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

Optimal. Leaf size=144 \[ -\frac{a \left (c+d x^2\right )^{5/2}}{5 b^2}-\frac{a \left (c+d x^2\right )^{3/2} (b c-a d)}{3 b^3}-\frac{a \sqrt{c+d x^2} (b c-a d)^2}{b^4}+\frac{a (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{b^{9/2}}+\frac{\left (c+d x^2\right )^{7/2}}{7 b d} \]

[Out]

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

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Rubi [A]  time = 0.15003, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 80, 50, 63, 208} \[ -\frac{a \left (c+d x^2\right )^{5/2}}{5 b^2}-\frac{a \left (c+d x^2\right )^{3/2} (b c-a d)}{3 b^3}-\frac{a \sqrt{c+d x^2} (b c-a d)^2}{b^4}+\frac{a (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{b^{9/2}}+\frac{\left (c+d x^2\right )^{7/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(b*c - a*d)^2*Sqrt[c + d*x^2])/b^4) - (a*(b*c - a*d)*(c + d*x^2)^(3/2))/(3*b^3) - (a*(c + d*x^2)^(5/2))/(
5*b^2) + (c + d*x^2)^(7/2)/(7*b*d) + (a*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/
b^(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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.278928, size = 136, normalized size = 0.94 \[ -\frac{a \left (c+d x^2\right )^{5/2}}{5 b^2}-\frac{a (b c-a d) \left (\sqrt{b} \sqrt{c+d x^2} \left (-3 a d+4 b c+b d x^2\right )-3 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )\right )}{3 b^{9/2}}+\frac{\left (c+d x^2\right )^{7/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.011, size = 3127, normalized size = 21.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88092, size = 1137, normalized size = 7.9 \begin{align*} \left [\frac{105 \,{\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \,{\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \,{\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \,{\left (15 \, b^{3} d^{3} x^{6} + 15 \, b^{3} c^{3} - 161 \, a b^{2} c^{2} d + 245 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + 3 \,{\left (15 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{4} +{\left (45 \, b^{3} c^{2} d - 77 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{420 \, b^{4} d}, \frac{105 \,{\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b c - a d}{b}}}{2 \,{\left (b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \,{\left (15 \, b^{3} d^{3} x^{6} + 15 \, b^{3} c^{3} - 161 \, a b^{2} c^{2} d + 245 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + 3 \,{\left (15 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{4} +{\left (45 \, b^{3} c^{2} d - 77 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{210 \, b^{4} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/420*(105*(a*b^2*c^2*d - 2*a^2*b*c*d^2 + a^3*d^3)*sqrt((b*c - a*d)/b)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c
*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 + 4*(b^2*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d
)/b))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(15*b^3*d^3*x^6 + 15*b^3*c^3 - 161*a*b^2*c^2*d + 245*a^2*b*c*d^2 - 105*
a^3*d^3 + 3*(15*b^3*c*d^2 - 7*a*b^2*d^3)*x^4 + (45*b^3*c^2*d - 77*a*b^2*c*d^2 + 35*a^2*b*d^3)*x^2)*sqrt(d*x^2
+ c))/(b^4*d), 1/210*(105*(a*b^2*c^2*d - 2*a^2*b*c*d^2 + a^3*d^3)*sqrt(-(b*c - a*d)/b)*arctan(-1/2*(b*d*x^2 +
2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)) + 2*(15*b^3*d^3*x^6 +
15*b^3*c^3 - 161*a*b^2*c^2*d + 245*a^2*b*c*d^2 - 105*a^3*d^3 + 3*(15*b^3*c*d^2 - 7*a*b^2*d^3)*x^4 + (45*b^3*c
^2*d - 77*a*b^2*c*d^2 + 35*a^2*b*d^3)*x^2)*sqrt(d*x^2 + c))/(b^4*d)]

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Sympy [A]  time = 62.0938, size = 144, normalized size = 1. \begin{align*} - \frac{a \left (c + d x^{2}\right )^{\frac{5}{2}}}{5 b^{2}} + \frac{a \left (a d - b c\right )^{3} \operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{b^{5} \sqrt{\frac{a d - b c}{b}}} + \frac{\left (c + d x^{2}\right )^{\frac{7}{2}}}{7 b d} + \frac{\left (c + d x^{2}\right )^{\frac{3}{2}} \left (a^{2} d - a b c\right )}{3 b^{3}} + \frac{\sqrt{c + d x^{2}} \left (- a^{3} d^{2} + 2 a^{2} b c d - a b^{2} c^{2}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(c + d*x**2)**(5/2)/(5*b**2) + a*(a*d - b*c)**3*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(b**5*sqrt((a*d
- b*c)/b)) + (c + d*x**2)**(7/2)/(7*b*d) + (c + d*x**2)**(3/2)*(a**2*d - a*b*c)/(3*b**3) + sqrt(c + d*x**2)*(-
a**3*d**2 + 2*a**2*b*c*d - a*b**2*c**2)/b**4

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Giac [A]  time = 1.13966, size = 308, normalized size = 2.14 \begin{align*} -\frac{{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} b^{4}} + \frac{15 \,{\left (d x^{2} + c\right )}^{\frac{7}{2}} b^{6} d^{6} - 21 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} a b^{5} d^{7} - 35 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b^{5} c d^{7} - 105 \, \sqrt{d x^{2} + c} a b^{5} c^{2} d^{7} + 35 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} b^{4} d^{8} + 210 \, \sqrt{d x^{2} + c} a^{2} b^{4} c d^{8} - 105 \, \sqrt{d x^{2} + c} a^{3} b^{3} d^{9}}{105 \, b^{7} d^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(
-b^2*c + a*b*d)*b^4) + 1/105*(15*(d*x^2 + c)^(7/2)*b^6*d^6 - 21*(d*x^2 + c)^(5/2)*a*b^5*d^7 - 35*(d*x^2 + c)^(
3/2)*a*b^5*c*d^7 - 105*sqrt(d*x^2 + c)*a*b^5*c^2*d^7 + 35*(d*x^2 + c)^(3/2)*a^2*b^4*d^8 + 210*sqrt(d*x^2 + c)*
a^2*b^4*c*d^8 - 105*sqrt(d*x^2 + c)*a^3*b^3*d^9)/(b^7*d^7)