}+\frac{\int \frac{x^4 \left (c (6 b c-5 a d)+d (7 b c-6 a d) x^2\right )}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{6 b}\\ &=\frac{(7 b c-6 a d) x^3 \sqrt{c+d x^2}}{24 b^2}+\frac{d x^5 \sqrt{c+d x^2}}{6 b}-\frac{\int \frac{x^2 \left (3 a c d (7 b c-6 a d)-3 d \left (b^2 c^2-10 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 d}\\ &=\frac{\left (b^2 c^2-10 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 b^3 d}+\frac{(7 b c-6 a d) x^3 \sqrt{c+d x^2}}{24 b^2}+\frac{d x^5 \sqrt{c+d x^2}}{6 b}+\frac{\int \frac{-3 a c d \left (b^2 c^2-10 a b c d+8 a^2 d^2\right )-3 d (b c-2 a d) \left (b^2 c^2+8 a b c d-8 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{48 b^3 d^2}\\ &=\frac{\left (b^2 c^2-10 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 b^3 d}+\frac{(7 b c-6 a d) x^3 \sqrt{c+d x^2}}{24 b^2}+\frac{d x^5 \sqrt{c+d x^2}}{6 b}+\frac{\left (a^2 (b c-a d)^2\right ) \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{b^4}-\frac{\left ((b c-2 a d) \left (b^2 c^2+8 a b c d-8 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{16 b^4 d}\\ &=\frac{\left (b^2 c^2-10 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 b^3 d}+\frac{(7 b c-6 a d) x^3 \sqrt{c+d x^2}}{24 b^2}+\frac{d x^5 \sqrt{c+d x^2}}{6 b}+\frac{\left (a^2 (b c-a d)^2\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 ((b c-2 a d) \left (b^2 c^2+8 a b c d-8 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{16 b^4 d}\\ &=\frac{\left (b^2 c^2-10 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{16 b^3 d}+\frac{(7 b c-6 a d) x^3 \sqrt{c+d x^2}}{24 b^2}+\frac{d x^5 \sqrt{c+d x^2}}{6 b}+\frac{a^{3/2} (b c-a d)^{3/2} \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^4}-\frac{(b c-2 a d) \left (b^2 c^2+8 a b c d-8 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 b^4 d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.209171, size = 196, normalized size = 0.93 \[ \frac{b \sqrt{d} x \sqrt{c+d x^2} \left (24 a^2 d^2-6 a b d \left (5 c+2 d x^2\right )+b^2 \left (3 c^2+14 c d x^2+8 d^2 x^4\right )\right )-3 \left (-24 a^2 b c d^2+16 a^3 d^3+6 a b^2 c^2 d+b^3 c^3\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )+48 a^{3/2} d^{3/2} (b c-a d)^{3/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{48 b^4 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*x*Sqrt[c + d*x^2]*(24*a^2*d^2 - 6*a*b*d*(5*c + 2*d*x^2) + b^2*(3*c^2 + 14*c*d*x^2 + 8*d^2*x^4)) + 4
8*a^(3/2)*d^(3/2)*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])] - 3*(b^3*c^3 + 6*a*b
^2*c^2*d - 24*a^2*b*c*d^2 + 16*a^3*d^3)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(48*b^4*d^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.019, size = 2081, normalized size = 9.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 13.03, size = 2427, normalized size = 11.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(b^3*c^3 + 6*a*b^2*c^2*d - 24*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*(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*(7*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(b^3*c^2*d - 10*a*b^2*c*d^2
+ 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d^2), 1/48*(3*(b^3*c^3 + 6*a*b^2*c^2*d - 24*a^2*b*c*d^2 + 16*a^3*d^3)*
sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 12*(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)) + (8*b^3*d^3*x^5 + 2*(7*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 +
3*(b^3*c^2*d - 10*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d^2), 1/96*(48*(a*b*c*d^2 - a^2*d^3)*sq
rt(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*(b^3*c^3 + 6*a*b^2*c^2*d - 24*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*(7*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(b^3*c^2*d -
10*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d^2), 1/48*(24*(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*(b^3*c^3 + 6*a*b^2*c^2*d - 24*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*(7*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(b^3*c^2*d - 10*a*b^2*c*d^2 + 8*a^2*b*d^3)*
x)*sqrt(d*x^2 + c))/(b^4*d^2)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \left (c + d x^{2}\right )^{\frac{3}{2}}}{a + b x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.15657, size = 358, normalized size = 1.7 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (\frac{4 \, d x^{2}}{b} + \frac{7 \, b^{9} c d^{4} - 6 \, a b^{8} d^{5}}{b^{10} d^{4}}\right )} x^{2} + \frac{3 \,{\left (b^{9} c^{2} d^{3} - 10 \, a b^{8} c d^{4} + 8 \, a^{2} b^{7} d^{5}\right )}}{b^{10} d^{4}}\right )} \sqrt{d x^{2} + c} x - \frac{{\left (a^{2} b^{2} c^{2} \sqrt{d} - 2 \, a^{3} b c d^{\frac{3}{2}} + a^{4} d^{\frac{5}{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 (b^{3} c^{3} \sqrt{d} + 6 \, a b^{2} c^{2} d^{\frac{3}{2}} - 24 \, a^{2} b c d^{\frac{5}{2}} + 16 \, a^{3} d^{\frac{7}{2}}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{32 \, b^{4} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*d*x^2/b + (7*b^9*c*d^4 - 6*a*b^8*d^5)/(b^10*d^4))*x^2 + 3*(b^9*c^2*d^3 - 10*a*b^8*c*d^4 + 8*a^2*b^7
*d^5)/(b^10*d^4))*sqrt(d*x^2 + c)*x - (a^2*b^2*c^2*sqrt(d) - 2*a^3*b*c*d^(3/2) + a^4*d^(5/2))*arctan(1/2*((sqr
t(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*(b^
3*c^3*sqrt(d) + 6*a*b^2*c^2*d^(3/2) - 24*a^2*b*c*d^(5/2) + 16*a^3*d^(7/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2
)/(b^4*d^2)