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

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

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Rubi [A]  time = 0.27, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 98, 149, 156, 63, 208} \begin {gather*} -\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 )} \end {gather*}

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 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 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 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]

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]]

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*}

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Mathematica [A]  time = 0.45, size = 175, normalized size = 0.97 \begin {gather*} -\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}}-\left (c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\right )}{2 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((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[(S
qrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/b^(3/2))/a^3

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IntegrateAlgebraic [A]  time = 0.52, size = 231, normalized size = 1.28 \begin {gather*} \frac {\left (4 b c^{5/2}-5 a c^{3/2} d\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3}+\frac {\sqrt {c+d x^2} \left (-a^2 d^2 x^2-a b c^2+2 a b c d x^2-2 b^2 c^2 x^2\right )}{2 a^2 b x^2 \left (a+b x^2\right )}+\frac {\left (-a^4 d^4-a^3 b c d^3+9 a^2 b^2 c^2 d^2-11 a b^3 c^3 d+4 b^4 c^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{2 a^3 b^{3/2} (a d-b c)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-(a*b*c^2) - 2*b^2*c^2*x^2 + 2*a*b*c*d*x^2 - a^2*d^2*x^2))/(2*a^2*b*x^2*(a + b*x^2)) + ((4*b
^4*c^4 - 11*a*b^3*c^3*d + 9*a^2*b^2*c^2*d^2 - a^3*b*c*d^3 - a^4*d^4)*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c
+ d*x^2])/(b*c - a*d)])/(2*a^3*b^(3/2)*(-(b*c) + a*d)^(3/2)) + ((4*b*c^(5/2) - 5*a*c^(3/2)*d)*ArcTanh[Sqrt[c
+ d*x^2]/Sqrt[c]])/(2*a^3)

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fricas [A]  time = 3.82, size = 1266, normalized size = 7.03

result too large to display

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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giac [A]  time = 0.35, size = 283, normalized size = 1.57 \begin {gather*} -\frac {{\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{3} \sqrt {-c}} + \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 )}{2 \, \sqrt {-b^{2} c + a b d} a^{3} b} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} d - 2 \, \sqrt {d x^{2} + c} b^{2} c^{3} d - 2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d^{2} + 3 \, \sqrt {d x^{2} + c} a b c^{2} d^{2} + {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} - \sqrt {d x^{2} + c} a^{2} c d^{3}}{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} \end {gather*}

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

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maple [B]  time = 0.02, size = 7590, normalized size = 42.17 \begin {gather*} \text {output too large to display} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{2} x^{3}}\,{d x} \end {gather*}

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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mupad [B]  time = 1.99, size = 1152, normalized size = 6.40 \begin {gather*} \frac {\frac {\sqrt {d\,x^2+c}\,\left (a^2\,c\,d^3-3\,a\,b\,c^2\,d^2+2\,b^2\,c^3\,d\right )}{2\,a^2\,b}-\frac {d\,{\left (d\,x^2+c\right )}^{3/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+2\,b^2\,c^2\right )}{2\,a^2\,b}}{\left (d\,x^2+c\right )\,\left (a\,d-2\,b\,c\right )+b\,{\left (d\,x^2+c\right )}^2+b\,c^2-a\,c\,d}-\frac {\mathrm {atanh}\left (\frac {5\,d^9\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{4\,\left (\frac {5\,c^2\,d^9}{4}+\frac {4\,b\,c^3\,d^8}{a}-\frac {33\,b^2\,c^4\,d^7}{2\,a^2}+\frac {65\,b^3\,c^5\,d^6}{4\,a^3}-\frac {5\,b^4\,c^6\,d^5}{a^4}\right )}+\frac {4\,c\,d^8\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{4\,c^3\,d^8+\frac {5\,a\,c^2\,d^9}{4\,b}-\frac {33\,b\,c^4\,d^7}{2\,a}+\frac {65\,b^2\,c^5\,d^6}{4\,a^2}-\frac {5\,b^3\,c^6\,d^5}{a^3}}+\frac {65\,b^2\,c^3\,d^6\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{4\,\left (4\,a^2\,c^3\,d^8+\frac {65\,b^2\,c^5\,d^6}{4}-\frac {5\,b^3\,c^6\,d^5}{a}+\frac {5\,a^3\,c^2\,d^9}{4\,b}-\frac {33\,a\,b\,c^4\,d^7}{2}\right )}-\frac {5\,b^3\,c^4\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{4\,a^3\,c^3\,d^8-5\,b^3\,c^6\,d^5+\frac {65\,a\,b^2\,c^5\,d^6}{4}-\frac {33\,a^2\,b\,c^4\,d^7}{2}+\frac {5\,a^4\,c^2\,d^9}{4\,b}}-\frac {33\,b\,c^2\,d^7\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{2\,\left (4\,a\,c^3\,d^8-\frac {33\,b\,c^4\,d^7}{2}+\frac {65\,b^2\,c^5\,d^6}{4\,a}+\frac {5\,a^2\,c^2\,d^9}{4\,b}-\frac {5\,b^3\,c^6\,d^5}{a^2}\right )}\right )\,\left (5\,a\,d-4\,b\,c\right )\,\sqrt {c^3}}{2\,a^3}-\frac {\mathrm {atanh}\left (\frac {15\,c^3\,d^6\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{4\,\left (\frac {7\,a^3\,c^2\,d^9}{4}+\frac {55\,b^3\,c^5\,d^6}{4}-\frac {41\,a\,b^2\,c^4\,d^7}{4}-\frac {a^2\,b\,c^3\,d^8}{2}+\frac {a^4\,c\,d^{10}}{4\,b}-\frac {5\,b^4\,c^6\,d^5}{a}\right )}+\frac {9\,c^2\,d^7\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{4\,\left (\frac {a^3\,c\,d^{10}}{4}-\frac {41\,b^3\,c^4\,d^7}{4}-\frac {a\,b^2\,c^3\,d^8}{2}+\frac {7\,a^2\,b\,c^2\,d^9}{4}+\frac {55\,b^4\,c^5\,d^6}{4\,a}-\frac {5\,b^5\,c^6\,d^5}{a^2}\right )}+\frac {5\,c^4\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{\frac {a^3\,c^3\,d^8}{2}+5\,b^3\,c^6\,d^5-\frac {55\,a\,b^2\,c^5\,d^6}{4}+\frac {41\,a^2\,b\,c^4\,d^7}{4}-\frac {a^5\,c\,d^{10}}{4\,b^2}-\frac {7\,a^4\,c^2\,d^9}{4\,b}}-\frac {c\,d^8\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{4\,\left (\frac {b^3\,c^3\,d^8}{2}-\frac {7\,a\,b^2\,c^2\,d^9}{4}+\frac {41\,b^4\,c^4\,d^7}{4\,a}-\frac {55\,b^5\,c^5\,d^6}{4\,a^2}+\frac {5\,b^6\,c^6\,d^5}{a^3}-\frac {a^2\,b\,c\,d^{10}}{4}\right )}\right )\,\sqrt {-b^3\,{\left (a\,d-b\,c\right )}^3}\,\left (a\,d+4\,b\,c\right )}{2\,a^3\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((c + d*x^2)^(1/2)*(a^2*c*d^3 + 2*b^2*c^3*d - 3*a*b*c^2*d^2))/(2*a^2*b) - (d*(c + d*x^2)^(3/2)*(a^2*d^2 + 2*b
^2*c^2 - 2*a*b*c*d))/(2*a^2*b))/((c + d*x^2)*(a*d - 2*b*c) + b*(c + d*x^2)^2 + b*c^2 - a*c*d) - (atanh((5*d^9*
(c + d*x^2)^(1/2)*(c^3)^(1/2))/(4*((5*c^2*d^9)/4 + (4*b*c^3*d^8)/a - (33*b^2*c^4*d^7)/(2*a^2) + (65*b^3*c^5*d^
6)/(4*a^3) - (5*b^4*c^6*d^5)/a^4)) + (4*c*d^8*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(4*c^3*d^8 + (5*a*c^2*d^9)/(4*b)
- (33*b*c^4*d^7)/(2*a) + (65*b^2*c^5*d^6)/(4*a^2) - (5*b^3*c^6*d^5)/a^3) + (65*b^2*c^3*d^6*(c + d*x^2)^(1/2)*(
c^3)^(1/2))/(4*(4*a^2*c^3*d^8 + (65*b^2*c^5*d^6)/4 - (5*b^3*c^6*d^5)/a + (5*a^3*c^2*d^9)/(4*b) - (33*a*b*c^4*d
^7)/2)) - (5*b^3*c^4*d^5*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(4*a^3*c^3*d^8 - 5*b^3*c^6*d^5 + (65*a*b^2*c^5*d^6)/4
- (33*a^2*b*c^4*d^7)/2 + (5*a^4*c^2*d^9)/(4*b)) - (33*b*c^2*d^7*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(2*(4*a*c^3*d^8
- (33*b*c^4*d^7)/2 + (65*b^2*c^5*d^6)/(4*a) + (5*a^2*c^2*d^9)/(4*b) - (5*b^3*c^6*d^5)/a^2)))*(5*a*d - 4*b*c)*
(c^3)^(1/2))/(2*a^3) - (atanh((15*c^3*d^6*(c + d*x^2)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d^2 - 3*a*b^5
*c^2*d)^(1/2))/(4*((7*a^3*c^2*d^9)/4 + (55*b^3*c^5*d^6)/4 - (41*a*b^2*c^4*d^7)/4 - (a^2*b*c^3*d^8)/2 + (a^4*c*
d^10)/(4*b) - (5*b^4*c^6*d^5)/a)) + (9*c^2*d^7*(c + d*x^2)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d^2 - 3*
a*b^5*c^2*d)^(1/2))/(4*((a^3*c*d^10)/4 - (41*b^3*c^4*d^7)/4 - (a*b^2*c^3*d^8)/2 + (7*a^2*b*c^2*d^9)/4 + (55*b^
4*c^5*d^6)/(4*a) - (5*b^5*c^6*d^5)/a^2)) + (5*c^4*d^5*(c + d*x^2)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d
^2 - 3*a*b^5*c^2*d)^(1/2))/((a^3*c^3*d^8)/2 + 5*b^3*c^6*d^5 - (55*a*b^2*c^5*d^6)/4 + (41*a^2*b*c^4*d^7)/4 - (a
^5*c*d^10)/(4*b^2) - (7*a^4*c^2*d^9)/(4*b)) - (c*d^8*(c + d*x^2)^(1/2)*(b^6*c^3 - a^3*b^3*d^3 + 3*a^2*b^4*c*d^
2 - 3*a*b^5*c^2*d)^(1/2))/(4*((b^3*c^3*d^8)/2 - (7*a*b^2*c^2*d^9)/4 + (41*b^4*c^4*d^7)/(4*a) - (55*b^5*c^5*d^6
)/(4*a^2) + (5*b^6*c^6*d^5)/a^3 - (a^2*b*c*d^10)/4)))*(-b^3*(a*d - b*c)^3)^(1/2)*(a*d + 4*b*c))/(2*a^3*b^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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