3.368 \(\int \frac{x^4 (a+b x^2)}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=161 \[ -\frac{x^3 \left (4 a d^2+5 b c^2\right )}{4 d^4 \sqrt{d x-c} \sqrt{c+d x}}+\frac{3 x \sqrt{d x-c} \sqrt{c+d x} \left (4 a d^2+5 b c^2\right )}{8 d^6}+\frac{3 c^2 \left (4 a d^2+5 b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{4 d^7}+\frac{b x^5}{4 d^2 \sqrt{d x-c} \sqrt{c+d x}} \]

[Out]

-((5*b*c^2 + 4*a*d^2)*x^3)/(4*d^4*Sqrt[-c + d*x]*Sqrt[c + d*x]) + (b*x^5)/(4*d^2*Sqrt[-c + d*x]*Sqrt[c + d*x])
+ (3*(5*b*c^2 + 4*a*d^2)*x*Sqrt[-c + d*x]*Sqrt[c + d*x])/(8*d^6) + (3*c^2*(5*b*c^2 + 4*a*d^2)*ArcTanh[Sqrt[-c
+ d*x]/Sqrt[c + d*x]])/(4*d^7)

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Rubi [A]  time = 0.122936, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {460, 98, 21, 90, 12, 63, 217, 206} \[ -\frac{x^3 \left (4 a d^2+5 b c^2\right )}{4 d^4 \sqrt{d x-c} \sqrt{c+d x}}+\frac{3 x \sqrt{d x-c} \sqrt{c+d x} \left (4 a d^2+5 b c^2\right )}{8 d^6}+\frac{3 c^2 \left (4 a d^2+5 b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{4 d^7}+\frac{b x^5}{4 d^2 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^2))/((-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]

[Out]

-((5*b*c^2 + 4*a*d^2)*x^3)/(4*d^4*Sqrt[-c + d*x]*Sqrt[c + d*x]) + (b*x^5)/(4*d^2*Sqrt[-c + d*x]*Sqrt[c + d*x])
+ (3*(5*b*c^2 + 4*a*d^2)*x*Sqrt[-c + d*x]*Sqrt[c + d*x])/(8*d^6) + (3*c^2*(5*b*c^2 + 4*a*d^2)*ArcTanh[Sqrt[-c
+ d*x]/Sqrt[c + d*x]])/(4*d^7)

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 98

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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^4 \left (a+b x^2\right )}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx &=\frac{b x^5}{4 d^2 \sqrt{-c+d x} \sqrt{c+d x}}-\frac{1}{4} \left (-4 a-\frac{5 b c^2}{d^2}\right ) \int \frac{x^4}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\\ &=-\frac{\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{b x^5}{4 d^2 \sqrt{-c+d x} \sqrt{c+d x}}-\frac{\left (4 a+\frac{5 b c^2}{d^2}\right ) \int \frac{x^2 \left (-3 c^2-3 c d x\right )}{\sqrt{-c+d x} (c+d x)^{3/2}} \, dx}{4 c d^2}\\ &=-\frac{\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{b x^5}{4 d^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{\left (3 \left (5 b c^2+4 a d^2\right )\right ) \int \frac{x^2}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{4 d^4}\\ &=-\frac{\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{b x^5}{4 d^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{3 \left (5 b c^2+4 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{8 d^6}+\frac{\left (3 \left (5 b c^2+4 a d^2\right )\right ) \int \frac{c^2}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{8 d^6}\\ &=-\frac{\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{b x^5}{4 d^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{3 \left (5 b c^2+4 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{8 d^6}+\frac{\left (3 c^2 \left (5 b c^2+4 a d^2\right )\right ) \int \frac{1}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{8 d^6}\\ &=-\frac{\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{b x^5}{4 d^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{3 \left (5 b c^2+4 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{8 d^6}+\frac{\left (3 c^2 \left (5 b c^2+4 a d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c+x^2}} \, dx,x,\sqrt{-c+d x}\right )}{4 d^7}\\ &=-\frac{\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{b x^5}{4 d^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{3 \left (5 b c^2+4 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{8 d^6}+\frac{\left (3 c^2 \left (5 b c^2+4 a d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt{-c+d x}}{\sqrt{c+d x}}\right )}{4 d^7}\\ &=-\frac{\left (5 b c^2+4 a d^2\right ) x^3}{4 d^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{b x^5}{4 d^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{3 \left (5 b c^2+4 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{8 d^6}+\frac{3 c^2 \left (5 b c^2+4 a d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{-c+d x}}{\sqrt{c+d x}}\right )}{4 d^7}\\ \end{align*}

Mathematica [A]  time = 0.141607, size = 119, normalized size = 0.74 \[ \frac{3 c^3 \sqrt{1-\frac{d^2 x^2}{c^2}} \left (4 a d^2+5 b c^2\right ) \sin ^{-1}\left (\frac{d x}{c}\right )+4 a d^3 x \left (d^2 x^2-3 c^2\right )+b d x \left (5 c^2 d^2 x^2-15 c^4+2 d^4 x^4\right )}{8 d^7 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*x^2))/((-c + d*x)^(3/2)*(c + d*x)^(3/2)),x]

[Out]

(4*a*d^3*x*(-3*c^2 + d^2*x^2) + b*d*x*(-15*c^4 + 5*c^2*d^2*x^2 + 2*d^4*x^4) + 3*c^3*(5*b*c^2 + 4*a*d^2)*Sqrt[1
- (d^2*x^2)/c^2]*ArcSin[(d*x)/c])/(8*d^7*Sqrt[-c + d*x]*Sqrt[c + d*x])

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Maple [C]  time = 0.03, size = 316, normalized size = 2. \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{8\,{d}^{7}} \left ( 2\,{\it csgn} \left ( d \right ){x}^{5}b{d}^{5}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+4\,{\it csgn} \left ( d \right ){x}^{3}a{d}^{5}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+5\,{\it csgn} \left ( d \right ){x}^{3}b{c}^{2}{d}^{3}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+12\,\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{c}^{2}{d}^{4}+15\,\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}^{4}{d}^{2}-12\,{\it csgn} \left ( d \right ){d}^{3}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}xa{c}^{2}-15\,{\it csgn} \left ( d \right ) d\sqrt{{d}^{2}{x}^{2}-{c}^{2}}xb{c}^{4}-12\,\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}^{4}{d}^{2}-15\,\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}^{6} \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^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x)

[Out]

1/8*(2*csgn(d)*x^5*b*d^5*(d^2*x^2-c^2)^(1/2)+4*csgn(d)*x^3*a*d^5*(d^2*x^2-c^2)^(1/2)+5*csgn(d)*x^3*b*c^2*d^3*(
d^2*x^2-c^2)^(1/2)+12*ln(((d^2*x^2-c^2)^(1/2)*csgn(d)+d*x)*csgn(d))*x^2*a*c^2*d^4+15*ln(((d^2*x^2-c^2)^(1/2)*c
sgn(d)+d*x)*csgn(d))*x^2*b*c^4*d^2-12*csgn(d)*d^3*(d^2*x^2-c^2)^(1/2)*x*a*c^2-15*csgn(d)*d*(d^2*x^2-c^2)^(1/2)
*x*b*c^4-12*ln(((d^2*x^2-c^2)^(1/2)*csgn(d)+d*x)*csgn(d))*a*c^4*d^2-15*ln(((d^2*x^2-c^2)^(1/2)*csgn(d)+d*x)*cs
gn(d))*b*c^6)*csgn(d)/(d^2*x^2-c^2)^(1/2)/d^7/(d*x+c)^(1/2)/(d*x-c)^(1/2)

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Maxima [A]  time = 0.951779, size = 289, normalized size = 1.8 \begin{align*} \frac{b x^{5}}{4 \, \sqrt{d^{2} x^{2} - c^{2}} d^{2}} + \frac{5 \, b c^{2} x^{3}}{8 \, \sqrt{d^{2} x^{2} - c^{2}} d^{4}} + \frac{a x^{3}}{2 \, \sqrt{d^{2} x^{2} - c^{2}} d^{2}} - \frac{15 \, b c^{4} x}{8 \, \sqrt{d^{2} x^{2} - c^{2}} d^{6}} - \frac{3 \, a c^{2} x}{2 \, \sqrt{d^{2} x^{2} - c^{2}} d^{4}} + \frac{15 \, b c^{4} \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - c^{2}} \sqrt{d^{2}}\right )}{8 \, \sqrt{d^{2}} d^{6}} + \frac{3 \, a 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51211, size = 390, normalized size = 2.42 \begin{align*} \frac{8 \, b c^{6} + 8 \, a c^{4} d^{2} - 8 \,{\left (b c^{4} d^{2} + a c^{2} d^{4}\right )} x^{2} +{\left (2 \, b d^{5} x^{5} +{\left (5 \, b c^{2} d^{3} + 4 \, a d^{5}\right )} x^{3} - 3 \,{\left (5 \, b c^{4} d + 4 \, a c^{2} d^{3}\right )} x\right )} \sqrt{d x + c} \sqrt{d x - c} + 3 \,{\left (5 \, b c^{6} + 4 \, a c^{4} d^{2} -{\left (5 \, b c^{4} d^{2} + 4 \, a c^{2} d^{4}\right )} x^{2}\right )} \log \left (-d x + \sqrt{d x + c} \sqrt{d x - c}\right )}{8 \,{\left (d^{9} x^{2} - c^{2} d^{7}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/8*(8*b*c^6 + 8*a*c^4*d^2 - 8*(b*c^4*d^2 + a*c^2*d^4)*x^2 + (2*b*d^5*x^5 + (5*b*c^2*d^3 + 4*a*d^5)*x^3 - 3*(5
*b*c^4*d + 4*a*c^2*d^3)*x)*sqrt(d*x + c)*sqrt(d*x - c) + 3*(5*b*c^6 + 4*a*c^4*d^2 - (5*b*c^4*d^2 + 4*a*c^2*d^4
)*x^2)*log(-d*x + sqrt(d*x + c)*sqrt(d*x - c)))/(d^9*x^2 - c^2*d^7)

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Sympy [C]  time = 99.7226, size = 233, normalized size = 1.45 \begin{align*} a \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 ) + b \left (\frac{c^{4}{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{9}{4}, - \frac{7}{4} & - \frac{5}{2}, - \frac{3}{2}, -1, 1 \\- \frac{9}{4}, -2, - \frac{7}{4}, - \frac{3}{2}, -1, 0 & \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} d^{7}} + \frac{i c^{4}{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{7}{2}, -3, - \frac{11}{4}, - \frac{5}{2}, - \frac{9}{4}, 1 & \\- \frac{11}{4}, - \frac{9}{4} & - \frac{7}{2}, -3, -2, 0 \end{matrix} \middle |{\frac{c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} d^{7}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**2+a)/(d*x-c)**(3/2)/(d*x+c)**(3/2),x)

[Out]

a*(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*p
i**(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)) + b*(c**4*meijerg(((-9/4, -7/4), (-5/2, -3/2, -1, 1)), ((-
9/4, -2, -7/4, -3/2, -1, 0), ()), c**2/(d**2*x**2))/(2*pi**(3/2)*d**7) + I*c**4*meijerg(((-7/2, -3, -11/4, -5/
2, -9/4, 1), ()), ((-11/4, -9/4), (-7/2, -3, -2, 0)), c**2*exp_polar(2*I*pi)/(d**2*x**2))/(2*pi**(3/2)*d**7))

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Giac [A]  time = 1.32636, size = 290, normalized size = 1.8 \begin{align*} -\frac{1}{688128} \,{\left (5 \, b c^{3} d^{35} + 4 \, a c d^{37}\right )} \log \left ({\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}\right ) - \frac{{\left ({\left ({\left (2 \,{\left (5 \, b d^{35} - \frac{{\left (d x + c\right )} b d^{35}}{c}\right )}{\left (d x + c\right )} - \frac{25 \, b c^{2} d^{35} + 4 \, a d^{37}}{c}\right )}{\left (d x + c\right )} + \frac{35 \, b c^{3} d^{35} + 12 \, a c d^{37}}{c}\right )}{\left (d x + c\right )} - \frac{2 \,{\left (7 \, b c^{4} d^{35} + 2 \, a c^{2} d^{37}\right )}}{c}\right )} \sqrt{d x + c}}{2064384 \, \sqrt{d x - c}} - \frac{2 \,{\left (b c^{5} + a c^{3} d^{2}\right )}}{{\left ({\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2} + 2 \, c\right )} d^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

-1/688128*(5*b*c^3*d^35 + 4*a*c*d^37)*log((sqrt(d*x + c) - sqrt(d*x - c))^2) - 1/2064384*(((2*(5*b*d^35 - (d*x
+ c)*b*d^35/c)*(d*x + c) - (25*b*c^2*d^35 + 4*a*d^37)/c)*(d*x + c) + (35*b*c^3*d^35 + 12*a*c*d^37)/c)*(d*x +
c) - 2*(7*b*c^4*d^35 + 2*a*c^2*d^37)/c)*sqrt(d*x + c)/sqrt(d*x - c) - 2*(b*c^5 + a*c^3*d^2)/(((sqrt(d*x + c) -
sqrt(d*x - c))^2 + 2*c)*d^7)