3.358 \(\int \frac{x^4 (a+b x^2)}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=164 \[ \frac{x^3 \sqrt{d x-c} \sqrt{c+d x} \left (6 a d^2+5 b c^2\right )}{24 d^4}+\frac{c^2 x \sqrt{d x-c} \sqrt{c+d x} \left (6 a d^2+5 b c^2\right )}{16 d^6}+\frac{c^4 \left (6 a d^2+5 b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{8 d^7}+\frac{b x^5 \sqrt{d x-c} \sqrt{c+d x}}{6 d^2} \]

[Out]

(c^2*(5*b*c^2 + 6*a*d^2)*x*Sqrt[-c + d*x]*Sqrt[c + d*x])/(16*d^6) + ((5*b*c^2 + 6*a*d^2)*x^3*Sqrt[-c + d*x]*Sq
rt[c + d*x])/(24*d^4) + (b*x^5*Sqrt[-c + d*x]*Sqrt[c + d*x])/(6*d^2) + (c^4*(5*b*c^2 + 6*a*d^2)*ArcTanh[Sqrt[-
c + d*x]/Sqrt[c + d*x]])/(8*d^7)

________________________________________________________________________________________

Rubi [A]  time = 0.119734, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {460, 100, 12, 90, 63, 217, 206} \[ \frac{x^3 \sqrt{d x-c} \sqrt{c+d x} \left (6 a d^2+5 b c^2\right )}{24 d^4}+\frac{c^2 x \sqrt{d x-c} \sqrt{c+d x} \left (6 a d^2+5 b c^2\right )}{16 d^6}+\frac{c^4 \left (6 a d^2+5 b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{8 d^7}+\frac{b x^5 \sqrt{d x-c} \sqrt{c+d x}}{6 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(5*b*c^2 + 6*a*d^2)*x*Sqrt[-c + d*x]*Sqrt[c + d*x])/(16*d^6) + ((5*b*c^2 + 6*a*d^2)*x^3*Sqrt[-c + d*x]*Sq
rt[c + d*x])/(24*d^4) + (b*x^5*Sqrt[-c + d*x]*Sqrt[c + d*x])/(6*d^2) + (c^4*(5*b*c^2 + 6*a*d^2)*ArcTanh[Sqrt[-
c + d*x]/Sqrt[c + d*x]])/(8*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 100

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 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 )}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx &=\frac{b x^5 \sqrt{-c+d x} \sqrt{c+d x}}{6 d^2}-\frac{1}{6} \left (-6 a-\frac{5 b c^2}{d^2}\right ) \int \frac{x^4}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx\\ &=\frac{\left (5 b c^2+6 a d^2\right ) x^3 \sqrt{-c+d x} \sqrt{c+d x}}{24 d^4}+\frac{b x^5 \sqrt{-c+d x} \sqrt{c+d x}}{6 d^2}+\frac{\left (5 b c^2+6 a d^2\right ) \int \frac{3 c^2 x^2}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{24 d^4}\\ &=\frac{\left (5 b c^2+6 a d^2\right ) x^3 \sqrt{-c+d x} \sqrt{c+d x}}{24 d^4}+\frac{b x^5 \sqrt{-c+d x} \sqrt{c+d x}}{6 d^2}+\frac{\left (c^2 \left (5 b c^2+6 a d^2\right )\right ) \int \frac{x^2}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{8 d^4}\\ &=\frac{c^2 \left (5 b c^2+6 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{16 d^6}+\frac{\left (5 b c^2+6 a d^2\right ) x^3 \sqrt{-c+d x} \sqrt{c+d x}}{24 d^4}+\frac{b x^5 \sqrt{-c+d x} \sqrt{c+d x}}{6 d^2}+\frac{\left (c^2 \left (5 b c^2+6 a d^2\right )\right ) \int \frac{c^2}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{16 d^6}\\ &=\frac{c^2 \left (5 b c^2+6 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{16 d^6}+\frac{\left (5 b c^2+6 a d^2\right ) x^3 \sqrt{-c+d x} \sqrt{c+d x}}{24 d^4}+\frac{b x^5 \sqrt{-c+d x} \sqrt{c+d x}}{6 d^2}+\frac{\left (c^4 \left (5 b c^2+6 a d^2\right )\right ) \int \frac{1}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{16 d^6}\\ &=\frac{c^2 \left (5 b c^2+6 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{16 d^6}+\frac{\left (5 b c^2+6 a d^2\right ) x^3 \sqrt{-c+d x} \sqrt{c+d x}}{24 d^4}+\frac{b x^5 \sqrt{-c+d x} \sqrt{c+d x}}{6 d^2}+\frac{\left (c^4 \left (5 b c^2+6 a d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c+x^2}} \, dx,x,\sqrt{-c+d x}\right )}{8 d^7}\\ &=\frac{c^2 \left (5 b c^2+6 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{16 d^6}+\frac{\left (5 b c^2+6 a d^2\right ) x^3 \sqrt{-c+d x} \sqrt{c+d x}}{24 d^4}+\frac{b x^5 \sqrt{-c+d x} \sqrt{c+d x}}{6 d^2}+\frac{\left (c^4 \left (5 b c^2+6 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 )}{8 d^7}\\ &=\frac{c^2 \left (5 b c^2+6 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{16 d^6}+\frac{\left (5 b c^2+6 a d^2\right ) x^3 \sqrt{-c+d x} \sqrt{c+d x}}{24 d^4}+\frac{b x^5 \sqrt{-c+d x} \sqrt{c+d x}}{6 d^2}+\frac{c^4 \left (5 b c^2+6 a d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{-c+d x}}{\sqrt{c+d x}}\right )}{8 d^7}\\ \end{align*}

Mathematica [A]  time = 0.115963, size = 148, normalized size = 0.9 \[ \frac{d x \left (d^2 x^2-c^2\right ) \left (6 a d^2 \left (3 c^2+2 d^2 x^2\right )+b \left (10 c^2 d^2 x^2+15 c^4+8 d^4 x^4\right )\right )+3 c^4 \sqrt{d^2 x^2-c^2} \left (6 a d^2+5 b c^2\right ) \tanh ^{-1}\left (\frac{d x}{\sqrt{d^2 x^2-c^2}}\right )}{48 d^7 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*(-c^2 + d^2*x^2)*(6*a*d^2*(3*c^2 + 2*d^2*x^2) + b*(15*c^4 + 10*c^2*d^2*x^2 + 8*d^4*x^4)) + 3*c^4*(5*b*c^2
+ 6*a*d^2)*Sqrt[-c^2 + d^2*x^2]*ArcTanh[(d*x)/Sqrt[-c^2 + d^2*x^2]])/(48*d^7*Sqrt[-c + d*x]*Sqrt[c + d*x])

________________________________________________________________________________________

Maple [C]  time = 0.028, size = 240, normalized size = 1.5 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{48\,{d}^{7}}\sqrt{dx-c}\sqrt{dx+c} \left ( 8\,{\it csgn} \left ( d \right ){x}^{5}b{d}^{5}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+12\,{\it csgn} \left ( d \right ){x}^{3}a{d}^{5}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+10\,{\it csgn} \left ( d \right ){x}^{3}b{c}^{2}{d}^{3}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+18\,{\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}+18\,\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}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x)

[Out]

1/48*(d*x-c)^(1/2)*(d*x+c)^(1/2)*(8*csgn(d)*x^5*b*d^5*(d^2*x^2-c^2)^(1/2)+12*csgn(d)*x^3*a*d^5*(d^2*x^2-c^2)^(
1/2)+10*csgn(d)*x^3*b*c^2*d^3*(d^2*x^2-c^2)^(1/2)+18*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+18*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)*csgn(d))*b*c^6)*csgn(d)/d^7/(d^2*x^2-c^2)^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 0.960679, size = 289, normalized size = 1.76 \begin{align*} \frac{\sqrt{d^{2} x^{2} - c^{2}} b x^{5}}{6 \, d^{2}} + \frac{5 \, \sqrt{d^{2} x^{2} - c^{2}} b c^{2} x^{3}}{24 \, d^{4}} + \frac{\sqrt{d^{2} x^{2} - c^{2}} a x^{3}}{4 \, d^{2}} + \frac{5 \, b c^{6} \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - c^{2}} \sqrt{d^{2}}\right )}{16 \, \sqrt{d^{2}} d^{6}} + \frac{3 \, a 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^{4}} + \frac{5 \, \sqrt{d^{2} x^{2} - c^{2}} b c^{4} x}{16 \, d^{6}} + \frac{3 \, \sqrt{d^{2} x^{2} - c^{2}} a c^{2} x}{8 \, 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)^(1/2)/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/6*sqrt(d^2*x^2 - c^2)*b*x^5/d^2 + 5/24*sqrt(d^2*x^2 - c^2)*b*c^2*x^3/d^4 + 1/4*sqrt(d^2*x^2 - c^2)*a*x^3/d^2
+ 5/16*b*c^6*log(2*d^2*x + 2*sqrt(d^2*x^2 - c^2)*sqrt(d^2))/(sqrt(d^2)*d^6) + 3/8*a*c^4*log(2*d^2*x + 2*sqrt(
d^2*x^2 - c^2)*sqrt(d^2))/(sqrt(d^2)*d^4) + 5/16*sqrt(d^2*x^2 - c^2)*b*c^4*x/d^6 + 3/8*sqrt(d^2*x^2 - c^2)*a*c
^2*x/d^4

________________________________________________________________________________________

Fricas [A]  time = 1.58638, size = 251, normalized size = 1.53 \begin{align*} \frac{{\left (8 \, b d^{5} x^{5} + 2 \,{\left (5 \, b c^{2} d^{3} + 6 \, a d^{5}\right )} x^{3} + 3 \,{\left (5 \, b c^{4} d + 6 \, a c^{2} d^{3}\right )} x\right )} \sqrt{d x + c} \sqrt{d x - c} - 3 \,{\left (5 \, b c^{6} + 6 \, a c^{4} d^{2}\right )} \log \left (-d x + \sqrt{d x + c} \sqrt{d x - c}\right )}{48 \, 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)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/48*((8*b*d^5*x^5 + 2*(5*b*c^2*d^3 + 6*a*d^5)*x^3 + 3*(5*b*c^4*d + 6*a*c^2*d^3)*x)*sqrt(d*x + c)*sqrt(d*x - c
) - 3*(5*b*c^6 + 6*a*c^4*d^2)*log(-d*x + sqrt(d*x + c)*sqrt(d*x - c)))/d^7

________________________________________________________________________________________

Sympy [C]  time = 87.1952, size = 240, normalized size = 1.46 \begin{align*} \frac{a c^{4}{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{7}{4}, - \frac{5}{4} & - \frac{3}{2}, - \frac{3}{2}, -1, 1 \\-2, - \frac{7}{4}, - \frac{3}{2}, - \frac{5}{4}, -1, 0 & \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{5}} - \frac{i a c^{4}{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{5}{2}, - \frac{9}{4}, -2, - \frac{7}{4}, - \frac{3}{2}, 1 & \\- \frac{9}{4}, - \frac{7}{4} & - \frac{5}{2}, -2, -2, 0 \end{matrix} \middle |{\frac{c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{5}} + \frac{b c^{6}{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{11}{4}, - \frac{9}{4} & - \frac{5}{2}, - \frac{5}{2}, -2, 1 \\-3, - \frac{11}{4}, - \frac{5}{2}, - \frac{9}{4}, -2, 0 & \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{7}} - \frac{i b c^{6}{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{7}{2}, - \frac{13}{4}, -3, - \frac{11}{4}, - \frac{5}{2}, 1 & \\- \frac{13}{4}, - \frac{11}{4} & - \frac{7}{2}, -3, -3, 0 \end{matrix} \middle |{\frac{c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} 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)**(1/2)/(d*x+c)**(1/2),x)

[Out]

a*c**4*meijerg(((-7/4, -5/4), (-3/2, -3/2, -1, 1)), ((-2, -7/4, -3/2, -5/4, -1, 0), ()), c**2/(d**2*x**2))/(4*
pi**(3/2)*d**5) - I*a*c**4*meijerg(((-5/2, -9/4, -2, -7/4, -3/2, 1), ()), ((-9/4, -7/4), (-5/2, -2, -2, 0)), c
**2*exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**5) + b*c**6*meijerg(((-11/4, -9/4), (-5/2, -5/2, -2, 1)), (
(-3, -11/4, -5/2, -9/4, -2, 0), ()), c**2/(d**2*x**2))/(4*pi**(3/2)*d**7) - I*b*c**6*meijerg(((-7/2, -13/4, -3
, -11/4, -5/2, 1), ()), ((-13/4, -11/4), (-7/2, -3, -3, 0)), c**2*exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*
d**7)

________________________________________________________________________________________

Giac [A]  time = 1.27013, size = 246, normalized size = 1.5 \begin{align*} -\frac{{\left (33 \, b c^{5} d^{36} + 30 \, a c^{3} d^{38} -{\left (85 \, b c^{4} d^{36} + 54 \, a c^{2} d^{38} - 2 \,{\left (55 \, b c^{3} d^{36} + 18 \, a c d^{38} -{\left (45 \, b c^{2} d^{36} + 6 \, a d^{38} + 4 \,{\left ({\left (d x + c\right )} b d^{36} - 5 \, b c d^{36}\right )}{\left (d x + c\right )}\right )}{\left (d x + c\right )}\right )}{\left (d x + c\right )}\right )}{\left (d x + c\right )}\right )} \sqrt{d x + c} \sqrt{d x - c} + 6 \,{\left (5 \, b c^{6} d^{36} + 6 \, a c^{4} d^{38}\right )} \log \left ({\left | -\sqrt{d x + c} + \sqrt{d x - c} \right |}\right )}{34603008 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

-1/34603008*((33*b*c^5*d^36 + 30*a*c^3*d^38 - (85*b*c^4*d^36 + 54*a*c^2*d^38 - 2*(55*b*c^3*d^36 + 18*a*c*d^38
- (45*b*c^2*d^36 + 6*a*d^38 + 4*((d*x + c)*b*d^36 - 5*b*c*d^36)*(d*x + c))*(d*x + c))*(d*x + c))*(d*x + c))*sq
rt(d*x + c)*sqrt(d*x - c) + 6*(5*b*c^6*d^36 + 6*a*c^4*d^38)*log(abs(-sqrt(d*x + c) + sqrt(d*x - c))))/d