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

Optimal. Leaf size=144 \[ -\frac {(b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 b^{3/2}}+\frac {c^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2}+\frac {d \sqrt {c+d x^2} (2 a d+b c)}{2 a b}-\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 144, 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)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 b^{3/2}}+\frac {c^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2}+\frac {d \sqrt {c+d x^2} (2 a d+b c)}{2 a b}-\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 125, normalized size = 0.87 \begin {gather*} \frac {-\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{3/2}}+c^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+\frac {a \sqrt {c+d x^2} \left (2 a d^2 x^2-b c^2\right )}{b x^2}}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.26, size = 145, normalized size = 1.01 \begin {gather*} \frac {(a d-b c)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{a^2 b^{3/2}}+\frac {\left (2 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^2}+\frac {\sqrt {c+d x^2} \left (2 a d^2 x^2-b c^2\right )}{2 a b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 4.39, size = 891, normalized size = 6.19 \begin {gather*} \left [\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \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 ) - {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (2 \, a^{2} d^{2} x^{2} - a b c^{2}\right )} \sqrt {d x^{2} + c}}{4 \, a^{2} b x^{2}}, -\frac {2 \, {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \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 ) - 2 \, {\left (2 \, a^{2} d^{2} x^{2} - a b c^{2}\right )} \sqrt {d x^{2} + c}}{4 \, a^{2} b x^{2}}, -\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \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 ) + {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (2 \, a^{2} d^{2} x^{2} - a b c^{2}\right )} \sqrt {d x^{2} + c}}{4 \, a^{2} b x^{2}}, -\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \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 ) + {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, a^{2} d^{2} x^{2} - a b c^{2}\right )} \sqrt {d x^{2} + c}}{2 \, a^{2} b x^{2}}\right ] \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),x, algorithm="fricas")

[Out]

[1/4*((b^2*c^2 - 2*a*b*c*d + a^2*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*b^2*c^2 - 5*a*b*c*d)*sqrt(c)*x^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/
x^2) + 2*(2*a^2*d^2*x^2 - a*b*c^2)*sqrt(d*x^2 + c))/(a^2*b*x^2), -1/4*(2*(2*b^2*c^2 - 5*a*b*c*d)*sqrt(-c)*x^2*
arctan(sqrt(-c)/sqrt(d*x^2 + c)) - (b^2*c^2 - 2*a*b*c*d + a^2*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*(2*a^2*d^2*x^2 - a*b*c^2)*sqrt(d*x^2 + c))/(a^2*b*x^
2), -1/4*(2*(b^2*c^2 - 2*a*b*c*d + a^2*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^2*c^2 - 5*a*b*c*d)*sqrt(c)*x^2*l
og(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(2*a^2*d^2*x^2 - a*b*c^2)*sqrt(d*x^2 + c))/(a^2*b*x^2),
-1/2*((b^2*c^2 - 2*a*b*c*d + a^2*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^2*c^2 - 5*a*b*c*d)*sqrt(-c)*x^2*arcta
n(sqrt(-c)/sqrt(d*x^2 + c)) - (2*a^2*d^2*x^2 - a*b*c^2)*sqrt(d*x^2 + c))/(a^2*b*x^2)]

________________________________________________________________________________________

giac [A]  time = 0.39, size = 158, normalized size = 1.10 \begin {gather*} \frac {\sqrt {d x^{2} + c} d^{2}}{b} - \frac {\sqrt {d x^{2} + c} c^{2}}{2 \, a x^{2}} - \frac {{\left (2 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{2} \sqrt {-c}} + \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, 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^{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),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.02, size = 3247, normalized size = 22.55 \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),x)

[Out]

5/4/a/b*d^(3/2)*(-a*b)^(1/2)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*
b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c+7/16/a^2*(-a*b)^(1/2)*d*c*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*
b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-1/4/a/b*(-a*b)^(1/2)*d^2*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)
^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-5/4/a/b*d^(3/2)*(-a*b)^(1/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b
)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c-7/16/
a^2*(-a*b)^(1/2)*d*c*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x+1/4/a/
b*(-a*b)^(1/2)*d^2*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-1/5/a^2*
b*(d*x^2+c)^(5/2)+1/10/a^2*b*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(5/2)-
1/6/a*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*d+1/2/b*((x-(-a*b)^(1/2
)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*d^2+1/6/a^2*b*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*
b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*c-1/a*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2
)/b)/b*d-(a*d-b*c)/b)^(1/2)*d*c+1/2/a^2*b*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b
*c)/b)^(1/2)*c^2+1/2/b^2*d^(5/2)*(-a*b)^(1/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1
/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))-1/3/a^2*b*c*(d*x^2+c)^(3/2)+1/a^2*b*c^(5/
2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)-1/a^2*b*(d*x^2+c)^(1/2)*c^2+1/2/a*d/c*(d*x^2+c)^(5/2)-5/2/a*d*c^(3/2)
*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+5/2/a*d*c*(d*x^2+c)^(1/2)-1/2/a/c/x^2*(d*x^2+c)^(7/2)+1/6/a^2*b*((x+(-a
*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*c-1/a*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*
b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*d*c+1/2/a^2*b*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a
*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*c^2-1/2/b^2*d^(5/2)*(-a*b)^(1/2)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/2)/b*d
)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))-1/8/a^2*(-a*b)^(1/
2)*d*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*x-15/16/a^2*d^(1/2)*(-a*
b)^(1/2)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(
1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c^2+1/2*a/b^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-
2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b
)^(1/2))/(x+(-a*b)^(1/2)/b))*d^3-3/2/b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-
b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)
)/(x+(-a*b)^(1/2)/b))*d^2*c+3/2/a/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/
b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+
(-a*b)^(1/2)/b))*d*c^2-1/2/a^2*b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b
+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x+(
-a*b)^(1/2)/b))*c^3+1/8/a^2*(-a*b)^(1/2)*d*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-
b*c)/b)^(3/2)*x+15/16/a^2*d^(1/2)*(-a*b)^(1/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(
1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c^2+1/2*a/b^2/(-(a*d-b*c)/b)^(1/2)*ln((2
*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/
2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*d^3-3/2/b/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^
(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-
a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*d^2*c+3/2/a/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*
(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(
1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*d*c^2-1/2/a^2*b/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x
-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/
2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*c^3+5/6/a*d*(d*x^2+c)^(3/2)+1/10/a^2*b*((x+(-a*b)^(1/2)/b)^2
*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(5/2)-1/6/a*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-
a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)*d+1/2/b*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*
d-b*c)/b)^(1/2)*d^2

________________________________________________________________________________________

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 )} 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),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.37, size = 1428, normalized size = 9.92

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2*(c + d*x^2)^(1/2))/b + (atan((a^3*d^9*(c + d*x^2)^(1/2)*(c^3)^(1/2)*5i)/(5*a^3*c^2*d^9 - (395*b^3*c^5*d^6
)/4 + 87*a*b^2*c^4*d^7 - 32*a^2*b*c^3*d^8 + (185*b^4*c^6*d^5)/(4*a) - (15*b^5*c^7*d^4)/(2*a^2)) + (a^2*c*d^8*(
c + d*x^2)^(1/2)*(c^3)^(1/2)*32i)/(32*a^2*c^3*d^8 + (395*b^2*c^5*d^6)/4 - (185*b^3*c^6*d^5)/(4*a) - (5*a^3*c^2
*d^9)/b + (15*b^4*c^7*d^4)/(2*a^2) - 87*a*b*c^4*d^7) + (b^2*c^3*d^6*(c + d*x^2)^(1/2)*(c^3)^(1/2)*395i)/(4*(32
*a^2*c^3*d^8 + (395*b^2*c^5*d^6)/4 - (185*b^3*c^6*d^5)/(4*a) - (5*a^3*c^2*d^9)/b + (15*b^4*c^7*d^4)/(2*a^2) -
87*a*b*c^4*d^7)) - (b^3*c^4*d^5*(c + d*x^2)^(1/2)*(c^3)^(1/2)*185i)/(4*(32*a^3*c^3*d^8 - (185*b^3*c^6*d^5)/4 +
(395*a*b^2*c^5*d^6)/4 - 87*a^2*b*c^4*d^7 + (15*b^4*c^7*d^4)/(2*a) - (5*a^4*c^2*d^9)/b)) + (b^4*c^5*d^4*(c + d
*x^2)^(1/2)*(c^3)^(1/2)*15i)/(2*(32*a^4*c^3*d^8 + (15*b^4*c^7*d^4)/2 - (185*a*b^3*c^6*d^5)/4 - 87*a^3*b*c^4*d^
7 + (395*a^2*b^2*c^5*d^6)/4 - (5*a^5*c^2*d^9)/b)) - (a*b*c^2*d^7*(c + d*x^2)^(1/2)*(c^3)^(1/2)*87i)/(32*a^2*c^
3*d^8 + (395*b^2*c^5*d^6)/4 - (185*b^3*c^6*d^5)/(4*a) - (5*a^3*c^2*d^9)/b + (15*b^4*c^7*d^4)/(2*a^2) - 87*a*b*
c^4*d^7))*(5*a*d - 2*b*c)*(c^3)^(1/2)*1i)/(2*a^2) - (atan((c^3*d^5*(c + d*x^2)^(1/2)*(b^8*c^5 - a^5*b^3*d^5 +
5*a^4*b^4*c*d^4 + 10*a^2*b^6*c^3*d^2 - 10*a^3*b^5*c^2*d^3 - 5*a*b^7*c^4*d)^(1/2)*20i)/((185*a*b^3*c^5*d^6)/2 -
(85*b^4*c^6*d^5)/2 - 16*a^4*c^2*d^9 + 56*a^3*b*c^3*d^8 + (2*a^5*c*d^10)/b - (199*a^2*b^2*c^4*d^7)/2 + (15*b^5
*c^7*d^4)/(2*a)) - (c^2*d^6*(c + d*x^2)^(1/2)*(b^8*c^5 - a^5*b^3*d^5 + 5*a^4*b^4*c*d^4 + 10*a^2*b^6*c^3*d^2 -
10*a^3*b^5*c^2*d^3 - 5*a*b^7*c^4*d)^(1/2)*10i)/(2*a^4*c*d^10 + (185*b^4*c^5*d^6)/2 - (199*a*b^3*c^4*d^7)/2 - 1
6*a^3*b*c^2*d^9 + 56*a^2*b^2*c^3*d^8 - (85*b^5*c^6*d^5)/(2*a) + (15*b^6*c^7*d^4)/(2*a^2)) - (c^4*d^4*(c + d*x^
2)^(1/2)*(b^8*c^5 - a^5*b^3*d^5 + 5*a^4*b^4*c*d^4 + 10*a^2*b^6*c^3*d^2 - 10*a^3*b^5*c^2*d^3 - 5*a*b^7*c^4*d)^(
1/2)*15i)/(2*(56*a^4*c^3*d^8 + (15*b^4*c^7*d^4)/2 - (85*a*b^3*c^6*d^5)/2 - (199*a^3*b*c^4*d^7)/2 + (2*a^6*c*d^
10)/b^2 + (185*a^2*b^2*c^5*d^6)/2 - (16*a^5*c^2*d^9)/b)) + (a*c*d^7*(c + d*x^2)^(1/2)*(b^8*c^5 - a^5*b^3*d^5 +
5*a^4*b^4*c*d^4 + 10*a^2*b^6*c^3*d^2 - 10*a^3*b^5*c^2*d^3 - 5*a*b^7*c^4*d)^(1/2)*2i)/((185*b^5*c^5*d^6)/2 - (
199*a*b^4*c^4*d^7)/2 + 56*a^2*b^3*c^3*d^8 - 16*a^3*b^2*c^2*d^9 - (85*b^6*c^6*d^5)/(2*a) + (15*b^7*c^7*d^4)/(2*
a^2) + 2*a^4*b*c*d^10))*(-b^3*(a*d - b*c)^5)^(1/2)*1i)/(a^2*b^3) - (b*c^2*d*(c + d*x^2)^(1/2))/(2*a*(b*(c + d*
x^2) - b*c))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________