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

Optimal. Leaf size=115 \[ -\frac{a \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac{a \sqrt{c+d x^2} (b c-a d)}{b^3}+\frac{a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{b^{7/2}}+\frac{\left (c+d x^2\right )^{5/2}}{5 b d} \]

[Out]

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

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Rubi [A]  time = 0.105579, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{a \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac{a \sqrt{c+d x^2} (b c-a d)}{b^3}+\frac{a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{b^{7/2}}+\frac{\left (c+d x^2\right )^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.091145, size = 108, normalized size = 0.94 \[ \frac{\sqrt{c+d x^2} \left (15 a^2 d^2-5 a b d \left (4 c+d x^2\right )+3 b^2 \left (c+d x^2\right )^2\right )}{15 b^3 d}+\frac{a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(15*a^2*d^2 + 3*b^2*(c + d*x^2)^2 - 5*a*b*d*(4*c + d*x^2)))/(15*b^3*d) + (a*(b*c - a*d)^(3/2)
*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/b^(7/2)

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Maple [B]  time = 0.013, size = 1897, normalized size = 16.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(d*x^2+c)^(5/2)/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)+1/4/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)^(1/2)*x+3/4/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+1/2/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-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-1/2/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
))+1/2/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^2-1/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+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^2-1/6/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)-1/4/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)^(
1/2)*x-3/4/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+1/2/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-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+1/2/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))+1/2
/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^2-1/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+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^2

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82952, size = 846, normalized size = 7.36 \begin{align*} \left [-\frac{15 \,{\left (a b c d - a^{2} d^{2}\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 (3 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} +{\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{60 \, b^{3} d}, \frac{15 \,{\left (a b c d - a^{2} d^{2}\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 (3 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} +{\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{30 \, b^{3} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/60*(15*(a*b*c*d - a^2*d^2)*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*(3*b^2*d^2*x^4 + 3*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2 + (6*b^2*c*d - 5*a*b*d^2)*x^2)*sqrt(d*x
^2 + c))/(b^3*d), 1/30*(15*(a*b*c*d - a^2*d^2)*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*(3*b^2*d^2*x^4 + 3*b^2*c^2 - 20*a*b*
c*d + 15*a^2*d^2 + (6*b^2*c*d - 5*a*b*d^2)*x^2)*sqrt(d*x^2 + c))/(b^3*d)]

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Sympy [A]  time = 30.7343, size = 104, normalized size = 0.9 \begin{align*} - \frac{a \left (c + d x^{2}\right )^{\frac{3}{2}}}{3 b^{2}} - \frac{a \left (a d - b c\right )^{2} \operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{b^{4} \sqrt{\frac{a d - b c}{b}}} + \frac{\left (c + d x^{2}\right )^{\frac{5}{2}}}{5 b d} + \frac{\sqrt{c + d x^{2}} \left (a^{2} d - a b c\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.61985, size = 204, normalized size = 1.77 \begin{align*} -\frac{{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\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^{3}} + \frac{3 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} b^{4} d^{4} - 5 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b^{3} d^{5} - 15 \, \sqrt{d x^{2} + c} a b^{3} c d^{5} + 15 \, \sqrt{d x^{2} + c} a^{2} b^{2} d^{6}}{15 \, b^{5} d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3)
+ 1/15*(3*(d*x^2 + c)^(5/2)*b^4*d^4 - 5*(d*x^2 + c)^(3/2)*a*b^3*d^5 - 15*sqrt(d*x^2 + c)*a*b^3*c*d^5 + 15*sqr
t(d*x^2 + c)*a^2*b^2*d^6)/(b^5*d^5)