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

Optimal. Leaf size=160 \[ \frac {(b c-a d)^{3/2} (3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^2 b^{5/2}}-\frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a^2}-\frac {d \sqrt {c+d x^2} (b c-3 a d)}{2 a b^2}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.22, antiderivative size = 160, 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, 154, 156, 63, 208} \begin {gather*} \frac {(b c-a d)^{3/2} (3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^2 b^{5/2}}-\frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a^2}-\frac {d \sqrt {c+d x^2} (b c-3 a d)}{2 a b^2}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

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

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Mathematica [A]  time = 0.21, size = 158, normalized size = 0.99 \begin {gather*} \frac {\frac {a \sqrt {c+d x^2} \left (3 a^2 d^2+2 a b d \left (d x^2-c\right )+b^2 c^2\right )}{b^2 \left (a+b x^2\right )}+\frac {\sqrt {b c-a d} \left (-3 a^2 d^2+a b c d+2 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}}-2 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.35, size = 164, normalized size = 1.02 \begin {gather*} \frac {(a d-b c)^{3/2} (3 a d+2 b c) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{2 a^2 b^{5/2}}+\frac {\sqrt {c+d x^2} \left (3 a^2 d^2-2 a b c d+2 a b d^2 x^2+b^2 c^2\right )}{2 a b^2 \left (a+b x^2\right )}-\frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 3.48, size = 1132, normalized size = 7.08 \begin {gather*} \left [-\frac {{\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (b^{3} c^{2} x^{2} + a b^{2} c^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} b d^{2} x^{2} + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}}, \frac {8 \, {\left (b^{3} c^{2} x^{2} + a b^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (2 \, a^{2} b d^{2} x^{2} + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}}, \frac {{\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (b^{3} c^{2} x^{2} + a b^{2} c^{2}\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (2 \, a^{2} b d^{2} x^{2} + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}}, \frac {{\left (2 \, a b^{2} c^{2} + a^{2} b c d - 3 \, a^{3} d^{2} + {\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 4 \, {\left (b^{3} c^{2} x^{2} + a b^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + 2 \, {\left (2 \, a^{2} b d^{2} x^{2} + a b^{2} c^{2} - 2 \, a^{2} b c d + 3 \, a^{3} d^{2}\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.34, size = 206, normalized size = 1.29 \begin {gather*} \frac {c^{3} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} + \frac {\sqrt {d x^{2} + c} d^{2}}{b^{2}} - \frac {{\left (2 \, b^{3} c^{3} - a b^{2} c^{2} d - 4 \, a^{2} b c d^{2} + 3 \, 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^{2} b^{2}} + \frac {\sqrt {d x^{2} + c} b^{2} c^{2} d - 2 \, \sqrt {d x^{2} + c} a b c d^{2} + \sqrt {d x^{2} + c} a^{2} d^{3}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} a b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 7477, normalized size = 46.73 \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/(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}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.31, size = 1321, normalized size = 8.26

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2*(c + d*x^2)^(1/2))/b^2 + (atan((a^2*d^8*(c + d*x^2)^(1/2)*(c^5)^(1/2)*9i)/(2*((9*a^2*c^3*d^8)/2 + 5*b^2*c
^5*d^6 + (10*b^3*c^6*d^5)/a - (15*b^4*c^7*d^4)/(2*a^2) - 12*a*b*c^4*d^7)) + (c^2*d^6*(c + d*x^2)^(1/2)*(c^5)^(
1/2)*5i)/(5*c^5*d^6 - (12*a*c^4*d^7)/b + (10*b*c^6*d^5)/a - (15*b^2*c^7*d^4)/(2*a^2) + (9*a^2*c^3*d^8)/(2*b^2)
) + (c^3*d^5*(c + d*x^2)^(1/2)*(c^5)^(1/2)*10i)/(10*c^6*d^5 + (5*a*c^5*d^6)/b - (15*b*c^7*d^4)/(2*a) - (12*a^2
*c^4*d^7)/b^2 + (9*a^3*c^3*d^8)/(2*b^3)) - (a*c*d^7*(c + d*x^2)^(1/2)*(c^5)^(1/2)*12i)/(5*b*c^5*d^6 - 12*a*c^4
*d^7 + (10*b^2*c^6*d^5)/a + (9*a^2*c^3*d^8)/(2*b) - (15*b^3*c^7*d^4)/(2*a^2)) - (b*c^4*d^4*(c + d*x^2)^(1/2)*(
c^5)^(1/2)*15i)/(2*(10*a*c^6*d^5 - (15*b*c^7*d^4)/2 + (5*a^2*c^5*d^6)/b - (12*a^3*c^4*d^7)/b^2 + (9*a^4*c^3*d^
8)/(2*b^3))))*(c^5)^(1/2)*1i)/a^2 + ((c + d*x^2)^(1/2)*(a^2*d^3 + b^2*c^2*d - 2*a*b*c*d^2))/(2*a*(b^3*(c + d*x
^2) - b^3*c + a*b^2*d)) - (atan((c^4*d^5*(c + d*x^2)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*
c^2*d)^(1/2)*35i)/(4*(9*a^3*b*c^3*d^8 - (25*b^4*c^6*d^5)/4 - (85*a*b^3*c^5*d^6)/4 - (81*a^4*c^2*d^9)/4 + (27*a
^5*c*d^10)/(4*b) + (49*a^2*b^2*c^4*d^7)/2 + (15*b^5*c^7*d^4)/(2*a))) - (c^3*d^6*(c + d*x^2)^(1/2)*(b^8*c^3 - a
^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1/2)*45i)/(4*((27*a^4*c*d^10)/4 - (85*b^4*c^5*d^6)/4 + (49*a*b^
3*c^4*d^7)/2 - (81*a^3*b*c^2*d^9)/4 + 9*a^2*b^2*c^3*d^8 - (25*b^5*c^6*d^5)/(4*a) + (15*b^6*c^7*d^4)/(2*a^2)))
+ (c^5*d^4*(c + d*x^2)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1/2)*15i)/(2*(9*a^4*c^
3*d^8 + (15*b^4*c^7*d^4)/2 - (25*a*b^3*c^6*d^5)/4 + (49*a^3*b*c^4*d^7)/2 + (27*a^6*c*d^10)/(4*b^2) - (85*a^2*b
^2*c^5*d^6)/4 - (81*a^5*c^2*d^9)/(4*b))) + (a^2*c*d^8*(c + d*x^2)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d
^2 - 3*a*b^7*c^2*d)^(1/2)*27i)/(4*((49*a*b^5*c^4*d^7)/2 - (85*b^6*c^5*d^6)/4 + (27*a^4*b^2*c*d^10)/4 + 9*a^2*b
^4*c^3*d^8 - (81*a^3*b^3*c^2*d^9)/4 - (25*b^7*c^6*d^5)/(4*a) + (15*b^8*c^7*d^4)/(2*a^2))) - (a*c^2*d^7*(c + d*
x^2)^(1/2)*(b^8*c^3 - a^3*b^5*d^3 + 3*a^2*b^6*c*d^2 - 3*a*b^7*c^2*d)^(1/2)*27i)/(4*((49*a*b^4*c^4*d^7)/2 - (85
*b^5*c^5*d^6)/4 + 9*a^2*b^3*c^3*d^8 - (81*a^3*b^2*c^2*d^9)/4 - (25*b^6*c^6*d^5)/(4*a) + (15*b^7*c^7*d^4)/(2*a^
2) + (27*a^4*b*c*d^10)/4)))*(-b^5*(a*d - b*c)^3)^(1/2)*(3*a*d + 2*b*c)*1i)/(2*a^2*b^5)

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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/(b*x**2+a)**2,x)

[Out]

Timed out

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