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

Optimal. Leaf size=129 \[ -\frac{a^2 \sqrt{c+d x^2}}{b^2 \sqrt{a+b x^2} (b c-a d)}-\frac{(3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{2 b^{5/2} d^{3/2}}+\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.164462, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {446, 89, 80, 63, 217, 206} \[ -\frac{a^2 \sqrt{c+d x^2}}{b^2 \sqrt{a+b x^2} (b c-a d)}-\frac{(3 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{2 b^{5/2} d^{3/2}}+\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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

Mathematica [A]  time = 0.345041, size = 185, normalized size = 1.43 \[ \frac{\sqrt{a+b x^2} \sqrt{b c-a d} \left (-3 a^2 d^2+2 a b c d+b^2 c^2\right ) \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )-b \sqrt{d} \left (c+d x^2\right ) \left (-3 a^2 d+a b \left (c-d x^2\right )+b^2 c x^2\right )}{2 b^3 d^{3/2} \sqrt{a+b x^2} \sqrt{c+d x^2} (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.038, size = 553, normalized size = 4.3 \begin{align*} -{\frac{1}{4\,{b}^{2}d \left ( ad-bc \right ) } \left ( 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 ){x}^{2}{a}^{2}b{d}^{2}-2\,\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 ){x}^{2}a{b}^{2}cd-\ln \left ({\frac{1}{2} \left ( 2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ){x}^{2}{b}^{3}{c}^{2}-2\,\sqrt{bd}\sqrt{ \left ( b{x}^{2}+a \right ) \left ( d{x}^{2}+c \right ) }{x}^{2}abd+2\,\sqrt{bd}\sqrt{ \left ( b{x}^{2}+a \right ) \left ( d{x}^{2}+c \right ) }{x}^{2}{b}^{2}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 ){a}^{3}{d}^{2}-2\,\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}bcd-\ln \left ({\frac{1}{2} \left ( 2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) a{b}^{2}{c}^{2}-6\,\sqrt{bd}\sqrt{ \left ( b{x}^{2}+a \right ) \left ( d{x}^{2}+c \right ) }{a}^{2}d+2\,\sqrt{bd}\sqrt{ \left ( b{x}^{2}+a \right ) \left ( d{x}^{2}+c \right ) }abc \right ) \sqrt{d{x}^{2}+c}{\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) \left ( d{x}^{2}+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.69498, size = 1049, normalized size = 8.13 \begin{align*} \left [\frac{{\left (a b^{2} c^{2} + 2 \, a^{2} b c d - 3 \, a^{3} d^{2} +{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{2}\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 (a b^{2} c d - 3 \, a^{2} b d^{2} +{\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{8 \,{\left (a b^{4} c d^{2} - a^{2} b^{3} d^{3} +{\left (b^{5} c d^{2} - a b^{4} d^{3}\right )} x^{2}\right )}}, \frac{{\left (a b^{2} c^{2} + 2 \, a^{2} b c d - 3 \, a^{3} d^{2} +{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{2}\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 (a b^{2} c d - 3 \, a^{2} b d^{2} +{\left (b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{4 \,{\left (a b^{4} c d^{2} - a^{2} b^{3} d^{3} +{\left (b^{5} c d^{2} - a b^{4} d^{3}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((a*b^2*c^2 + 2*a^2*b*c*d - 3*a^3*d^2 + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*x^2)*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*(a*b^2*c*d - 3*a^2*b*d^2 + (b^3*c*d - a*b^2*d^2)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^
2 + c))/(a*b^4*c*d^2 - a^2*b^3*d^3 + (b^5*c*d^2 - a*b^4*d^3)*x^2), 1/4*((a*b^2*c^2 + 2*a^2*b*c*d - 3*a^3*d^2 +
(b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*x^2)*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*(a*b^2*c*d - 3*a^2*b*d^2 + (b^3*
c*d - a*b^2*d^2)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a*b^4*c*d^2 - a^2*b^3*d^3 + (b^5*c*d^2 - a*b^4*d^3)*x^
2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24991, size = 254, normalized size = 1.97 \begin{align*} -\frac{\frac{8 \, \sqrt{b d} a^{2}}{b^{2} c - a b d -{\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 )}^{2}} - \frac{2 \, \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \sqrt{b x^{2} + a}}{b d} - \frac{{\left (\sqrt{b d} b c + 3 \, \sqrt{b d} a d\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 )}^{2}\right )}{b d^{2}}}{4 \, b{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*sqrt(b*d)*a^2/(b^2*c - a*b*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)
- 2*sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)/(b*d) - (sqrt(b*d)*b*c + 3*sqrt(b*d)*a*d)*log((sqrt(
b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)/(b*d^2))/(b*abs(b))