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

Optimal. Leaf size=175 \[ -\frac{a^2 \left (c+d x^2\right )^{5/2}}{c x}-\frac{c \left (b^2 c^2-12 a d (2 a d+b c)\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 d^{3/2}}-\frac{x \left (c+d x^2\right )^{3/2} \left (b^2 c^2-12 a d (2 a d+b c)\right )}{24 c d}-\frac{x \sqrt{c+d x^2} \left (b^2 c^2-12 a d (2 a d+b c)\right )}{16 d}+\frac{b^2 x \left (c+d x^2\right )^{5/2}}{6 d} \]

[Out]

-((b^2*c^2 - 12*a*d*(b*c + 2*a*d))*x*Sqrt[c + d*x^2])/(16*d) - ((b^2*c^2 - 12*a*d*(b*c + 2*a*d))*x*(c + d*x^2)
^(3/2))/(24*c*d) - (a^2*(c + d*x^2)^(5/2))/(c*x) + (b^2*x*(c + d*x^2)^(5/2))/(6*d) - (c*(b^2*c^2 - 12*a*d*(b*c
+ 2*a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*d^(3/2))

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Rubi [A]  time = 0.118555, antiderivative size = 172, 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, 388, 195, 217, 206} \[ -\frac{a^2 \left (c+d x^2\right )^{5/2}}{c x}-\frac{c \left (b^2 c^2-12 a d (2 a d+b c)\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 d^{3/2}}-\frac{x \sqrt{c+d x^2} \left (b^2 c^2-12 a d (2 a d+b c)\right )}{16 d}-\frac{1}{24} x \left (c+d x^2\right )^{3/2} \left (\frac{b^2 c}{d}-\frac{12 a (2 a d+b c)}{c}\right )+\frac{b^2 x \left (c+d x^2\right )^{5/2}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*c^2 - 12*a*d*(b*c + 2*a*d))*x*Sqrt[c + d*x^2])/(16*d) - (((b^2*c)/d - (12*a*(b*c + 2*a*d))/c)*x*(c + d*
x^2)^(3/2))/24 - (a^2*(c + d*x^2)^(5/2))/(c*x) + (b^2*x*(c + d*x^2)^(5/2))/(6*d) - (c*(b^2*c^2 - 12*a*d*(b*c +
2*a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*d^(3/2))

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 388

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

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^2} \, dx &=-\frac{a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac{\int \left (2 a (b c+2 a d)+b^2 c x^2\right ) \left (c+d x^2\right )^{3/2} \, dx}{c}\\ &=-\frac{a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac{b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac{\left (b^2 c^2-12 a d (b c+2 a d)\right ) \int \left (c+d x^2\right )^{3/2} \, dx}{6 c d}\\ &=-\frac{1}{24} \left (\frac{b^2 c}{d}-\frac{12 a (b c+2 a d)}{c}\right ) x \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac{b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac{\left (b^2 c^2-12 a d (b c+2 a d)\right ) \int \sqrt{c+d x^2} \, dx}{8 d}\\ &=-\frac{\left (b^2 c^2-12 a d (b c+2 a d)\right ) x \sqrt{c+d x^2}}{16 d}-\frac{1}{24} \left (\frac{b^2 c}{d}-\frac{12 a (b c+2 a d)}{c}\right ) x \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac{b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac{\left (c \left (b^2 c^2-12 a d (b c+2 a d)\right )\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{16 d}\\ &=-\frac{\left (b^2 c^2-12 a d (b c+2 a d)\right ) x \sqrt{c+d x^2}}{16 d}-\frac{1}{24} \left (\frac{b^2 c}{d}-\frac{12 a (b c+2 a d)}{c}\right ) x \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac{b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac{\left (c \left (b^2 c^2-12 a d (b c+2 a d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{16 d}\\ &=-\frac{\left (b^2 c^2-12 a d (b c+2 a d)\right ) x \sqrt{c+d x^2}}{16 d}-\frac{1}{24} \left (\frac{b^2 c}{d}-\frac{12 a (b c+2 a d)}{c}\right ) x \left (c+d x^2\right )^{3/2}-\frac{a^2 \left (c+d x^2\right )^{5/2}}{c x}+\frac{b^2 x \left (c+d x^2\right )^{5/2}}{6 d}-\frac{c \left (b^2 c^2-12 a d (b c+2 a d)\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{16 d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.14645, size = 135, normalized size = 0.77 \[ \sqrt{c+d x^2} \left (\frac{x \left (8 a^2 d^2+20 a b c d+b^2 c^2\right )}{16 d}-\frac{a^2 c}{x}+\frac{1}{24} b x^3 (12 a d+7 b c)+\frac{1}{6} b^2 d x^5\right )-\frac{c \left (-24 a^2 d^2-12 a b c d+b^2 c^2\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{16 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + d*x^2]*(-((a^2*c)/x) + ((b^2*c^2 + 20*a*b*c*d + 8*a^2*d^2)*x)/(16*d) + (b*(7*b*c + 12*a*d)*x^3)/24 +
(b^2*d*x^5)/6) - (c*(b^2*c^2 - 12*a*b*c*d - 24*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(16*d^(3/2))

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Maple [A]  time = 0.012, size = 221, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}x}{6\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{{b}^{2}cx}{24\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}{c}^{2}x}{16\,d}\sqrt{d{x}^{2}+c}}-{\frac{{b}^{2}{c}^{3}}{16}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{3}{2}}}}+{\frac{abx}{2} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{3\,abcx}{4}\sqrt{d{x}^{2}+c}}+{\frac{3\,ab{c}^{2}}{4}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){\frac{1}{\sqrt{d}}}}-{\frac{{a}^{2}}{cx} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}dx}{c} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{a}^{2}dx}{2}\sqrt{d{x}^{2}+c}}+{\frac{3\,{a}^{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^2,x)

[Out]

1/6*b^2*x*(d*x^2+c)^(5/2)/d-1/24*b^2*c/d*x*(d*x^2+c)^(3/2)-1/16*b^2*c^2/d*x*(d*x^2+c)^(1/2)-1/16*b^2*c^3/d^(3/
2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+1/2*a*b*x*(d*x^2+c)^(3/2)+3/4*a*b*c*x*(d*x^2+c)^(1/2)+3/4*a*b*c^2/d^(1/2)*ln(
x*d^(1/2)+(d*x^2+c)^(1/2))-a^2*(d*x^2+c)^(5/2)/c/x+a^2*d/c*x*(d*x^2+c)^(3/2)+3/2*a^2*d*x*(d*x^2+c)^(1/2)+3/2*a
^2*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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38773, size = 660, normalized size = 3.77 \begin{align*} \left [-\frac{3 \,{\left (b^{2} c^{3} - 12 \, a b c^{2} d - 24 \, a^{2} c d^{2}\right )} \sqrt{d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) - 2 \,{\left (8 \, b^{2} d^{3} x^{6} - 48 \, a^{2} c d^{2} + 2 \,{\left (7 \, b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{4} + 3 \,{\left (b^{2} c^{2} d + 20 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{96 \, d^{2} x}, \frac{3 \,{\left (b^{2} c^{3} - 12 \, a b c^{2} d - 24 \, a^{2} c d^{2}\right )} \sqrt{-d} x \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) +{\left (8 \, b^{2} d^{3} x^{6} - 48 \, a^{2} c d^{2} + 2 \,{\left (7 \, b^{2} c d^{2} + 12 \, a b d^{3}\right )} x^{4} + 3 \,{\left (b^{2} c^{2} d + 20 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{48 \, d^{2} x}\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^2,x, algorithm="fricas")

[Out]

[-1/96*(3*(b^2*c^3 - 12*a*b*c^2*d - 24*a^2*c*d^2)*sqrt(d)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) -
2*(8*b^2*d^3*x^6 - 48*a^2*c*d^2 + 2*(7*b^2*c*d^2 + 12*a*b*d^3)*x^4 + 3*(b^2*c^2*d + 20*a*b*c*d^2 + 8*a^2*d^3)*
x^2)*sqrt(d*x^2 + c))/(d^2*x), 1/48*(3*(b^2*c^3 - 12*a*b*c^2*d - 24*a^2*c*d^2)*sqrt(-d)*x*arctan(sqrt(-d)*x/sq
rt(d*x^2 + c)) + (8*b^2*d^3*x^6 - 48*a^2*c*d^2 + 2*(7*b^2*c*d^2 + 12*a*b*d^3)*x^4 + 3*(b^2*c^2*d + 20*a*b*c*d^
2 + 8*a^2*d^3)*x^2)*sqrt(d*x^2 + c))/(d^2*x)]

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Sympy [B]  time = 16.6259, size = 367, normalized size = 2.1 \begin{align*} - \frac{a^{2} c^{\frac{3}{2}}}{x \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{a^{2} \sqrt{c} d x \sqrt{1 + \frac{d x^{2}}{c}}}{2} - \frac{a^{2} \sqrt{c} d x}{\sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 a^{2} c \sqrt{d} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{2} + a b c^{\frac{3}{2}} x \sqrt{1 + \frac{d x^{2}}{c}} + \frac{a b c^{\frac{3}{2}} x}{4 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 a b \sqrt{c} d x^{3}}{4 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 a b c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{4 \sqrt{d}} + \frac{a b d^{2} x^{5}}{2 \sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{b^{2} c^{\frac{5}{2}} x}{16 d \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{17 b^{2} c^{\frac{3}{2}} x^{3}}{48 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{11 b^{2} \sqrt{c} d x^{5}}{24 \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{b^{2} c^{3} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{16 d^{\frac{3}{2}}} + \frac{b^{2} d^{2} x^{7}}{6 \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**2,x)

[Out]

-a**2*c**(3/2)/(x*sqrt(1 + d*x**2/c)) + a**2*sqrt(c)*d*x*sqrt(1 + d*x**2/c)/2 - a**2*sqrt(c)*d*x/sqrt(1 + d*x*
*2/c) + 3*a**2*c*sqrt(d)*asinh(sqrt(d)*x/sqrt(c))/2 + a*b*c**(3/2)*x*sqrt(1 + d*x**2/c) + a*b*c**(3/2)*x/(4*sq
rt(1 + d*x**2/c)) + 3*a*b*sqrt(c)*d*x**3/(4*sqrt(1 + d*x**2/c)) + 3*a*b*c**2*asinh(sqrt(d)*x/sqrt(c))/(4*sqrt(
d)) + a*b*d**2*x**5/(2*sqrt(c)*sqrt(1 + d*x**2/c)) + b**2*c**(5/2)*x/(16*d*sqrt(1 + d*x**2/c)) + 17*b**2*c**(3
/2)*x**3/(48*sqrt(1 + d*x**2/c)) + 11*b**2*sqrt(c)*d*x**5/(24*sqrt(1 + d*x**2/c)) - b**2*c**3*asinh(sqrt(d)*x/
sqrt(c))/(16*d**(3/2)) + b**2*d**2*x**7/(6*sqrt(c)*sqrt(1 + d*x**2/c))

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Giac [A]  time = 1.13756, size = 234, normalized size = 1.34 \begin{align*} \frac{2 \, a^{2} c^{2} \sqrt{d}}{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c} + \frac{1}{48} \,{\left (2 \,{\left (4 \, b^{2} d x^{2} + \frac{7 \, b^{2} c d^{4} + 12 \, a b d^{5}}{d^{4}}\right )} x^{2} + \frac{3 \,{\left (b^{2} c^{2} d^{3} + 20 \, a b c d^{4} + 8 \, a^{2} d^{5}\right )}}{d^{4}}\right )} \sqrt{d x^{2} + c} x + \frac{{\left (b^{2} c^{3} \sqrt{d} - 12 \, a b c^{2} d^{\frac{3}{2}} - 24 \, a^{2} c d^{\frac{5}{2}}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{32 \, d^{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^2,x, algorithm="giac")

[Out]

2*a^2*c^2*sqrt(d)/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c) + 1/48*(2*(4*b^2*d*x^2 + (7*b^2*c*d^4 + 12*a*b*d^5)/d^
4)*x^2 + 3*(b^2*c^2*d^3 + 20*a*b*c*d^4 + 8*a^2*d^5)/d^4)*sqrt(d*x^2 + c)*x + 1/32*(b^2*c^3*sqrt(d) - 12*a*b*c^
2*d^(3/2) - 24*a^2*c*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/d^2