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

Optimal. Leaf size=152 \[ -\frac{\sqrt{d x-c} \left (2 a d^2+3 b c^2\right )}{2 d^5 \sqrt{c+d x}}-\frac{c \left (2 a d^2+3 b c^2\right )}{2 d^5 \sqrt{d x-c} \sqrt{c+d x}}+\frac{\left (2 a d^2+3 b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{d^5}+\frac{b x^3}{2 d^2 \sqrt{d x-c} \sqrt{c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.11468, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {460, 89, 12, 78, 63, 217, 206} \[ -\frac{\sqrt{d x-c} \left (2 a d^2+3 b c^2\right )}{2 d^5 \sqrt{c+d x}}-\frac{c \left (2 a d^2+3 b c^2\right )}{2 d^5 \sqrt{d x-c} \sqrt{c+d x}}+\frac{\left (2 a d^2+3 b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{d^5}+\frac{b x^3}{2 d^2 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 460

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1))/(b1*b2*e*
(m + n*(p + 1) + 1)), x] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(b1*b2*(m + n*(p + 1) + 1)), I
nt[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] &&
EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 78

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

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

Mathematica [A]  time = 0.107926, size = 90, normalized size = 0.59 \[ \frac{c \sqrt{1-\frac{d^2 x^2}{c^2}} \left (2 a d^2+3 b c^2\right ) \sin ^{-1}\left (\frac{d x}{c}\right )-2 a d^3 x-3 b c^2 d x+b d^3 x^3}{2 d^5 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*c^2*d*x - 2*a*d^3*x + b*d^3*x^3 + c*(3*b*c^2 + 2*a*d^2)*Sqrt[1 - (d^2*x^2)/c^2]*ArcSin[(d*x)/c])/(2*d^5*
Sqrt[-c + d*x]*Sqrt[c + d*x])

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Maple [C]  time = 0.022, size = 254, normalized size = 1.7 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{2\,{d}^{5}} \left ({\it csgn} \left ( d \right ){x}^{3}b{d}^{3}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+2\,\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ){x}^{2}a{d}^{4}+3\,\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ){x}^{2}b{c}^{2}{d}^{2}-2\,{\it csgn} \left ( d \right ){d}^{3}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}xa-3\,{\it csgn} \left ( d \right ) d\sqrt{{d}^{2}{x}^{2}-{c}^{2}}xb{c}^{2}-2\,\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ) a{c}^{2}{d}^{2}-3\,\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ) b{c}^{4} \right ){\frac{1}{\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{dx-c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(csgn(d)*x^3*b*d^3*(d^2*x^2-c^2)^(1/2)+2*ln(((d^2*x^2-c^2)^(1/2)*csgn(d)+d*x)*csgn(d))*x^2*a*d^4+3*ln(((d^
2*x^2-c^2)^(1/2)*csgn(d)+d*x)*csgn(d))*x^2*b*c^2*d^2-2*csgn(d)*d^3*(d^2*x^2-c^2)^(1/2)*x*a-3*csgn(d)*d*(d^2*x^
2-c^2)^(1/2)*x*b*c^2-2*ln(((d^2*x^2-c^2)^(1/2)*csgn(d)+d*x)*csgn(d))*a*c^2*d^2-3*ln(((d^2*x^2-c^2)^(1/2)*csgn(
d)+d*x)*csgn(d))*b*c^4)*csgn(d)/(d^2*x^2-c^2)^(1/2)/d^5/(d*x+c)^(1/2)/(d*x-c)^(1/2)

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Maxima [A]  time = 0.955846, size = 211, normalized size = 1.39 \begin{align*} \frac{b x^{3}}{2 \, \sqrt{d^{2} x^{2} - c^{2}} d^{2}} - \frac{3 \, b c^{2} x}{2 \, \sqrt{d^{2} x^{2} - c^{2}} d^{4}} - \frac{a x}{\sqrt{d^{2} x^{2} - c^{2}} d^{2}} + \frac{3 \, b c^{2} \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - c^{2}} \sqrt{d^{2}}\right )}{2 \, \sqrt{d^{2}} d^{4}} + \frac{a \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - c^{2}} \sqrt{d^{2}}\right )}{\sqrt{d^{2}} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.54815, size = 325, normalized size = 2.14 \begin{align*} \frac{2 \, b c^{4} + 2 \, a c^{2} d^{2} - 2 \,{\left (b c^{2} d^{2} + a d^{4}\right )} x^{2} +{\left (b d^{3} x^{3} -{\left (3 \, b c^{2} d + 2 \, a d^{3}\right )} x\right )} \sqrt{d x + c} \sqrt{d x - c} +{\left (3 \, b c^{4} + 2 \, a c^{2} d^{2} -{\left (3 \, b c^{2} d^{2} + 2 \, a d^{4}\right )} x^{2}\right )} \log \left (-d x + \sqrt{d x + c} \sqrt{d x - c}\right )}{2 \,{\left (d^{7} x^{2} - c^{2} d^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*c^4 + 2*a*c^2*d^2 - 2*(b*c^2*d^2 + a*d^4)*x^2 + (b*d^3*x^3 - (3*b*c^2*d + 2*a*d^3)*x)*sqrt(d*x + c)*s
qrt(d*x - c) + (3*b*c^4 + 2*a*c^2*d^2 - (3*b*c^2*d^2 + 2*a*d^4)*x^2)*log(-d*x + sqrt(d*x + c)*sqrt(d*x - c)))/
(d^7*x^2 - c^2*d^5)

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Sympy [C]  time = 66.7475, size = 212, normalized size = 1.39 \begin{align*} a \left (\frac{{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{4} & - \frac{1}{2}, \frac{1}{2}, 1, 1 \\- \frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}, 1, 0 & \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} d^{3}} + \frac{i{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{3}{2}, -1, - \frac{3}{4}, - \frac{1}{2}, - \frac{1}{4}, 1 & \\- \frac{3}{4}, - \frac{1}{4} & - \frac{3}{2}, -1, 0, 0 \end{matrix} \middle |{\frac{c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} d^{3}}\right ) + b \left (\frac{c^{2}{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{5}{4}, - \frac{3}{4} & - \frac{3}{2}, - \frac{1}{2}, 0, 1 \\- \frac{5}{4}, -1, - \frac{3}{4}, - \frac{1}{2}, 0, 0 & \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} d^{5}} + \frac{i c^{2}{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{5}{2}, -2, - \frac{7}{4}, - \frac{3}{2}, - \frac{5}{4}, 1 & \\- \frac{7}{4}, - \frac{5}{4} & - \frac{5}{2}, -2, -1, 0 \end{matrix} \middle |{\frac{c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} d^{5}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(meijerg(((-1/4, 1/4), (-1/2, 1/2, 1, 1)), ((-1/4, 0, 1/4, 1/2, 1, 0), ()), c**2/(d**2*x**2))/(2*pi**(3/2)*d
**3) + I*meijerg(((-3/2, -1, -3/4, -1/2, -1/4, 1), ()), ((-3/4, -1/4), (-3/2, -1, 0, 0)), c**2*exp_polar(2*I*p
i)/(d**2*x**2))/(2*pi**(3/2)*d**3)) + b*(c**2*meijerg(((-5/4, -3/4), (-3/2, -1/2, 0, 1)), ((-5/4, -1, -3/4, -1
/2, 0, 0), ()), c**2/(d**2*x**2))/(2*pi**(3/2)*d**5) + I*c**2*meijerg(((-5/2, -2, -7/4, -3/2, -5/4, 1), ()), (
(-7/4, -5/4), (-5/2, -2, -1, 0)), c**2*exp_polar(2*I*pi)/(d**2*x**2))/(2*pi**(3/2)*d**5))

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Giac [A]  time = 1.28983, size = 208, normalized size = 1.37 \begin{align*} -\frac{{\left ({\left (3 \, b d^{15} - \frac{{\left (d x + c\right )} b d^{15}}{c}\right )}{\left (d x + c\right )} - \frac{b c^{2} d^{15} - a d^{17}}{c}\right )} \sqrt{d x + c}}{768 \, \sqrt{d x - c}} - \frac{{\left (3 \, b c^{2} d^{15} + 2 \, a d^{17}\right )} \log \left ({\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}\right )}{768 \, c} - \frac{2 \,{\left (b c^{3} + a c d^{2}\right )}}{{\left ({\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2} + 2 \, c\right )} d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*((3*b*d^15 - (d*x + c)*b*d^15/c)*(d*x + c) - (b*c^2*d^15 - a*d^17)/c)*sqrt(d*x + c)/sqrt(d*x - c) - 1/7
68*(3*b*c^2*d^15 + 2*a*d^17)*log((sqrt(d*x + c) - sqrt(d*x - c))^2)/c - 2*(b*c^3 + a*c*d^2)/(((sqrt(d*x + c) -
sqrt(d*x - c))^2 + 2*c)*d^5)