3.623 \(\int \frac{(a+b x^2)^2 (c+d x^2)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=147 \[ -\frac{a^2 \left (c+d x^2\right )^{5/2}}{5 c x^5}-\frac{2 a b \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac{b \left (c+d x^2\right )^{3/2} (4 a d+3 b c)}{3 c x}+\frac{b d x \sqrt{c+d x^2} (4 a d+3 b c)}{2 c}+\frac{1}{2} b \sqrt{d} (4 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0897115, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {462, 453, 277, 195, 217, 206} \[ -\frac{a^2 \left (c+d x^2\right )^{5/2}}{5 c x^5}-\frac{2 a b \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac{b \left (c+d x^2\right )^{3/2} (4 a d+3 b c)}{3 c x}+\frac{b d x \sqrt{c+d x^2} (4 a d+3 b c)}{2 c}+\frac{1}{2} b \sqrt{d} (4 a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^6,x]

[Out]

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

Rule 462

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]

Rule 453

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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 277

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 \frac{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{x^6} \, dx &=-\frac{a^2 \left (c+d x^2\right )^{5/2}}{5 c x^5}+\frac{\int \frac{\left (10 a b c+5 b^2 c x^2\right ) \left (c+d x^2\right )^{3/2}}{x^4} \, dx}{5 c}\\ &=-\frac{a^2 \left (c+d x^2\right )^{5/2}}{5 c x^5}-\frac{2 a b \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac{(b (3 b c+4 a d)) \int \frac{\left (c+d x^2\right )^{3/2}}{x^2} \, dx}{3 c}\\ &=-\frac{b (3 b c+4 a d) \left (c+d x^2\right )^{3/2}}{3 c x}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{5 c x^5}-\frac{2 a b \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac{(b d (3 b c+4 a d)) \int \sqrt{c+d x^2} \, dx}{c}\\ &=\frac{b d (3 b c+4 a d) x \sqrt{c+d x^2}}{2 c}-\frac{b (3 b c+4 a d) \left (c+d x^2\right )^{3/2}}{3 c x}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{5 c x^5}-\frac{2 a b \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac{1}{2} (b d (3 b c+4 a d)) \int \frac{1}{\sqrt{c+d x^2}} \, dx\\ &=\frac{b d (3 b c+4 a d) x \sqrt{c+d x^2}}{2 c}-\frac{b (3 b c+4 a d) \left (c+d x^2\right )^{3/2}}{3 c x}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{5 c x^5}-\frac{2 a b \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac{1}{2} (b d (3 b c+4 a d)) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )\\ &=\frac{b d (3 b c+4 a d) x \sqrt{c+d x^2}}{2 c}-\frac{b (3 b c+4 a d) \left (c+d x^2\right )^{3/2}}{3 c x}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{5 c x^5}-\frac{2 a b \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac{1}{2} b \sqrt{d} (3 b c+4 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.104643, size = 113, normalized size = 0.77 \[ \frac{1}{2} b \sqrt{d} (4 a d+3 b c) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )-\frac{\sqrt{c+d x^2} \left (6 a^2 \left (c+d x^2\right )^2+20 a b c x^2 \left (c+4 d x^2\right )+15 b^2 c x^4 \left (2 c-d x^2\right )\right )}{30 c x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(3/2))/x^6,x]

[Out]

-(Sqrt[c + d*x^2]*(15*b^2*c*x^4*(2*c - d*x^2) + 6*a^2*(c + d*x^2)^2 + 20*a*b*c*x^2*(c + 4*d*x^2)))/(30*c*x^5)
+ (b*Sqrt[d]*(3*b*c + 4*a*d)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/2

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 203, normalized size = 1.4 \begin{align*} -{\frac{2\,ab}{3\,c{x}^{3}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{4\,dab}{3\,{c}^{2}x} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{4\,ab{d}^{2}x}{3\,{c}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+2\,{\frac{ab{d}^{2}x\sqrt{d{x}^{2}+c}}{c}}+2\,ab{d}^{3/2}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) -{\frac{{a}^{2}}{5\,c{x}^{5}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{{b}^{2}}{cx} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{{b}^{2}dx}{c} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{b}^{2}dx}{2}\sqrt{d{x}^{2}+c}}+{\frac{3\,{b}^{2}c}{2}\sqrt{d}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^6,x)

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.50456, size = 609, normalized size = 4.14 \begin{align*} \left [\frac{15 \,{\left (3 \, b^{2} c^{2} + 4 \, a b c d\right )} \sqrt{d} x^{5} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (15 \, b^{2} c d x^{6} - 2 \,{\left (15 \, b^{2} c^{2} + 40 \, a b c d + 3 \, a^{2} d^{2}\right )} x^{4} - 6 \, a^{2} c^{2} - 4 \,{\left (5 \, a b c^{2} + 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{60 \, c x^{5}}, -\frac{15 \,{\left (3 \, b^{2} c^{2} + 4 \, a b c d\right )} \sqrt{-d} x^{5} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) -{\left (15 \, b^{2} c d x^{6} - 2 \,{\left (15 \, b^{2} c^{2} + 40 \, a b c d + 3 \, a^{2} d^{2}\right )} x^{4} - 6 \, a^{2} c^{2} - 4 \,{\left (5 \, a b c^{2} + 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{30 \, c x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^6,x, algorithm="fricas")

[Out]

[1/60*(15*(3*b^2*c^2 + 4*a*b*c*d)*sqrt(d)*x^5*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(15*b^2*c*d*
x^6 - 2*(15*b^2*c^2 + 40*a*b*c*d + 3*a^2*d^2)*x^4 - 6*a^2*c^2 - 4*(5*a*b*c^2 + 3*a^2*c*d)*x^2)*sqrt(d*x^2 + c)
)/(c*x^5), -1/30*(15*(3*b^2*c^2 + 4*a*b*c*d)*sqrt(-d)*x^5*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (15*b^2*c*d*x^6
- 2*(15*b^2*c^2 + 40*a*b*c*d + 3*a^2*d^2)*x^4 - 6*a^2*c^2 - 4*(5*a*b*c^2 + 3*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(
c*x^5)]

________________________________________________________________________________________

Sympy [B]  time = 7.24789, size = 304, normalized size = 2.07 \begin{align*} - \frac{a^{2} c \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{5 x^{4}} - \frac{2 a^{2} d^{\frac{3}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{5 x^{2}} - \frac{a^{2} d^{\frac{5}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{5 c} - \frac{2 a b \sqrt{c} d}{x \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{2 a b c \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{3 x^{2}} - \frac{2 a b d^{\frac{3}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{3} + 2 a b d^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )} - \frac{2 a b d^{2} x}{\sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{b^{2} c^{\frac{3}{2}}}{x \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{b^{2} \sqrt{c} d x \sqrt{1 + \frac{d x^{2}}{c}}}{2} - \frac{b^{2} \sqrt{c} d x}{\sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 b^{2} c \sqrt{d} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(3/2)/x**6,x)

[Out]

-a**2*c*sqrt(d)*sqrt(c/(d*x**2) + 1)/(5*x**4) - 2*a**2*d**(3/2)*sqrt(c/(d*x**2) + 1)/(5*x**2) - a**2*d**(5/2)*
sqrt(c/(d*x**2) + 1)/(5*c) - 2*a*b*sqrt(c)*d/(x*sqrt(1 + d*x**2/c)) - 2*a*b*c*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*
x**2) - 2*a*b*d**(3/2)*sqrt(c/(d*x**2) + 1)/3 + 2*a*b*d**(3/2)*asinh(sqrt(d)*x/sqrt(c)) - 2*a*b*d**2*x/(sqrt(c
)*sqrt(1 + d*x**2/c)) - b**2*c**(3/2)/(x*sqrt(1 + d*x**2/c)) + b**2*sqrt(c)*d*x*sqrt(1 + d*x**2/c)/2 - b**2*sq
rt(c)*d*x/sqrt(1 + d*x**2/c) + 3*b**2*c*sqrt(d)*asinh(sqrt(d)*x/sqrt(c))/2

________________________________________________________________________________________

Giac [B]  time = 1.21198, size = 549, normalized size = 3.73 \begin{align*} \frac{1}{2} \, \sqrt{d x^{2} + c} b^{2} d x - \frac{1}{4} \,{\left (3 \, b^{2} c \sqrt{d} + 4 \, a b d^{\frac{3}{2}}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right ) + \frac{2 \,{\left (15 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} b^{2} c^{2} \sqrt{d} + 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} a b c d^{\frac{3}{2}} + 15 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} a^{2} d^{\frac{5}{2}} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} b^{2} c^{3} \sqrt{d} - 180 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} a b c^{2} d^{\frac{3}{2}} + 90 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b^{2} c^{4} \sqrt{d} + 220 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b c^{3} d^{\frac{3}{2}} + 30 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{2} c^{2} d^{\frac{5}{2}} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{2} c^{5} \sqrt{d} - 140 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c^{4} d^{\frac{3}{2}} + 15 \, b^{2} c^{6} \sqrt{d} + 40 \, a b c^{5} d^{\frac{3}{2}} + 3 \, a^{2} c^{4} d^{\frac{5}{2}}\right )}}{15 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/x^6,x, algorithm="giac")

[Out]

1/2*sqrt(d*x^2 + c)*b^2*d*x - 1/4*(3*b^2*c*sqrt(d) + 4*a*b*d^(3/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2) + 2/1
5*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c^2*sqrt(d) + 60*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a*b*c*d^(3/2) + 15*
(sqrt(d)*x - sqrt(d*x^2 + c))^8*a^2*d^(5/2) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^2*c^3*sqrt(d) - 180*(sqrt(d
)*x - sqrt(d*x^2 + c))^6*a*b*c^2*d^(3/2) + 90*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2*c^4*sqrt(d) + 220*(sqrt(d)*x
- sqrt(d*x^2 + c))^4*a*b*c^3*d^(3/2) + 30*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c^2*d^(5/2) - 60*(sqrt(d)*x - s
qrt(d*x^2 + c))^2*b^2*c^5*sqrt(d) - 140*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^4*d^(3/2) + 15*b^2*c^6*sqrt(d) +
40*a*b*c^5*d^(3/2) + 3*a^2*c^4*d^(5/2))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^5