3.4.70 \(\int \frac {x^2 (a+b x^2)}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\) [370]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 152, normalized size of antiderivative = 1.00, 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 = {471, 91, 12, 79, 65, 223, 212} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 65

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 79

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] || L
tQ[p, n]))))

Rule 91

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 471

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]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.18, size = 108, normalized size = 0.71 \begin {gather*} \frac {-3 b c^2 d x-2 a d^3 x+b d^3 x^3+2 \left (3 b c^2+2 a d^2\right ) \sqrt {-c+d x} \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {-c+d x}}\right )}{2 d^5 \sqrt {-c+d x} \sqrt {c+d x}} \end {gather*}

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.32, size = 263, normalized size = 1.73

method	result	size
default	\(\frac {\sqrt {d x -c}\, \left (-\mathrm {csgn}\left (d \right ) b \,d^{3} x^{3} \sqrt {d^{2} x^{2}-c^{2}}-2 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) a \,d^{4} x^{2}-3 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) b \,c^{2} d^{2} x^{2}+2 \sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right ) d^{3} a x +3 \sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right ) d b \,c^{2} x +2 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) a \,c^{2} d^{2}+3 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) b \,c^{4}\right ) \mathrm {csgn}\left (d \right )}{2 \left (-d x +c \right ) \sqrt {d^{2} x^{2}-c^{2}}\, d^{5} \sqrt {d x +c}}\)	\(263\)
risch	\(-\frac {b x \left (-d x +c \right ) \sqrt {d x +c}}{2 d^{4} \sqrt {d x -c}}-\frac {\left (-\frac {\ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right ) a}{d^{2} \sqrt {d^{2}}}-\frac {3 \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right ) b \,c^{2}}{2 d^{4} \sqrt {d^{2}}}+\frac {\sqrt {d^{2} \left (x +\frac {c}{d}\right )^{2}-2 c d \left (x +\frac {c}{d}\right )}\, a}{2 d^{4} \left (x +\frac {c}{d}\right )}+\frac {\sqrt {d^{2} \left (x +\frac {c}{d}\right )^{2}-2 c d \left (x +\frac {c}{d}\right )}\, b \,c^{2}}{2 d^{6} \left (x +\frac {c}{d}\right )}+\frac {\sqrt {d^{2} \left (x -\frac {c}{d}\right )^{2}+2 c d \left (x -\frac {c}{d}\right )}\, a}{2 d^{4} \left (x -\frac {c}{d}\right )}+\frac {\sqrt {d^{2} \left (x -\frac {c}{d}\right )^{2}+2 c d \left (x -\frac {c}{d}\right )}\, b \,c^{2}}{2 d^{6} \left (x -\frac {c}{d}\right )}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{\sqrt {d x -c}\, \sqrt {d x +c}}\)	\(324\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 138, normalized size = 0.91 \begin {gather*} \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}} d\right )}{2 \, d^{5}} + \frac {a \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{d^{3}} \end {gather*}

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)*d)/d^5 + a*log(2*d^2*x + 2*sqrt(d^2*x^2 - c^2)*d)/d^3

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Fricas [A]
time = 3.71, size = 159, normalized size = 1.05 \begin {gather*} \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 {gather*}

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 [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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Giac [A]
time = 0.60, size = 147, normalized size = 0.97 \begin {gather*} \frac {\sqrt {d x + c} {\left ({\left (d x + c\right )} {\left (\frac {{\left (d x + c\right )} b}{d^{5}} - \frac {3 \, b c}{d^{5}}\right )} + \frac {b c^{2} d^{15} - a d^{17}}{d^{20}}\right )}}{2 \, \sqrt {d x - c}} - \frac {{\left (3 \, b c^{2} + 2 \, a d^{2}\right )} \log \left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}\right )}{2 \, d^{5}} - \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 {gather*}

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]

1/2*sqrt(d*x + c)*((d*x + c)*((d*x + c)*b/d^5 - 3*b*c/d^5) + (b*c^2*d^15 - a*d^17)/d^20)/sqrt(d*x - c) - 1/2*(
3*b*c^2 + 2*a*d^2)*log((sqrt(d*x + c) - sqrt(d*x - c))^2)/d^5 - 2*(b*c^3 + a*c*d^2)/(((sqrt(d*x + c) - sqrt(d*
x - c))^2 + 2*c)*d^5)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (b\,x^2+a\right )}{{\left (c+d\,x\right )}^{3/2}\,{\left (d\,x-c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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