3.4.33 \(\int x (a+\frac {b}{c+d x^2})^{3/2} \, dx\) [333]

Optimal. Leaf size=94 \[ -\frac {3 b \sqrt {a+\frac {b}{c+d x^2}}}{2 d}+\frac {\left (c+d x^2\right ) \left (a+\frac {b}{c+d x^2}\right )^{3/2}}{2 d}+\frac {3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1605, 248, 43, 52, 65, 214} \begin {gather*} \frac {\left (c+d x^2\right ) \left (a+\frac {b}{c+d x^2}\right )^{3/2}}{2 d}-\frac {3 b \sqrt {a+\frac {b}{c+d x^2}}}{2 d}+\frac {3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b/(c + d*x^2))^(3/2),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 214

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

Rule 248

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 1605

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 87, normalized size = 0.93 \begin {gather*} \frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (-2 b+a \left (c+d x^2\right )\right )+3 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(335\) vs. \(2(78)=156\).
time = 0.12, size = 336, normalized size = 3.57

method	result	size
derivativedivides	\(\frac {\sqrt {\frac {a \left (d \,x^{2}+c \right )+b}{d \,x^{2}+c}}\, \left (3 \ln \left (\frac {2 \sqrt {a \left (d \,x^{2}+c \right )^{2}+b \left (d \,x^{2}+c \right )}\, \sqrt {a}+2 a \left (d \,x^{2}+c \right )+b}{2 \sqrt {a}}\right ) a b \left (d \,x^{2}+c \right )^{2}+6 a^{\frac {3}{2}} \sqrt {a \left (d \,x^{2}+c \right )^{2}+b \left (d \,x^{2}+c \right )}\, \left (d \,x^{2}+c \right )^{2}-4 \left (a \left (d \,x^{2}+c \right )^{2}+b \left (d \,x^{2}+c \right )\right )^{\frac {3}{2}} \sqrt {a}\right )}{4 d \left (d \,x^{2}+c \right ) \sqrt {\left (d \,x^{2}+c \right ) \left (a \left (d \,x^{2}+c \right )+b \right )}\, \sqrt {a}}\)	\(187\)
risch	\(\frac {\left (d \,x^{2}+c \right ) a \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{2 d}+\frac {\left (\frac {3 b a \ln \left (\frac {a c d +\frac {1}{2} b d +a \,d^{2} x^{2}}{\sqrt {a \,d^{2}}}+\sqrt {c^{2} a +b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}\right )}{4 \sqrt {a \,d^{2}}}-\frac {b \,x^{2} a}{\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}-\frac {b a c}{d \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}-\frac {b^{2}}{d \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}}\right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}}{a d \,x^{2}+a c +b}\)	\(289\)
default	\(\frac {\left (3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b \,d^{2} x^{2}+2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}\, a d \,x^{2}+3 \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b c d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+c^{2} a +b c}\, \sqrt {a \,d^{2}}\, a c -4 \sqrt {a \,d^{2}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, b \right ) \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}{4 d \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {a \,d^{2}}}\)	\(336\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b/(d*x^2+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 156, normalized size = 1.66 \begin {gather*} -\frac {a b \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a d - \frac {{\left (a d x^{2} + a c + b\right )} d}{d x^{2} + c}\right )}} - \frac {3 \, \sqrt {a} b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{4 \, d} - \frac {b \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.41, size = 269, normalized size = 2.86 \begin {gather*} \left [\frac {3 \, \sqrt {a} b \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} + 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (a d x^{2} + a c - 2 \, b\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, d}, -\frac {3 \, \sqrt {-a} b \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) - 2 \, {\left (a d x^{2} + a c - 2 \, b\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*sqrt(a)*b*log(8*a^2*d^2*x^4 + 8*a^2*c^2 + 8*(2*a^2*c + a*b)*d*x^2 + 8*a*b*c + b^2 + 4*(2*a*d^2*x^4 + (
4*a*c + b)*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))) + 4*(a*d*x^2 + a*c - 2*b)*sqr
t((a*d*x^2 + a*c + b)/(d*x^2 + c)))/d, -1/4*(3*sqrt(-a)*b*arctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqrt(-a)*sqrt((a*
d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c + a*b)) - 2*(a*d*x^2 + a*c - 2*b)*sqrt((a*d*x^2 + a*c + b)/(d
*x^2 + c)))/d]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (\frac {a c + a d x^{2} + b}{c + d x^{2}}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (78) = 156\).
time = 5.26, size = 387, normalized size = 4.12 \begin {gather*} -\frac {\sqrt {a} b \log \left ({\left | -2 \, a^{\frac {5}{2}} c^{3} d - 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a^{2} c^{2} {\left | d \right |} - 6 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} a^{\frac {3}{2}} c d - a^{\frac {3}{2}} b c^{2} d - 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{3} a {\left | d \right |} - 2 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a b c {\left | d \right |} - {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )}^{2} \sqrt {a} b d \right |}\right ) \mathrm {sgn}\left (d x^{2} + c\right )}{4 \, {\left | d \right |}} - \frac {\sqrt {a} b {\left | d \right |} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (d x^{2} + c\right )}{4 \, d^{2}} + \frac {\sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c} a \mathrm {sgn}\left (d x^{2} + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.74, size = 61, normalized size = 0.65 \begin {gather*} -\frac {{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}\,\left (d\,x^2+c\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {a\,\left (d\,x^2+c\right )}{b}\right )}{d\,{\left (\frac {a\,\left (d\,x^2+c\right )}{b}+1\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/(c + d*x^2))^(3/2),x)

[Out]

-((a + b/(c + d*x^2))^(3/2)*(c + d*x^2)*hypergeom([-3/2, -1/2], 1/2, -(a*(c + d*x^2))/b))/(d*((a*(c + d*x^2))/
b + 1)^(3/2))

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