3.7.44 \(\int \frac {(a+b x^2)^2}{x^4 \sqrt {c+d x^2}} \, dx\) [644]

Optimal. Leaf size=84 \[ -\frac {a^2 \sqrt {c+d x^2}}{3 c x^3}-\frac {2 a (3 b c-a d) \sqrt {c+d x^2}}{3 c^2 x}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{\sqrt {d}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {473, 462, 223, 212} \begin {gather*} -\frac {a^2 \sqrt {c+d x^2}}{3 c x^3}-\frac {2 a \sqrt {c+d x^2} (3 b c-a d)}{3 c^2 x}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{\sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 462

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

Rule 473

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

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 73, normalized size = 0.87 \begin {gather*} \frac {a \sqrt {c+d x^2} \left (-a c-6 b c x^2+2 a d x^2\right )}{3 c^2 x^3}-\frac {b^2 \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{\sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[c + d*x^2]*(-(a*c) - 6*b*c*x^2 + 2*a*d*x^2))/(3*c^2*x^3) - (b^2*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/S
qrt[d]

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Maple [A]
time = 0.10, size = 84, normalized size = 1.00

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, a \left (-2 a d \,x^{2}+6 c \,x^{2} b +a c \right )}{3 c^{2} x^{3}}+\frac {b^{2} \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{\sqrt {d}}\)	\(61\)
default	\(\frac {b^{2} \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{\sqrt {d}}+a^{2} \left (-\frac {\sqrt {d \,x^{2}+c}}{3 c \,x^{3}}+\frac {2 d \sqrt {d \,x^{2}+c}}{3 c^{2} x}\right )-\frac {2 a b \sqrt {d \,x^{2}+c}}{c x}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 77, normalized size = 0.92 \begin {gather*} \frac {b^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {d}} - \frac {2 \, \sqrt {d x^{2} + c} a b}{c x} + \frac {2 \, \sqrt {d x^{2} + c} a^{2} d}{3 \, c^{2} x} - \frac {\sqrt {d x^{2} + c} a^{2}}{3 \, c x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.61, size = 173, normalized size = 2.06 \begin {gather*} \left [\frac {3 \, b^{2} c^{2} \sqrt {d} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (a^{2} c d + 2 \, {\left (3 \, a b c d - a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, c^{2} d x^{3}}, -\frac {3 \, b^{2} c^{2} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (a^{2} c d + 2 \, {\left (3 \, a b c d - a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, c^{2} d x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*b^2*c^2*sqrt(d)*x^3*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(a^2*c*d + 2*(3*a*b*c*d - a^2*
d^2)*x^2)*sqrt(d*x^2 + c))/(c^2*d*x^3), -1/3*(3*b^2*c^2*sqrt(-d)*x^3*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (a^2
*c*d + 2*(3*a*b*c*d - a^2*d^2)*x^2)*sqrt(d*x^2 + c))/(c^2*d*x^3)]

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Sympy [A]
time = 1.43, size = 158, normalized size = 1.88 \begin {gather*} - \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 c x^{2}} + \frac {2 a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3 c^{2}} - \frac {2 a b \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{c} + b^{2} \left (\begin {cases} \frac {\sqrt {- \frac {c}{d}} \operatorname {asin}{\left (x \sqrt {- \frac {d}{c}} \right )}}{\sqrt {c}} & \text {for}\: c > 0 \wedge d < 0 \\\frac {\sqrt {\frac {c}{d}} \operatorname {asinh}{\left (x \sqrt {\frac {d}{c}} \right )}}{\sqrt {c}} & \text {for}\: c > 0 \wedge d > 0 \\\frac {\sqrt {- \frac {c}{d}} \operatorname {acosh}{\left (x \sqrt {- \frac {d}{c}} \right )}}{\sqrt {- c}} & \text {for}\: d > 0 \wedge c < 0 \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**4/(d*x**2+c)**(1/2),x)

[Out]

-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*c*x**2) + 2*a**2*d**(3/2)*sqrt(c/(d*x**2) + 1)/(3*c**2) - 2*a*b*sqrt(d)*
sqrt(c/(d*x**2) + 1)/c + b**2*Piecewise((sqrt(-c/d)*asin(x*sqrt(-d/c))/sqrt(c), (c > 0) & (d < 0)), (sqrt(c/d)
*asinh(x*sqrt(d/c))/sqrt(c), (c > 0) & (d > 0)), (sqrt(-c/d)*acosh(x*sqrt(-d/c))/sqrt(-c), (d > 0) & (c < 0)))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (70) = 140\).
time = 0.93, size = 156, normalized size = 1.86 \begin {gather*} -\frac {b^{2} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{2 \, \sqrt {d}} + \frac {4 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b \sqrt {d} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} d^{\frac {3}{2}} + 3 \, a b c^{2} \sqrt {d} - a^{2} c d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^2}{x^4\,\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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