3.756 \(\int \frac{(c+d x^2)^{5/2}}{x^3 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=180 \[ -\frac{(b c-a d)^{3/2} (a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 a^3 b^{3/2}}+\frac{c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^3}-\frac{\sqrt{c+d x^2} (b c-a d) (2 b c-a d)}{2 a^2 b \left (a+b x^2\right )}-\frac{c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.269985, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 98, 149, 156, 63, 208} \[ -\frac{(b c-a d)^{3/2} (a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 a^3 b^{3/2}}+\frac{c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^3}-\frac{\sqrt{c+d x^2} (b c-a d) (2 b c-a d)}{2 a^2 b \left (a+b x^2\right )}-\frac{c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^(5/2)/(x^3*(a + b*x^2)^2),x]

[Out]

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

Rule 446

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

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.432793, size = 175, normalized size = 0.97 \[ -\frac{\frac{a \sqrt{c+d x^2} \left (a^2 d^2 x^2+a b c \left (c-2 d x^2\right )+2 b^2 c^2 x^2\right )}{b x^2 \left (a+b x^2\right )}+\frac{\sqrt{b c-a d} \left (-a^2 d^2-3 a b c d+4 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{b^{3/2}}+c^{3/2} (-(4 b c-5 a d)) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^(5/2)/(x^3*(a + b*x^2)^2),x]

[Out]

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

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Maple [B]  time = 0.017, size = 7590, normalized size = 42.2 \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^3/(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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^3/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^(5/2)/((b*x^2 + a)^2*x^3), x)

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Fricas [A]  time = 12.609, size = 2638, normalized size = 14.66 \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^3/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(5/2)/x**3/(b*x**2+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.15505, size = 390, normalized size = 2.17 \begin{align*} -\frac{1}{2} \, d^{3}{\left (\frac{{\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{a^{3} \sqrt{-c} d^{3}} + \frac{2 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{2} c^{2} - 2 \, \sqrt{d x^{2} + c} b^{2} c^{3} - 2 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b c d + 3 \, \sqrt{d x^{2} + c} a b c^{2} d +{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} d^{2} - \sqrt{d x^{2} + c} a^{2} c d^{2}}{{\left ({\left (d x^{2} + c\right )}^{2} b - 2 \,{\left (d x^{2} + c\right )} b c + b c^{2} +{\left (d x^{2} + c\right )} a d - a c d\right )} a^{2} b d^{2}} - \frac{{\left (4 \, b^{3} c^{3} - 7 \, a b^{2} c^{2} d + 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a^{3} b d^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(5/2)/x^3/(b*x^2+a)^2,x, algorithm="giac")

[Out]

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