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

Optimal. Leaf size=159 \[ \frac{c^2 x \sqrt{d x-c} \sqrt{c+d x} \left (2 a d^2+b c^2\right )}{16 d^4}+\frac{x (d x-c)^{3/2} (c+d x)^{3/2} \left (2 a d^2+b c^2\right )}{8 d^4}-\frac{c^4 \left (2 a d^2+b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{8 d^5}+\frac{b x^3 (d x-c)^{3/2} (c+d x)^{3/2}}{6 d^2} \]

[Out]

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

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Rubi [A]  time = 0.123454, antiderivative size = 159, 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, 90, 12, 38, 63, 217, 206} \[ \frac{c^2 x \sqrt{d x-c} \sqrt{c+d x} \left (2 a d^2+b c^2\right )}{16 d^4}+\frac{x (d x-c)^{3/2} (c+d x)^{3/2} \left (2 a d^2+b c^2\right )}{8 d^4}-\frac{c^4 \left (2 a d^2+b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{8 d^5}+\frac{b x^3 (d x-c)^{3/2} (c+d x)^{3/2}}{6 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(b*c^2 + 2*a*d^2)*x*Sqrt[-c + d*x]*Sqrt[c + d*x])/(16*d^4) + ((b*c^2 + 2*a*d^2)*x*(-c + d*x)^(3/2)*(c + d
*x)^(3/2))/(8*d^4) + (b*x^3*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(6*d^2) - (c^4*(b*c^2 + 2*a*d^2)*ArcTanh[Sqrt[-c
+ d*x]/Sqrt[c + d*x]])/(8*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 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 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 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 x^2 \sqrt{-c+d x} \sqrt{c+d x} \left (a+b x^2\right ) \, dx &=\frac{b x^3 (-c+d x)^{3/2} (c+d x)^{3/2}}{6 d^2}+\frac{1}{2} \left (2 a+\frac{b c^2}{d^2}\right ) \int x^2 \sqrt{-c+d x} \sqrt{c+d x} \, dx\\ &=\frac{\left (b c^2+2 a d^2\right ) x (-c+d x)^{3/2} (c+d x)^{3/2}}{8 d^4}+\frac{b x^3 (-c+d x)^{3/2} (c+d x)^{3/2}}{6 d^2}+\frac{\left (b c^2+2 a d^2\right ) \int c^2 \sqrt{-c+d x} \sqrt{c+d x} \, dx}{8 d^4}\\ &=\frac{\left (b c^2+2 a d^2\right ) x (-c+d x)^{3/2} (c+d x)^{3/2}}{8 d^4}+\frac{b x^3 (-c+d x)^{3/2} (c+d x)^{3/2}}{6 d^2}+\frac{\left (c^2 \left (b c^2+2 a d^2\right )\right ) \int \sqrt{-c+d x} \sqrt{c+d x} \, dx}{8 d^4}\\ &=\frac{c^2 \left (b c^2+2 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{16 d^4}+\frac{\left (b c^2+2 a d^2\right ) x (-c+d x)^{3/2} (c+d x)^{3/2}}{8 d^4}+\frac{b x^3 (-c+d x)^{3/2} (c+d x)^{3/2}}{6 d^2}-\frac{\left (c^4 \left (b c^2+2 a d^2\right )\right ) \int \frac{1}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{16 d^4}\\ &=\frac{c^2 \left (b c^2+2 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{16 d^4}+\frac{\left (b c^2+2 a d^2\right ) x (-c+d x)^{3/2} (c+d x)^{3/2}}{8 d^4}+\frac{b x^3 (-c+d x)^{3/2} (c+d x)^{3/2}}{6 d^2}-\frac{\left (c^4 \left (b c^2+2 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^5}\\ &=\frac{c^2 \left (b c^2+2 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{16 d^4}+\frac{\left (b c^2+2 a d^2\right ) x (-c+d x)^{3/2} (c+d x)^{3/2}}{8 d^4}+\frac{b x^3 (-c+d x)^{3/2} (c+d x)^{3/2}}{6 d^2}-\frac{\left (c^4 \left (b c^2+2 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^5}\\ &=\frac{c^2 \left (b c^2+2 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{16 d^4}+\frac{\left (b c^2+2 a d^2\right ) x (-c+d x)^{3/2} (c+d x)^{3/2}}{8 d^4}+\frac{b x^3 (-c+d x)^{3/2} (c+d x)^{3/2}}{6 d^2}-\frac{c^4 \left (b c^2+2 a d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{-c+d x}}{\sqrt{c+d x}}\right )}{8 d^5}\\ \end{align*}

Mathematica [A]  time = 0.163397, size = 135, normalized size = 0.85 \[ \frac{\sqrt{d x-c} \sqrt{c+d x} \left (d x \sqrt{1-\frac{d^2 x^2}{c^2}} \left (b \left (-2 c^2 d^2 x^2-3 c^4+8 d^4 x^4\right )-6 a d^2 \left (c^2-2 d^2 x^2\right )\right )+3 \left (2 a c^3 d^2+b c^5\right ) \sin ^{-1}\left (\frac{d x}{c}\right )\right )}{48 d^5 \sqrt{1-\frac{d^2 x^2}{c^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-c + d*x]*Sqrt[c + d*x]*(d*x*Sqrt[1 - (d^2*x^2)/c^2]*(-6*a*d^2*(c^2 - 2*d^2*x^2) + b*(-3*c^4 - 2*c^2*d^2
*x^2 + 8*d^4*x^4)) + 3*(b*c^5 + 2*a*c^3*d^2)*ArcSin[(d*x)/c]))/(48*d^5*Sqrt[1 - (d^2*x^2)/c^2])

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Maple [C]  time = 0.013, size = 240, normalized size = 1.5 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{48\,{d}^{5}}\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}}-2\,{\it csgn} \left ( d \right ){x}^{3}b{c}^{2}{d}^{3}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}-6\,{\it csgn} \left ( d \right ){d}^{3}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}xa{c}^{2}-3\,{\it csgn} \left ( d \right ) d\sqrt{{d}^{2}{x}^{2}-{c}^{2}}xb{c}^{4}-6\,\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}-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}^{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^2*(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)-2*csgn(d)*x^3*b*c^2*d^3*(d^2*x^2-c^2)^(1/2)-6*csgn(d)*d^3*(d^2*x^2-c^2)^(1/2)*x*a*c^2-3*csgn(d)*d*(d^2*x^
2-c^2)^(1/2)*x*b*c^4-6*ln(((d^2*x^2-c^2)^(1/2)*csgn(d)+d*x)*csgn(d))*a*c^4*d^2-3*ln(((d^2*x^2-c^2)^(1/2)*csgn(
d)+d*x)*csgn(d))*b*c^6)*csgn(d)/(d^2*x^2-c^2)^(1/2)/d^5

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Maxima [A]  time = 0.969889, size = 284, normalized size = 1.79 \begin{align*} \frac{{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} b x^{3}}{6 \, d^{2}} - \frac{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^{4}} - \frac{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^{2}} + \frac{\sqrt{d^{2} x^{2} - c^{2}} b c^{4} x}{16 \, d^{4}} + \frac{\sqrt{d^{2} x^{2} - c^{2}} a c^{2} x}{8 \, d^{2}} + \frac{{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} b c^{2} x}{8 \, d^{4}} + \frac{{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} a x}{4 \, 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)^(1/2)*(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.64964, size = 243, normalized size = 1.53 \begin{align*} \frac{{\left (8 \, b d^{5} x^{5} - 2 \,{\left (b c^{2} d^{3} - 6 \, a d^{5}\right )} x^{3} - 3 \,{\left (b c^{4} d + 2 \, a c^{2} d^{3}\right )} x\right )} \sqrt{d x + c} \sqrt{d x - c} + 3 \,{\left (b c^{6} + 2 \, a c^{4} d^{2}\right )} \log \left (-d x + \sqrt{d x + c} \sqrt{d x - c}\right )}{48 \, 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)^(1/2)*(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*x**2)*sqrt(-c + d*x)*sqrt(c + d*x), x)

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Giac [A]  time = 1.31216, size = 311, normalized size = 1.96 \begin{align*} \frac{6 \,{\left (\frac{2 \, c^{4} \log \left ({\left | -\sqrt{d x + c} + \sqrt{d x - c} \right |}\right )}{d^{2}} +{\left ({\left (d x + c\right )}{\left (2 \,{\left (d x + c\right )}{\left (\frac{d x + c}{d^{2}} - \frac{3 \, c}{d^{2}}\right )} + \frac{5 \, c^{2}}{d^{2}}\right )} - \frac{c^{3}}{d^{2}}\right )} \sqrt{d x + c} \sqrt{d x - c}\right )} a +{\left (\frac{6 \, c^{6} \log \left ({\left | -\sqrt{d x + c} + \sqrt{d x - c} \right |}\right )}{d^{4}} +{\left ({\left (2 \,{\left ({\left (d x + c\right )}{\left (4 \,{\left (d x + c\right )}{\left (\frac{d x + c}{d^{4}} - \frac{5 \, c}{d^{4}}\right )} + \frac{39 \, c^{2}}{d^{4}}\right )} - \frac{37 \, c^{3}}{d^{4}}\right )}{\left (d x + c\right )} + \frac{31 \, c^{4}}{d^{4}}\right )}{\left (d x + c\right )} - \frac{3 \, c^{5}}{d^{4}}\right )} \sqrt{d x + c} \sqrt{d x - c}\right )} b}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(6*(2*c^4*log(abs(-sqrt(d*x + c) + sqrt(d*x - c)))/d^2 + ((d*x + c)*(2*(d*x + c)*((d*x + c)/d^2 - 3*c/d^2
) + 5*c^2/d^2) - c^3/d^2)*sqrt(d*x + c)*sqrt(d*x - c))*a + (6*c^6*log(abs(-sqrt(d*x + c) + sqrt(d*x - c)))/d^4
+ ((2*((d*x + c)*(4*(d*x + c)*((d*x + c)/d^4 - 5*c/d^4) + 39*c^2/d^4) - 37*c^3/d^4)*(d*x + c) + 31*c^4/d^4)*(
d*x + c) - 3*c^5/d^4)*sqrt(d*x + c)*sqrt(d*x - c))*b)/d