3.755 \(\int \frac{(c+d x^2)^{5/2}}{x^2 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=168 \[ -\frac{(b c-a d)^{3/2} (2 a d+3 b c) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{5/2} b^2}-\frac{c \sqrt{c+d x^2} (3 b c-a d)}{2 a^2 b x}+\frac{\left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b x \left (a+b x^2\right )}+\frac{d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.188748, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {468, 580, 523, 217, 206, 377, 205} \[ -\frac{(b c-a d)^{3/2} (2 a d+3 b c) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{5/2} b^2}-\frac{c \sqrt{c+d x^2} (3 b c-a d)}{2 a^2 b x}+\frac{\left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b x \left (a+b x^2\right )}+\frac{d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^(5/2)/(x^2*(a + b*x^2)^2),x]

[Out]

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

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[((c*b -
a*d)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*b*e*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), I
nt[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c*b - a*d)*(m + 1)) + d*(c*b*n*(p
+ 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, 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])

Rule 377

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.14581, size = 150, normalized size = 0.89 \[ -\frac{(b c-a d)^{3/2} (2 a d+3 b c) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{5/2} b^2}+\sqrt{c+d x^2} \left (-\frac{x (b c-a d)^2}{2 a^2 b \left (a+b x^2\right )}-\frac{c^2}{a^2 x}\right )+\frac{d^{5/2} \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^(5/2)/(x^2*(a + b*x^2)^2),x]

[Out]

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

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Maple [B]  time = 0.014, size = 7529, normalized size = 44.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^(5/2)/x^2/(b*x^2+a)^2,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )}^{\frac{5}{2}}}{{\left (b x^{2} + a\right )}^{2} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^2/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^(5/2)/((b*x^2 + a)^2*x^2), x)

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Fricas [A]  time = 4.51749, size = 2458, normalized size = 14.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^2/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/8*(4*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - ((3*b^3*c^2 - a*
b^2*c*d - 2*a^2*b*d^2)*x^3 + (3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*d^2)*x)*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a
*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d
*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(2*a*b^2*c^2 + (3*b^3*c^2 - 2*a*b^2*c*d + a^2
*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a^2*b^3*x^3 + a^3*b^2*x), -1/8*(8*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(-d)*arctan(s
qrt(-d)*x/sqrt(d*x^2 + c)) + ((3*b^3*c^2 - a*b^2*c*d - 2*a^2*b*d^2)*x^3 + (3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*d^2
)*x)*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2
- 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*
(2*a*b^2*c^2 + (3*b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a^2*b^3*x^3 + a^3*b^2*x), -1/4*(((
3*b^3*c^2 - a*b^2*c*d - 2*a^2*b*d^2)*x^3 + (3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*d^2)*x)*sqrt((b*c - a*d)/a)*arctan
(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x))
- 2*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(2*a*b^2*c^2 + (3*
b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a^2*b^3*x^3 + a^3*b^2*x), -1/4*(4*(a^2*b*d^2*x^3 + a
^3*d^2*x)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + ((3*b^3*c^2 - a*b^2*c*d - 2*a^2*b*d^2)*x^3 + (3*a*b^2*
c^2 - a^2*b*c*d - 2*a^3*d^2)*x)*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt(
(b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)) + 2*(2*a*b^2*c^2 + (3*b^3*c^2 - 2*a*b^2*c*d + a^2*b*
d^2)*x^2)*sqrt(d*x^2 + c))/(a^2*b^3*x^3 + a^3*b^2*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x^{2}\right )^{\frac{5}{2}}}{x^{2} \left (a + b x^{2}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(5/2)/x**2/(b*x**2+a)**2,x)

[Out]

Integral((c + d*x**2)**(5/2)/(x**2*(a + b*x**2)**2), x)

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Giac [B]  time = 1.21701, size = 740, normalized size = 4.4 \begin{align*} -\frac{d^{\frac{5}{2}} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{2 \, b^{2}} + \frac{{\left (3 \, b^{3} c^{3} \sqrt{d} - 4 \, a b^{2} c^{2} d^{\frac{3}{2}} - a^{2} b c d^{\frac{5}{2}} + 2 \, a^{3} d^{\frac{7}{2}}\right )} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt{a b c d - a^{2} d^{2}} a^{2} b^{2}} + \frac{3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b^{3} c^{3} \sqrt{d} - 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b^{2} c^{2} d^{\frac{3}{2}} + 5 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{2} b c d^{\frac{5}{2}} - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{3} d^{\frac{7}{2}} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{3} c^{4} \sqrt{d} + 14 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b^{2} c^{3} d^{\frac{3}{2}} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} b c^{2} d^{\frac{5}{2}} + 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{3} c d^{\frac{7}{2}} + 3 \, b^{3} c^{5} \sqrt{d} - 2 \, a b^{2} c^{4} d^{\frac{3}{2}} + a^{2} b c^{3} d^{\frac{5}{2}}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} b - 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a d + 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c^{2} - 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a c d - b c^{3}\right )} a^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^2/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-1/2*d^(5/2)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/b^2 + 1/2*(3*b^3*c^3*sqrt(d) - 4*a*b^2*c^2*d^(3/2) - a^2*b*c
*d^(5/2) + 2*a^3*d^(7/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2)
)/(sqrt(a*b*c*d - a^2*d^2)*a^2*b^2) + (3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^3*c^3*sqrt(d) - 4*(sqrt(d)*x - sqrt
(d*x^2 + c))^4*a*b^2*c^2*d^(3/2) + 5*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*b*c*d^(5/2) - 2*(sqrt(d)*x - sqrt(d*x
^2 + c))^4*a^3*d^(7/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b^3*c^4*sqrt(d) + 14*(sqrt(d)*x - sqrt(d*x^2 + c))^
2*a*b^2*c^3*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*b*c^2*d^(5/2) + 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*
a^3*c*d^(7/2) + 3*b^3*c^5*sqrt(d) - 2*a*b^2*c^4*d^(3/2) + a^2*b*c^3*d^(5/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^6
*b - 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*d + 3*(sqrt(d)*x - sqrt(d*x^2
+ c))^2*b*c^2 - 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*c*d - b*c^3)*a^2*b^2)