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

Optimal. Leaf size=184 \[ -\frac{a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac{x \left (c+d x^2\right )^{3/2} \left (8 a d (a d+3 b c)+3 b^2 c^2\right )}{12 c^2}+\frac{x \sqrt{c+d x^2} \left (8 a d (a d+3 b c)+3 b^2 c^2\right )}{8 c}+\frac{\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 \sqrt{d}}-\frac{2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{3 c^2 x} \]

[Out]

((3*b^2*c^2 + 8*a*d*(3*b*c + a*d))*x*Sqrt[c + d*x^2])/(8*c) + ((3*b^2*c^2 + 8*a*d*(3*b*c + a*d))*x*(c + d*x^2)
^(3/2))/(12*c^2) - (a^2*(c + d*x^2)^(5/2))/(3*c*x^3) - (2*a*(3*b*c + a*d)*(c + d*x^2)^(5/2))/(3*c^2*x) + ((3*b
^2*c^2 + 8*a*d*(3*b*c + a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*Sqrt[d])

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Rubi [A]  time = 0.129846, antiderivative size = 181, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {462, 453, 195, 217, 206} \[ -\frac{a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac{1}{12} x \left (c+d x^2\right )^{3/2} \left (\frac{8 a d (a d+3 b c)}{c^2}+3 b^2\right )+\frac{x \sqrt{c+d x^2} \left (8 a d (a d+3 b c)+3 b^2 c^2\right )}{8 c}+\frac{\left (8 a d (a d+3 b c)+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 \sqrt{d}}-\frac{2 a \left (c+d x^2\right )^{5/2} (a d+3 b c)}{3 c^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b^2*c^2 + 8*a*d*(3*b*c + a*d))*x*Sqrt[c + d*x^2])/(8*c) + ((3*b^2 + (8*a*d*(3*b*c + a*d))/c^2)*x*(c + d*x^
2)^(3/2))/12 - (a^2*(c + d*x^2)^(5/2))/(3*c*x^3) - (2*a*(3*b*c + a*d)*(c + d*x^2)^(5/2))/(3*c^2*x) + ((3*b^2*c
^2 + 8*a*d*(3*b*c + a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*Sqrt[d])

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 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^4} \, dx &=-\frac{a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}+\frac{\int \frac{\left (2 a (3 b c+a d)+3 b^2 c x^2\right ) \left (c+d x^2\right )^{3/2}}{x^2} \, dx}{3 c}\\ &=-\frac{a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac{2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}-\frac{1}{3} \left (-3 b^2-\frac{8 a d (3 b c+a d)}{c^2}\right ) \int \left (c+d x^2\right )^{3/2} \, dx\\ &=\frac{1}{12} \left (3 b^2+\frac{8 a d (3 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac{2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}-\frac{1}{4} \left (c \left (-3 b^2-\frac{8 a d (3 b c+a d)}{c^2}\right )\right ) \int \sqrt{c+d x^2} \, dx\\ &=\frac{1}{8} c \left (3 b^2+\frac{8 a d (3 b c+a d)}{c^2}\right ) x \sqrt{c+d x^2}+\frac{1}{12} \left (3 b^2+\frac{8 a d (3 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac{2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}-\frac{1}{8} \left (-3 b^2 c^2-24 a b c d-8 a^2 d^2\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx\\ &=\frac{1}{8} c \left (3 b^2+\frac{8 a d (3 b c+a d)}{c^2}\right ) x \sqrt{c+d x^2}+\frac{1}{12} \left (3 b^2+\frac{8 a d (3 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac{2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}-\frac{1}{8} \left (-3 b^2 c^2-24 a b c d-8 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )\\ &=\frac{1}{8} c \left (3 b^2+\frac{8 a d (3 b c+a d)}{c^2}\right ) x \sqrt{c+d x^2}+\frac{1}{12} \left (3 b^2+\frac{8 a d (3 b c+a d)}{c^2}\right ) x \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{3 c x^3}-\frac{2 a (3 b c+a d) \left (c+d x^2\right )^{5/2}}{3 c^2 x}+\frac{\left (3 b^2 c^2+24 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 0.0916018, size = 118, normalized size = 0.64 \[ \frac{1}{24} \left (\frac{3 \left (8 a^2 d^2+24 a b c d+3 b^2 c^2\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{\sqrt{d}}+\frac{\sqrt{c+d x^2} \left (-8 a^2 c+3 b x^4 (8 a d+5 b c)-16 a x^2 (2 a d+3 b c)+6 b^2 d x^6\right )}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[c + d*x^2]*(-8*a^2*c - 16*a*(3*b*c + 2*a*d)*x^2 + 3*b*(5*b*c + 8*a*d)*x^4 + 6*b^2*d*x^6))/x^3 + (3*(3*b
^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/Sqrt[d])/24

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Maple [A]  time = 0.011, size = 241, normalized size = 1.3 \begin{align*}{\frac{x{b}^{2}}{4} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{b}^{2}cx}{8}\sqrt{d{x}^{2}+c}}+{\frac{3\,{b}^{2}{c}^{2}}{8}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){\frac{1}{\sqrt{d}}}}-{\frac{{a}^{2}}{3\,c{x}^{3}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{2\,{a}^{2}d}{3\,{c}^{2}x} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{a}^{2}{d}^{2}x}{3\,{c}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{{a}^{2}{d}^{2}x}{c}\sqrt{d{x}^{2}+c}}+{a}^{2}{d}^{{\frac{3}{2}}}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) -2\,{\frac{ab \left ( d{x}^{2}+c \right ) ^{5/2}}{cx}}+2\,{\frac{abdx \left ( d{x}^{2}+c \right ) ^{3/2}}{c}}+3\,abdx\sqrt{d{x}^{2}+c}+3\,ab\sqrt{d}c\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^4,x)

[Out]

1/4*x*b^2*(d*x^2+c)^(3/2)+3/8*b^2*c*x*(d*x^2+c)^(1/2)+3/8*b^2*c^2/d^(1/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))-1/3*a^
2*(d*x^2+c)^(5/2)/c/x^3-2/3*a^2*d/c^2/x*(d*x^2+c)^(5/2)+2/3*a^2*d^2/c^2*x*(d*x^2+c)^(3/2)+a^2*d^2/c*x*(d*x^2+c
)^(1/2)+a^2*d^(3/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))-2*a*b/c/x*(d*x^2+c)^(5/2)+2*a*b*d/c*x*(d*x^2+c)^(3/2)+3*a*b*
d*x*(d*x^2+c)^(1/2)+3*a*b*d^(1/2)*c*ln(x*d^(1/2)+(d*x^2+c)^(1/2))

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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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38383, size = 603, normalized size = 3.28 \begin{align*} \left [\frac{3 \,{\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt{d} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (6 \, b^{2} d^{2} x^{6} + 3 \,{\left (5 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{4} - 8 \, a^{2} c d - 16 \,{\left (3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{48 \, d x^{3}}, -\frac{3 \,{\left (3 \, b^{2} c^{2} + 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt{-d} x^{3} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) -{\left (6 \, b^{2} d^{2} x^{6} + 3 \,{\left (5 \, b^{2} c d + 8 \, a b d^{2}\right )} x^{4} - 8 \, a^{2} c d - 16 \,{\left (3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{24 \, d x^{3}}\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^4,x, algorithm="fricas")

[Out]

[1/48*(3*(3*b^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*sqrt(d)*x^3*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*
(6*b^2*d^2*x^6 + 3*(5*b^2*c*d + 8*a*b*d^2)*x^4 - 8*a^2*c*d - 16*(3*a*b*c*d + 2*a^2*d^2)*x^2)*sqrt(d*x^2 + c))/
(d*x^3), -1/24*(3*(3*b^2*c^2 + 24*a*b*c*d + 8*a^2*d^2)*sqrt(-d)*x^3*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (6*b^
2*d^2*x^6 + 3*(5*b^2*c*d + 8*a*b*d^2)*x^4 - 8*a^2*c*d - 16*(3*a*b*c*d + 2*a^2*d^2)*x^2)*sqrt(d*x^2 + c))/(d*x^
3)]

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Sympy [B]  time = 10.932, size = 352, normalized size = 1.91 \begin{align*} - \frac{a^{2} \sqrt{c} d}{x \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{a^{2} c \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{3 x^{2}} - \frac{a^{2} d^{\frac{3}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{3} + a^{2} d^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )} - \frac{a^{2} d^{2} x}{\sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{2 a b c^{\frac{3}{2}}}{x \sqrt{1 + \frac{d x^{2}}{c}}} + a b \sqrt{c} d x \sqrt{1 + \frac{d x^{2}}{c}} - \frac{2 a b \sqrt{c} d x}{\sqrt{1 + \frac{d x^{2}}{c}}} + 3 a b c \sqrt{d} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )} + \frac{b^{2} c^{\frac{3}{2}} x \sqrt{1 + \frac{d x^{2}}{c}}}{2} + \frac{b^{2} c^{\frac{3}{2}} x}{8 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 b^{2} \sqrt{c} d x^{3}}{8 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 b^{2} c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{8 \sqrt{d}} + \frac{b^{2} d^{2} x^{5}}{4 \sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} \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**4,x)

[Out]

-a**2*sqrt(c)*d/(x*sqrt(1 + d*x**2/c)) - a**2*c*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - a**2*d**(3/2)*sqrt(c/(
d*x**2) + 1)/3 + a**2*d**(3/2)*asinh(sqrt(d)*x/sqrt(c)) - a**2*d**2*x/(sqrt(c)*sqrt(1 + d*x**2/c)) - 2*a*b*c**
(3/2)/(x*sqrt(1 + d*x**2/c)) + a*b*sqrt(c)*d*x*sqrt(1 + d*x**2/c) - 2*a*b*sqrt(c)*d*x/sqrt(1 + d*x**2/c) + 3*a
*b*c*sqrt(d)*asinh(sqrt(d)*x/sqrt(c)) + b**2*c**(3/2)*x*sqrt(1 + d*x**2/c)/2 + b**2*c**(3/2)*x/(8*sqrt(1 + d*x
**2/c)) + 3*b**2*sqrt(c)*d*x**3/(8*sqrt(1 + d*x**2/c)) + 3*b**2*c**2*asinh(sqrt(d)*x/sqrt(c))/(8*sqrt(d)) + b*
*2*d**2*x**5/(4*sqrt(c)*sqrt(1 + d*x**2/c))

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Giac [A]  time = 1.15275, size = 354, normalized size = 1.92 \begin{align*} \frac{1}{8} \,{\left (2 \, b^{2} d x^{2} + \frac{5 \, b^{2} c d^{2} + 8 \, a b d^{3}}{d^{2}}\right )} \sqrt{d x^{2} + c} x - \frac{{\left (3 \, b^{2} c^{2} \sqrt{d} + 24 \, a b c d^{\frac{3}{2}} + 8 \, a^{2} d^{\frac{5}{2}}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{16 \, d} + \frac{4 \,{\left (3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b c^{2} \sqrt{d} + 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{2} c d^{\frac{3}{2}} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c^{3} \sqrt{d} - 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} c^{2} d^{\frac{3}{2}} + 3 \, a b c^{4} \sqrt{d} + 2 \, a^{2} c^{3} d^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{3}} \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^4,x, algorithm="giac")

[Out]

1/8*(2*b^2*d*x^2 + (5*b^2*c*d^2 + 8*a*b*d^3)/d^2)*sqrt(d*x^2 + c)*x - 1/16*(3*b^2*c^2*sqrt(d) + 24*a*b*c*d^(3/
2) + 8*a^2*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/d + 4/3*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^2*sq
rt(d) + 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^3*sqrt(d) -
3*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*c^2*d^(3/2) + 3*a*b*c^4*sqrt(d) + 2*a^2*c^3*d^(3/2))/((sqrt(d)*x - sqrt(
d*x^2 + c))^2 - c)^3