3.609 \(\int \frac{(a+b x^2)^2 \sqrt{c+d x^2}}{x^6} \, dx\)

Optimal. Leaf size=103 \[ -\frac{a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac{2 a \left (c+d x^2\right )^{3/2} (5 b c-a d)}{15 c^2 x^3}-\frac{b^2 \sqrt{c+d x^2}}{x}+b^2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right ) \]

[Out]

-((b^2*Sqrt[c + d*x^2])/x) - (a^2*(c + d*x^2)^(3/2))/(5*c*x^5) - (2*a*(5*b*c - a*d)*(c + d*x^2)^(3/2))/(15*c^2
*x^3) + b^2*Sqrt[d]*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]]

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Rubi [A]  time = 0.0580408, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {462, 451, 277, 217, 206} \[ -\frac{a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac{2 a \left (c+d x^2\right )^{3/2} (5 b c-a d)}{15 c^2 x^3}-\frac{b^2 \sqrt{c+d x^2}}{x}+b^2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*Sqrt[c + d*x^2])/x) - (a^2*(c + d*x^2)^(3/2))/(5*c*x^5) - (2*a*(5*b*c - a*d)*(c + d*x^2)^(3/2))/(15*c^2
*x^3) + b^2*Sqrt[d]*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^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 451

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 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 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 \sqrt{c+d x^2}}{x^6} \, dx &=-\frac{a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}+\frac{\int \frac{\left (2 a (5 b c-a d)+5 b^2 c x^2\right ) \sqrt{c+d x^2}}{x^4} \, dx}{5 c}\\ &=-\frac{a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac{2 a (5 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^3}+b^2 \int \frac{\sqrt{c+d x^2}}{x^2} \, dx\\ &=-\frac{b^2 \sqrt{c+d x^2}}{x}-\frac{a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac{2 a (5 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^3}+\left (b^2 d\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx\\ &=-\frac{b^2 \sqrt{c+d x^2}}{x}-\frac{a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac{2 a (5 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^3}+\left (b^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )\\ &=-\frac{b^2 \sqrt{c+d x^2}}{x}-\frac{a^2 \left (c+d x^2\right )^{3/2}}{5 c x^5}-\frac{2 a (5 b c-a d) \left (c+d x^2\right )^{3/2}}{15 c^2 x^3}+b^2 \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.083774, size = 104, normalized size = 1.01 \[ b^2 \sqrt{d} \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )-\frac{\sqrt{c+d x^2} \left (a^2 \left (3 c^2+c d x^2-2 d^2 x^4\right )+10 a b c x^2 \left (c+d x^2\right )+15 b^2 c^2 x^4\right )}{15 c^2 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d*x^2]*(15*b^2*c^2*x^4 + 10*a*b*c*x^2*(c + d*x^2) + a^2*(3*c^2 + c*d*x^2 - 2*d^2*x^4)))/(15*c^2*x^5
) + b^2*Sqrt[d]*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]]

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Maple [A]  time = 0.012, size = 123, normalized size = 1.2 \begin{align*} -{\frac{2\,ab}{3\,c{x}^{3}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{2}}{5\,c{x}^{5}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{2\,{a}^{2}d}{15\,{c}^{2}{x}^{3}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}}{cx} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{{b}^{2}dx}{c}\sqrt{d{x}^{2}+c}}+{b}^{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)^(1/2)/x^6,x)

[Out]

-2/3*a*b/c/x^3*(d*x^2+c)^(3/2)-1/5*a^2*(d*x^2+c)^(3/2)/c/x^5+2/15*a^2*d/c^2/x^3*(d*x^2+c)^(3/2)-b^2/c/x*(d*x^2
+c)^(3/2)+b^2*d/c*x*(d*x^2+c)^(1/2)+b^2*d^(1/2)*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)^(1/2)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65963, size = 512, normalized size = 4.97 \begin{align*} \left [\frac{15 \, b^{2} c^{2} \sqrt{d} x^{5} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) - 2 \,{\left ({\left (15 \, b^{2} c^{2} + 10 \, a b c d - 2 \, a^{2} d^{2}\right )} x^{4} + 3 \, a^{2} c^{2} +{\left (10 \, a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{30 \, c^{2} x^{5}}, -\frac{15 \, b^{2} c^{2} \sqrt{-d} x^{5} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) +{\left ({\left (15 \, b^{2} c^{2} + 10 \, a b c d - 2 \, a^{2} d^{2}\right )} x^{4} + 3 \, a^{2} c^{2} +{\left (10 \, a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{15 \, c^{2} 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)^(1/2)/x^6,x, algorithm="fricas")

[Out]

[1/30*(15*b^2*c^2*sqrt(d)*x^5*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*((15*b^2*c^2 + 10*a*b*c*d -
2*a^2*d^2)*x^4 + 3*a^2*c^2 + (10*a*b*c^2 + a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(c^2*x^5), -1/15*(15*b^2*c^2*sqrt(-d
)*x^5*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + ((15*b^2*c^2 + 10*a*b*c*d - 2*a^2*d^2)*x^4 + 3*a^2*c^2 + (10*a*b*c^
2 + a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(c^2*x^5)]

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Sympy [B]  time = 3.44587, size = 199, normalized size = 1.93 \begin{align*} - \frac{a^{2} \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{5 x^{4}} - \frac{a^{2} d^{\frac{3}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{15 c x^{2}} + \frac{2 a^{2} d^{\frac{5}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{15 c^{2}} - \frac{2 a b \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 c} - \frac{b^{2} \sqrt{c}}{x \sqrt{1 + \frac{d x^{2}}{c}}} + b^{2} \sqrt{d} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )} - \frac{b^{2} d x}{\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)**(1/2)/x**6,x)

[Out]

-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(5*x**4) - a**2*d**(3/2)*sqrt(c/(d*x**2) + 1)/(15*c*x**2) + 2*a**2*d**(5/2)
*sqrt(c/(d*x**2) + 1)/(15*c**2) - 2*a*b*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - 2*a*b*d**(3/2)*sqrt(c/(d*x**2)
+ 1)/(3*c) - b**2*sqrt(c)/(x*sqrt(1 + d*x**2/c)) + b**2*sqrt(d)*asinh(sqrt(d)*x/sqrt(c)) - b**2*d*x/(sqrt(c)*
sqrt(1 + d*x**2/c))

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Giac [B]  time = 1.16319, size = 544, normalized size = 5.28 \begin{align*} -\frac{1}{2} \, b^{2} \sqrt{d} \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 \sqrt{d} + 30 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} a b d^{\frac{3}{2}} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} b^{2} c^{2} \sqrt{d} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} a b c d^{\frac{3}{2}} + 30 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} a^{2} d^{\frac{5}{2}} + 90 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b^{2} c^{3} \sqrt{d} + 40 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b c^{2} d^{\frac{3}{2}} + 10 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{2} c d^{\frac{5}{2}} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{2} c^{4} \sqrt{d} - 20 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c^{3} d^{\frac{3}{2}} + 10 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} c^{2} d^{\frac{5}{2}} + 15 \, b^{2} c^{5} \sqrt{d} + 10 \, a b c^{4} d^{\frac{3}{2}} - 2 \, a^{2} c^{3} 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)^(1/2)/x^6,x, algorithm="giac")

[Out]

-1/2*b^2*sqrt(d)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2) + 2/15*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c*sqrt(d)
+ 30*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a*b*d^(3/2) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^2*c^2*sqrt(d) - 60*(s
qrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c*d^(3/2) + 30*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a^2*d^(5/2) + 90*(sqrt(d)*x -
sqrt(d*x^2 + c))^4*b^2*c^3*sqrt(d) + 40*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^2*d^(3/2) + 10*(sqrt(d)*x - sqr
t(d*x^2 + c))^4*a^2*c*d^(5/2) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c^4*sqrt(d) - 20*(sqrt(d)*x - sqrt(d*x^
2 + c))^2*a*b*c^3*d^(3/2) + 10*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*c^2*d^(5/2) + 15*b^2*c^5*sqrt(d) + 10*a*b*c
^4*d^(3/2) - 2*a^2*c^3*d^(5/2))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^5