3.955 \(\int \frac{x^3 (a+b x^2)^{5/2}}{\sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=237 \[ \frac{5 (b c-a d)^3 (a d+7 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{128 b^{3/2} d^{9/2}}-\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2} (a d+7 b c)}{48 b d^2}+\frac{5 \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (b c-a d) (a d+7 b c)}{192 b d^3}-\frac{5 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)^2 (a d+7 b c)}{128 b d^4}+\frac{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}}{8 b d} \]

[Out]

(-5*(b*c - a*d)^2*(7*b*c + a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(128*b*d^4) + (5*(b*c - a*d)*(7*b*c + a*d)*(a
+ b*x^2)^(3/2)*Sqrt[c + d*x^2])/(192*b*d^3) - ((7*b*c + a*d)*(a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(48*b*d^2) +
((a + b*x^2)^(7/2)*Sqrt[c + d*x^2])/(8*b*d) + (5*(b*c - a*d)^3*(7*b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])
/(Sqrt[b]*Sqrt[c + d*x^2])])/(128*b^(3/2)*d^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.229626, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {446, 80, 50, 63, 217, 206} \[ \frac{5 (b c-a d)^3 (a d+7 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{128 b^{3/2} d^{9/2}}-\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2} (a d+7 b c)}{48 b d^2}+\frac{5 \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (b c-a d) (a d+7 b c)}{192 b d^3}-\frac{5 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)^2 (a d+7 b c)}{128 b d^4}+\frac{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}}{8 b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*x^2)^(5/2))/Sqrt[c + d*x^2],x]

[Out]

(-5*(b*c - a*d)^2*(7*b*c + a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(128*b*d^4) + (5*(b*c - a*d)*(7*b*c + a*d)*(a
+ b*x^2)^(3/2)*Sqrt[c + d*x^2])/(192*b*d^3) - ((7*b*c + a*d)*(a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(48*b*d^2) +
((a + b*x^2)^(7/2)*Sqrt[c + d*x^2])/(8*b*d) + (5*(b*c - a*d)^3*(7*b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])
/(Sqrt[b]*Sqrt[c + d*x^2])])/(128*b^(3/2)*d^(9/2))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, 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])

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b x^2\right )^{5/2}}{\sqrt{c+d x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x (a+b x)^{5/2}}{\sqrt{c+d x}} \, dx,x,x^2\right )\\ &=\frac{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}}{8 b d}-\frac{(7 b c+a d) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{16 b d}\\ &=-\frac{(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{48 b d^2}+\frac{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}}{8 b d}+\frac{(5 (b c-a d) (7 b c+a d)) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{96 b d^2}\\ &=\frac{5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{192 b d^3}-\frac{(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{48 b d^2}+\frac{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}}{8 b d}-\frac{\left (5 (b c-a d)^2 (7 b c+a d)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx,x,x^2\right )}{128 b d^3}\\ &=-\frac{5 (b c-a d)^2 (7 b c+a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{128 b d^4}+\frac{5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{192 b d^3}-\frac{(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{48 b d^2}+\frac{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}}{8 b d}+\frac{\left (5 (b c-a d)^3 (7 b c+a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )}{256 b d^4}\\ &=-\frac{5 (b c-a d)^2 (7 b c+a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{128 b d^4}+\frac{5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{192 b d^3}-\frac{(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{48 b d^2}+\frac{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}}{8 b d}+\frac{\left (5 (b c-a d)^3 (7 b c+a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x^2}\right )}{128 b^2 d^4}\\ &=-\frac{5 (b c-a d)^2 (7 b c+a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{128 b d^4}+\frac{5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{192 b d^3}-\frac{(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{48 b d^2}+\frac{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}}{8 b d}+\frac{\left (5 (b c-a d)^3 (7 b c+a d)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x^2}}{\sqrt{c+d x^2}}\right )}{128 b^2 d^4}\\ &=-\frac{5 (b c-a d)^2 (7 b c+a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{128 b d^4}+\frac{5 (b c-a d) (7 b c+a d) \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{192 b d^3}-\frac{(7 b c+a d) \left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{48 b d^2}+\frac{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}}{8 b d}+\frac{5 (b c-a d)^3 (7 b c+a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{128 b^{3/2} d^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.574439, size = 214, normalized size = 0.9 \[ \frac{b \sqrt{d} \sqrt{a+b x^2} \left (c+d x^2\right ) \left (a^2 b d^2 \left (118 d x^2-191 c\right )+15 a^3 d^3+a b^2 d \left (265 c^2-172 c d x^2+136 d^2 x^4\right )+b^3 \left (70 c^2 d x^2-105 c^3-56 c d^2 x^4+48 d^3 x^6\right )\right )+15 (a d+7 b c) (b c-a d)^{7/2} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{384 b^2 d^{9/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*x^2)^(5/2))/Sqrt[c + d*x^2],x]

[Out]

(b*Sqrt[d]*Sqrt[a + b*x^2]*(c + d*x^2)*(15*a^3*d^3 + a^2*b*d^2*(-191*c + 118*d*x^2) + a*b^2*d*(265*c^2 - 172*c
*d*x^2 + 136*d^2*x^4) + b^3*(-105*c^3 + 70*c^2*d*x^2 - 56*c*d^2*x^4 + 48*d^3*x^6)) + 15*(b*c - a*d)^(7/2)*(7*b
*c + a*d)*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/(384*b^2*d^(9/
2)*Sqrt[c + d*x^2])

________________________________________________________________________________________

Maple [B]  time = 0.019, size = 770, normalized size = 3.3 \begin{align*} -{\frac{1}{768\,b{d}^{4}}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( -96\,{x}^{6}{b}^{3}{d}^{3}\sqrt{bd}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}-272\,{x}^{4}a{b}^{2}{d}^{3}\sqrt{bd}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+112\,{x}^{4}{b}^{3}c{d}^{2}\sqrt{bd}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}-236\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}{a}^{2}b{d}^{3}\sqrt{bd}+344\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}ac{b}^{2}{d}^{2}\sqrt{bd}-140\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}{c}^{2}{b}^{3}d\sqrt{bd}+15\,{d}^{4}\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{4}+60\,{a}^{3}c\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) b{d}^{3}-270\,\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}{c}^{2}{b}^{2}{d}^{2}+300\,a{c}^{3}\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{3}d-105\,{b}^{4}\ln \left ( 1/2\,{\frac{2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){c}^{4}-30\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{a}^{3}{d}^{3}\sqrt{bd}+382\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{a}^{2}cb{d}^{2}\sqrt{bd}-530\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}a{c}^{2}{b}^{2}d\sqrt{bd}+210\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{c}^{3}{b}^{3}\sqrt{bd} \right ){\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x)

[Out]

-1/768*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-96*x^6*b^3*d^3*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)-272*x^
4*a*b^2*d^3*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+112*x^4*b^3*c*d^2*(b*d)^(1/2)*(b*d*x^4+a*d*x^2+b*c
*x^2+a*c)^(1/2)-236*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*a^2*b*d^3*(b*d)^(1/2)+344*(b*d*x^4+a*d*x^2+b*c*x^2
+a*c)^(1/2)*x^2*a*c*b^2*d^2*(b*d)^(1/2)-140*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*c^2*b^3*d*(b*d)^(1/2)+15*d
^4*ln(1/2*(2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^4+60*a^3*c*ln(1
/2*(2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b*d^3-270*ln(1/2*(2*d*x^
2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*c^2*b^2*d^2+300*a*c^3*ln(1/2*(
2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*d-105*b^4*ln(1/2*(2*d*x^
2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*c^4-30*(b*d*x^4+a*d*x^2+b*c*x^2+a*
c)^(1/2)*a^3*d^3*(b*d)^(1/2)+382*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*a^2*c*b*d^2*(b*d)^(1/2)-530*(b*d*x^4+a*d*
x^2+b*c*x^2+a*c)^(1/2)*a*c^2*b^2*d*(b*d)^(1/2)+210*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*c^3*b^3*(b*d)^(1/2))/b/
d^4/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/(b*d)^(1/2)

________________________________________________________________________________________

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^3*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.27101, size = 1272, normalized size = 5.37 \begin{align*} \left [-\frac{15 \,{\left (7 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - a^{4} d^{4}\right )} \sqrt{b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \,{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{b d}\right ) - 4 \,{\left (48 \, b^{4} d^{4} x^{6} - 105 \, b^{4} c^{3} d + 265 \, a b^{3} c^{2} d^{2} - 191 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \,{\left (7 \, b^{4} c d^{3} - 17 \, a b^{3} d^{4}\right )} x^{4} + 2 \,{\left (35 \, b^{4} c^{2} d^{2} - 86 \, a b^{3} c d^{3} + 59 \, a^{2} b^{2} d^{4}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{1536 \, b^{2} d^{5}}, -\frac{15 \,{\left (7 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - a^{4} d^{4}\right )} \sqrt{-b d} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{-b d}}{2 \,{\left (b^{2} d^{2} x^{4} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) - 2 \,{\left (48 \, b^{4} d^{4} x^{6} - 105 \, b^{4} c^{3} d + 265 \, a b^{3} c^{2} d^{2} - 191 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} - 8 \,{\left (7 \, b^{4} c d^{3} - 17 \, a b^{3} d^{4}\right )} x^{4} + 2 \,{\left (35 \, b^{4} c^{2} d^{2} - 86 \, a b^{3} c d^{3} + 59 \, a^{2} b^{2} d^{4}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{768 \, b^{2} d^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/1536*(15*(7*b^4*c^4 - 20*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - a^4*d^4)*sqrt(b*d)*log(8*b^2*d
^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 - 4*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)
*sqrt(d*x^2 + c)*sqrt(b*d)) - 4*(48*b^4*d^4*x^6 - 105*b^4*c^3*d + 265*a*b^3*c^2*d^2 - 191*a^2*b^2*c*d^3 + 15*a
^3*b*d^4 - 8*(7*b^4*c*d^3 - 17*a*b^3*d^4)*x^4 + 2*(35*b^4*c^2*d^2 - 86*a*b^3*c*d^3 + 59*a^2*b^2*d^4)*x^2)*sqrt
(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^5), -1/768*(15*(7*b^4*c^4 - 20*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 - 4*a^3*b*
c*d^3 - a^4*d^4)*sqrt(-b*d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d)/(b^2
*d^2*x^4 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^2)) - 2*(48*b^4*d^4*x^6 - 105*b^4*c^3*d + 265*a*b^3*c^2*d^2 - 191*a
^2*b^2*c*d^3 + 15*a^3*b*d^4 - 8*(7*b^4*c*d^3 - 17*a*b^3*d^4)*x^4 + 2*(35*b^4*c^2*d^2 - 86*a*b^3*c*d^3 + 59*a^2
*b^2*d^4)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^5)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (a + b x^{2}\right )^{\frac{5}{2}}}{\sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**2+a)**(5/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x**3*(a + b*x**2)**(5/2)/sqrt(c + d*x**2), x)

________________________________________________________________________________________

Giac [A]  time = 1.23741, size = 397, normalized size = 1.68 \begin{align*} \frac{\sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \sqrt{b x^{2} + a}{\left (2 \,{\left (b x^{2} + a\right )}{\left (4 \,{\left (b x^{2} + a\right )}{\left (\frac{6 \,{\left (b x^{2} + a\right )}}{b d} - \frac{7 \, b^{2} c d^{5} + a b d^{6}}{b^{2} d^{7}}\right )} + \frac{5 \,{\left (7 \, b^{3} c^{2} d^{4} - 6 \, a b^{2} c d^{5} - a^{2} b d^{6}\right )}}{b^{2} d^{7}}\right )} - \frac{15 \,{\left (7 \, b^{4} c^{3} d^{3} - 13 \, a b^{3} c^{2} d^{4} + 5 \, a^{2} b^{2} c d^{5} + a^{3} b d^{6}\right )}}{b^{2} d^{7}}\right )} - \frac{15 \,{\left (7 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} - a^{4} d^{4}\right )} \log \left ({\left | -\sqrt{b x^{2} + a} \sqrt{b d} + \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} d^{4}}}{384 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/384*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a)*(4*(b*x^2 + a)*(6*(b*x^2 + a)/(b*d
) - (7*b^2*c*d^5 + a*b*d^6)/(b^2*d^7)) + 5*(7*b^3*c^2*d^4 - 6*a*b^2*c*d^5 - a^2*b*d^6)/(b^2*d^7)) - 15*(7*b^4*
c^3*d^3 - 13*a*b^3*c^2*d^4 + 5*a^2*b^2*c*d^5 + a^3*b*d^6)/(b^2*d^7)) - 15*(7*b^4*c^4 - 20*a*b^3*c^3*d + 18*a^2
*b^2*c^2*d^2 - 4*a^3*b*c*d^3 - a^4*d^4)*log(abs(-sqrt(b*x^2 + a)*sqrt(b*d) + sqrt(b^2*c + (b*x^2 + a)*b*d - a*
b*d)))/(sqrt(b*d)*d^4))/abs(b)