3.7.15 \(\int \frac {(a+\frac {b}{x^2}) (c+\frac {d}{x^2})^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=159 \[ -\frac {c^3 (3 b c-8 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{128 d^{5/2}}+\frac {c^2 \sqrt {c+\frac {d}{x^2}} (3 b c-8 a d)}{128 d^2 x}+\frac {c \sqrt {c+\frac {d}{x^2}} (3 b c-8 a d)}{64 d x^3}+\frac {\left (c+\frac {d}{x^2}\right )^{3/2} (3 b c-8 a d)}{48 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3} \]

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Rubi [A]  time = 0.09, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {459, 335, 279, 321, 217, 206} \begin {gather*} \frac {c^2 \sqrt {c+\frac {d}{x^2}} (3 b c-8 a d)}{128 d^2 x}-\frac {c^3 (3 b c-8 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{128 d^{5/2}}+\frac {c \sqrt {c+\frac {d}{x^2}} (3 b c-8 a d)}{64 d x^3}+\frac {\left (c+\frac {d}{x^2}\right )^{3/2} (3 b c-8 a d)}{48 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b/x^2)*(c + d/x^2)^(3/2))/x^4,x]

[Out]

(c*(3*b*c - 8*a*d)*Sqrt[c + d/x^2])/(64*d*x^3) + ((3*b*c - 8*a*d)*(c + d/x^2)^(3/2))/(48*d*x^3) - (b*(c + d/x^
2)^(5/2))/(8*d*x^3) + (c^2*(3*b*c - 8*a*d)*Sqrt[c + d/x^2])/(128*d^2*x) - (c^3*(3*b*c - 8*a*d)*ArcTanh[Sqrt[d]
/(Sqrt[c + d/x^2]*x)])/(128*d^(5/2))

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])

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 279

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

Rule 321

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

Rule 335

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

Rule 459

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

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx &=-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}+\frac {(-3 b c+8 a d) \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{x^4} \, dx}{8 d}\\ &=-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}-\frac {(-3 b c+8 a d) \operatorname {Subst}\left (\int x^2 \left (c+d x^2\right )^{3/2} \, dx,x,\frac {1}{x}\right )}{8 d}\\ &=\frac {(3 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{48 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}+\frac {(c (3 b c-8 a d)) \operatorname {Subst}\left (\int x^2 \sqrt {c+d x^2} \, dx,x,\frac {1}{x}\right )}{16 d}\\ &=\frac {c (3 b c-8 a d) \sqrt {c+\frac {d}{x^2}}}{64 d x^3}+\frac {(3 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{48 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}+\frac {\left (c^2 (3 b c-8 a d)\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{64 d}\\ &=\frac {c (3 b c-8 a d) \sqrt {c+\frac {d}{x^2}}}{64 d x^3}+\frac {(3 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{48 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}+\frac {c^2 (3 b c-8 a d) \sqrt {c+\frac {d}{x^2}}}{128 d^2 x}-\frac {\left (c^3 (3 b c-8 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{128 d^2}\\ &=\frac {c (3 b c-8 a d) \sqrt {c+\frac {d}{x^2}}}{64 d x^3}+\frac {(3 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{48 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}+\frac {c^2 (3 b c-8 a d) \sqrt {c+\frac {d}{x^2}}}{128 d^2 x}-\frac {\left (c^3 (3 b c-8 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {1}{\sqrt {c+\frac {d}{x^2}} x}\right )}{128 d^2}\\ &=\frac {c (3 b c-8 a d) \sqrt {c+\frac {d}{x^2}}}{64 d x^3}+\frac {(3 b c-8 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{48 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{8 d x^3}+\frac {c^2 (3 b c-8 a d) \sqrt {c+\frac {d}{x^2}}}{128 d^2 x}-\frac {c^3 (3 b c-8 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{128 d^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 71, normalized size = 0.45 \begin {gather*} \frac {\sqrt {c+\frac {d}{x^2}} \left (c x^2+d\right )^2 \left (c^3 x^8 (8 a d-3 b c) \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};\frac {c x^2}{d}+1\right )-5 b d^4\right )}{40 d^5 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b/x^2)*(c + d/x^2)^(3/2))/x^4,x]

[Out]

(Sqrt[c + d/x^2]*(d + c*x^2)^2*(-5*b*d^4 + c^3*(-3*b*c + 8*a*d)*x^8*Hypergeometric2F1[5/2, 4, 7/2, 1 + (c*x^2)
/d]))/(40*d^5*x^7)

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IntegrateAlgebraic [A]  time = 0.31, size = 152, normalized size = 0.96 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {\left (8 a c^3 d-3 b c^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{128 d^{5/2}}+\frac {\sqrt {c x^2+d} \left (-24 a c^2 d x^6-112 a c d^2 x^4-64 a d^3 x^2+9 b c^3 x^6-6 b c^2 d x^4-72 b c d^2 x^2-48 b d^3\right )}{384 d^2 x^8}\right )}{\sqrt {c x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b/x^2)*(c + d/x^2)^(3/2))/x^4,x]

[Out]

(Sqrt[c + d/x^2]*x*((Sqrt[d + c*x^2]*(-48*b*d^3 - 72*b*c*d^2*x^2 - 64*a*d^3*x^2 - 6*b*c^2*d*x^4 - 112*a*c*d^2*
x^4 + 9*b*c^3*x^6 - 24*a*c^2*d*x^6))/(384*d^2*x^8) + ((-3*b*c^4 + 8*a*c^3*d)*ArcTanh[Sqrt[d + c*x^2]/Sqrt[d]])
/(128*d^(5/2))))/Sqrt[d + c*x^2]

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fricas [A]  time = 0.47, size = 298, normalized size = 1.87 \begin {gather*} \left [-\frac {3 \, {\left (3 \, b c^{4} - 8 \, a c^{3} d\right )} \sqrt {d} x^{7} \log \left (-\frac {c x^{2} + 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (3 \, b c^{3} d - 8 \, a c^{2} d^{2}\right )} x^{6} - 48 \, b d^{4} - 2 \, {\left (3 \, b c^{2} d^{2} + 56 \, a c d^{3}\right )} x^{4} - 8 \, {\left (9 \, b c d^{3} + 8 \, a d^{4}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{768 \, d^{3} x^{7}}, \frac {3 \, {\left (3 \, b c^{4} - 8 \, a c^{3} d\right )} \sqrt {-d} x^{7} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (3 \, {\left (3 \, b c^{3} d - 8 \, a c^{2} d^{2}\right )} x^{6} - 48 \, b d^{4} - 2 \, {\left (3 \, b c^{2} d^{2} + 56 \, a c d^{3}\right )} x^{4} - 8 \, {\left (9 \, b c d^{3} + 8 \, a d^{4}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{384 \, d^{3} x^{7}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^4,x, algorithm="fricas")

[Out]

[-1/768*(3*(3*b*c^4 - 8*a*c^3*d)*sqrt(d)*x^7*log(-(c*x^2 + 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2) + 2*d)/x^2) - 2*(
3*(3*b*c^3*d - 8*a*c^2*d^2)*x^6 - 48*b*d^4 - 2*(3*b*c^2*d^2 + 56*a*c*d^3)*x^4 - 8*(9*b*c*d^3 + 8*a*d^4)*x^2)*s
qrt((c*x^2 + d)/x^2))/(d^3*x^7), 1/384*(3*(3*b*c^4 - 8*a*c^3*d)*sqrt(-d)*x^7*arctan(sqrt(-d)*x*sqrt((c*x^2 + d
)/x^2)/(c*x^2 + d)) + (3*(3*b*c^3*d - 8*a*c^2*d^2)*x^6 - 48*b*d^4 - 2*(3*b*c^2*d^2 + 56*a*c*d^3)*x^4 - 8*(9*b*
c*d^3 + 8*a*d^4)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^3*x^7)]

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giac [A]  time = 0.37, size = 214, normalized size = 1.35 \begin {gather*} \frac {\frac {3 \, {\left (3 \, b c^{5} \mathrm {sgn}\relax (x) - 8 \, a c^{4} d \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d} d^{2}} + \frac {9 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} b c^{5} \mathrm {sgn}\relax (x) - 24 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} a c^{4} d \mathrm {sgn}\relax (x) - 33 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} b c^{5} d \mathrm {sgn}\relax (x) - 40 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a c^{4} d^{2} \mathrm {sgn}\relax (x) - 33 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c^{5} d^{2} \mathrm {sgn}\relax (x) + 88 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a c^{4} d^{3} \mathrm {sgn}\relax (x) + 9 \, \sqrt {c x^{2} + d} b c^{5} d^{3} \mathrm {sgn}\relax (x) - 24 \, \sqrt {c x^{2} + d} a c^{4} d^{4} \mathrm {sgn}\relax (x)}{c^{4} d^{2} x^{8}}}{384 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^4,x, algorithm="giac")

[Out]

1/384*(3*(3*b*c^5*sgn(x) - 8*a*c^4*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/(sqrt(-d)*d^2) + (9*(c*x^2 + d)^
(7/2)*b*c^5*sgn(x) - 24*(c*x^2 + d)^(7/2)*a*c^4*d*sgn(x) - 33*(c*x^2 + d)^(5/2)*b*c^5*d*sgn(x) - 40*(c*x^2 + d
)^(5/2)*a*c^4*d^2*sgn(x) - 33*(c*x^2 + d)^(3/2)*b*c^5*d^2*sgn(x) + 88*(c*x^2 + d)^(3/2)*a*c^4*d^3*sgn(x) + 9*s
qrt(c*x^2 + d)*b*c^5*d^3*sgn(x) - 24*sqrt(c*x^2 + d)*a*c^4*d^4*sgn(x))/(c^4*d^2*x^8))/c

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maple [B]  time = 0.07, size = 302, normalized size = 1.90 \begin {gather*} \frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (24 a \,c^{3} d^{\frac {5}{2}} x^{8} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-9 b \,c^{4} d^{\frac {3}{2}} x^{8} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-24 \sqrt {c \,x^{2}+d}\, a \,c^{3} d^{2} x^{8}+9 \sqrt {c \,x^{2}+d}\, b \,c^{4} d \,x^{8}-8 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \,c^{3} d \,x^{8}+3 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,c^{4} x^{8}+8 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,c^{2} d \,x^{6}-3 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,c^{3} x^{6}+16 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a c \,d^{2} x^{4}-6 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,c^{2} d \,x^{4}-64 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,d^{3} x^{2}+24 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b c \,d^{2} x^{2}-48 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,d^{3}\right )}{384 \left (c \,x^{2}+d \right )^{\frac {3}{2}} d^{4} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)*(c+d/x^2)^(3/2)/x^4,x)

[Out]

1/384*((c*x^2+d)/x^2)^(3/2)/x^5*(24*d^(5/2)*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*x^8*a*c^3-9*d^(3/2)*ln(2*(d+(c
*x^2+d)^(1/2)*d^(1/2))/x)*x^8*b*c^4-8*(c*x^2+d)^(3/2)*x^8*a*c^3*d+3*(c*x^2+d)^(3/2)*x^8*b*c^4+8*(c*x^2+d)^(5/2
)*x^6*a*c^2*d-3*(c*x^2+d)^(5/2)*x^6*b*c^3-24*(c*x^2+d)^(1/2)*x^8*a*c^3*d^2+9*(c*x^2+d)^(1/2)*x^8*b*c^4*d+16*(c
*x^2+d)^(5/2)*x^4*a*c*d^2-6*(c*x^2+d)^(5/2)*x^4*b*c^2*d-64*(c*x^2+d)^(5/2)*x^2*a*d^3+24*(c*x^2+d)^(5/2)*x^2*b*
c*d^2-48*(c*x^2+d)^(5/2)*b*d^3)/(c*x^2+d)^(3/2)/d^4

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maxima [B]  time = 1.57, size = 354, normalized size = 2.23 \begin {gather*} -\frac {1}{96} \, {\left (\frac {3 \, c^{3} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{3} x^{5} + 8 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{3} d x^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{3} d^{2} x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{3} d x^{6} - 3 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{2} x^{4} + 3 \, {\left (c + \frac {d}{x^{2}}\right )} d^{3} x^{2} - d^{4}}\right )} a + \frac {1}{256} \, {\left (\frac {3 \, c^{4} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} c^{4} x^{7} - 11 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{4} d x^{5} - 11 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{4} d^{2} x^{3} + 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{4} d^{3} x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{4} d^{2} x^{8} - 4 \, {\left (c + \frac {d}{x^{2}}\right )}^{3} d^{3} x^{6} + 6 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{4} x^{4} - 4 \, {\left (c + \frac {d}{x^{2}}\right )} d^{5} x^{2} + d^{6}}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^4,x, algorithm="maxima")

[Out]

-1/96*(3*c^3*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d)))/d^(3/2) + 2*(3*(c + d/x^2)^(5/2)
*c^3*x^5 + 8*(c + d/x^2)^(3/2)*c^3*d*x^3 - 3*sqrt(c + d/x^2)*c^3*d^2*x)/((c + d/x^2)^3*d*x^6 - 3*(c + d/x^2)^2
*d^2*x^4 + 3*(c + d/x^2)*d^3*x^2 - d^4))*a + 1/256*(3*c^4*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x
+ sqrt(d)))/d^(5/2) + 2*(3*(c + d/x^2)^(7/2)*c^4*x^7 - 11*(c + d/x^2)^(5/2)*c^4*d*x^5 - 11*(c + d/x^2)^(3/2)*
c^4*d^2*x^3 + 3*sqrt(c + d/x^2)*c^4*d^3*x)/((c + d/x^2)^4*d^2*x^8 - 4*(c + d/x^2)^3*d^3*x^6 + 6*(c + d/x^2)^2*
d^4*x^4 - 4*(c + d/x^2)*d^5*x^2 + d^6))*b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+\frac {b}{x^2}\right )\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b/x^2)*(c + d/x^2)^(3/2))/x^4,x)

[Out]

int(((a + b/x^2)*(c + d/x^2)^(3/2))/x^4, x)

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sympy [B]  time = 29.16, size = 287, normalized size = 1.81 \begin {gather*} - \frac {a c^{\frac {5}{2}}}{16 d x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {17 a c^{\frac {3}{2}}}{48 x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {11 a \sqrt {c} d}{24 x^{5} \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {a c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{16 d^{\frac {3}{2}}} - \frac {a d^{2}}{6 \sqrt {c} x^{7} \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {3 b c^{\frac {7}{2}}}{128 d^{2} x \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b c^{\frac {5}{2}}}{128 d x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {13 b c^{\frac {3}{2}}}{64 x^{5} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {5 b \sqrt {c} d}{16 x^{7} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 b c^{4} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{128 d^{\frac {5}{2}}} - \frac {b d^{2}}{8 \sqrt {c} x^{9} \sqrt {1 + \frac {d}{c x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)*(c+d/x**2)**(3/2)/x**4,x)

[Out]

-a*c**(5/2)/(16*d*x*sqrt(1 + d/(c*x**2))) - 17*a*c**(3/2)/(48*x**3*sqrt(1 + d/(c*x**2))) - 11*a*sqrt(c)*d/(24*
x**5*sqrt(1 + d/(c*x**2))) + a*c**3*asinh(sqrt(d)/(sqrt(c)*x))/(16*d**(3/2)) - a*d**2/(6*sqrt(c)*x**7*sqrt(1 +
d/(c*x**2))) + 3*b*c**(7/2)/(128*d**2*x*sqrt(1 + d/(c*x**2))) + b*c**(5/2)/(128*d*x**3*sqrt(1 + d/(c*x**2)))
- 13*b*c**(3/2)/(64*x**5*sqrt(1 + d/(c*x**2))) - 5*b*sqrt(c)*d/(16*x**7*sqrt(1 + d/(c*x**2))) - 3*b*c**4*asinh
(sqrt(d)/(sqrt(c)*x))/(128*d**(5/2)) - b*d**2/(8*sqrt(c)*x**9*sqrt(1 + d/(c*x**2)))

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