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

Optimal. Leaf size=124 \[ \frac {(b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a b^{5/2}}+\frac {d \sqrt {c+d x^2} (2 b c-a d)}{b^2}-\frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {d \left (c+d x^2\right )^{3/2}}{3 b} \]

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Rubi [A]  time = 0.19, antiderivative size = 124, 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, 84, 154, 156, 63, 208} \begin {gather*} \frac {d \sqrt {c+d x^2} (2 b c-a d)}{b^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 b^{5/2}}-\frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {d \left (c+d x^2\right )^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(2*b*c - a*d)*Sqrt[c + d*x^2])/b^2 + (d*(c + d*x^2)^(3/2))/(3*b) - (c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]
])/a + ((b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(a*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 84

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

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

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Mathematica [A]  time = 0.12, size = 114, normalized size = 0.92 \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 b^{5/2}}+\frac {d \sqrt {c+d x^2} \left (-3 a d+7 b c+b d x^2\right )}{3 b^2}-\frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[c + d*x^2]*(7*b*c - 3*a*d + b*d*x^2))/(3*b^2) - (c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/a + ((b*c -
a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(a*b^(5/2))

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IntegrateAlgebraic [A]  time = 0.17, size = 129, normalized size = 1.04 \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 b^{5/2}}+\frac {\sqrt {c+d x^2} \left (-3 a d^2+7 b c d+b d^2 x^2\right )}{3 b^2}-\frac {c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(7*b*c*d - 3*a*d^2 + b*d^2*x^2))/(3*b^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*b^(5/2)) - (c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/a

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fricas [A]  time = 3.47, size = 837, normalized size = 6.75 \begin {gather*} \left [\frac {6 \, b^{2} c^{\frac {5}{2}} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{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 (a b d^{2} x^{2} + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{12 \, a b^{2}}, \frac {12 \, b^{2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{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 (a b d^{2} x^{2} + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{12 \, a b^{2}}, \frac {3 \, b^{2} c^{\frac {5}{2}} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{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 (a b d^{2} x^{2} + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{6 \, a b^{2}}, \frac {6 \, b^{2} \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{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 (a b d^{2} x^{2} + 7 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{6 \, a b^{2}}\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),x, algorithm="fricas")

[Out]

[1/12*(6*b^2*c^(5/2)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*s
qrt((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*b*d^2*x^2 +
7*a*b*c*d - 3*a^2*d^2)*sqrt(d*x^2 + c))/(a*b^2), 1/12*(12*b^2*sqrt(-c)*c^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) +
3*(b^2*c^2 - 2*a*b*c*d + a^2*d^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*b*d^2*x^2 + 7*a*b*c*d - 3*a^2*d^2)*sqrt(d*x^2 + c))/(a*b^2), 1/6*(3*b^2*c^(5/2)*log(-
(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^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*
(a*b*d^2*x^2 + 7*a*b*c*d - 3*a^2*d^2)*sqrt(d*x^2 + c))/(a*b^2), 1/6*(6*b^2*sqrt(-c)*c^2*arctan(sqrt(-c)/sqrt(d
*x^2 + c)) + 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^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*(a*b*d^2*x^2 + 7*a*b*c*d - 3*a^2*d^2
)*sqrt(d*x^2 + c))/(a*b^2)]

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giac [A]  time = 0.38, size = 163, normalized size = 1.31 \begin {gather*} \frac {c^{3} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a \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 b^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d + 6 \, \sqrt {d x^{2} + c} b^{2} c d - 3 \, \sqrt {d x^{2} + c} a b d^{2}}{3 \, b^{3}} \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),x, algorithm="giac")

[Out]

c^3*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(a*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*b^2) + 1/3*((d*x^2 + c)^(3/2)*b^2*d + 6*sqrt(
d*x^2 + c)*b^2*c*d - 3*sqrt(d*x^2 + c)*a*b*d^2)/b^3

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maple [B]  time = 0.02, size = 3148, normalized size = 25.39 \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),x)

[Out]

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

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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 )} 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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.00, size = 2094, normalized size = 16.89

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)),x)

[Out]

(atan((((c^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 20
*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 + (((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*
b^4*c^2*d^4)/b^3 + ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(c + d*x^2)^(1/2)*(c^5)^(1/2))/(a*b^3))*(c^5)^(1/2))/(2*
a))*1i)/(2*a) + ((c^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4
*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 - (((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3
- 12*a^3*b^4*c^2*d^4)/b^3 - ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(c + d*x^2)^(1/2)*(c^5)^(1/2))/(a*b^3))*(c^5)^(
1/2))/(2*a))*1i)/(2*a))/((2*(a^5*c^3*d^8 - 3*b^5*c^8*d^3 + 12*a*b^4*c^7*d^4 - 6*a^4*b*c^4*d^7 - 19*a^2*b^3*c^6
*d^5 + 15*a^3*b^2*c^5*d^6))/b^3 - ((c^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^
3 + 15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 + (((4*a^4*b^3*c*d^5 +
8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 + ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(c + d*x^2)^(1/2)*(c^5)^(1/2)
)/(a*b^3))*(c^5)^(1/2))/(2*a)))/(2*a) + ((c^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*
c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 - (((4*a^4*b^3*c*
d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 - ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(c + d*x^2)^(1/2)*(c^5)
^(1/2))/(a*b^3))*(c^5)^(1/2))/(2*a)))/(2*a)))*(c^5)^(1/2)*1i)/a + (d*(c + d*x^2)^(3/2))/(3*b) + (atan((((-b^5*
(a*d - b*c)^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 2
0*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 + ((-b^5*(a*d - b*c)^5)^(1/2)*((4*a^4*b^3*c*d^5 +
8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 + ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(-b^5*(a*d - b*c)^5)^(1/2)*(
c + d*x^2)^(1/2))/(a*b^8)))/(2*a*b^5))*1i)/(2*a*b^5) + ((-b^5*(a*d - b*c)^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*
d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b
*c*d^7))/b^3 - ((-b^5*(a*d - b*c)^5)^(1/2)*((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 - (
(4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(-b^5*(a*d - b*c)^5)^(1/2)*(c + d*x^2)^(1/2))/(a*b^8)))/(2*a*b^5))*1i)/(2*a*
b^5))/((2*(a^5*c^3*d^8 - 3*b^5*c^8*d^3 + 12*a*b^4*c^7*d^4 - 6*a^4*b*c^4*d^7 - 19*a^2*b^3*c^6*d^5 + 15*a^3*b^2*
c^5*d^6))/b^3 - ((-b^5*(a*d - b*c)^5)^(1/2)*((2*(c + d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 +
15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 + ((-b^5*(a*d - b*c)^5)^(1
/2)*((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*b^4*c^2*d^4)/b^3 + ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(-b^5
*(a*d - b*c)^5)^(1/2)*(c + d*x^2)^(1/2))/(a*b^8)))/(2*a*b^5)))/(2*a*b^5) + ((-b^5*(a*d - b*c)^5)^(1/2)*((2*(c
+ d*x^2)^(1/2)*(a^6*d^8 + 2*b^6*c^6*d^2 - 6*a*b^5*c^5*d^3 + 15*a^2*b^4*c^4*d^4 - 20*a^3*b^3*c^3*d^5 + 15*a^4*b
^2*c^2*d^6 - 6*a^5*b*c*d^7))/b^3 - ((-b^5*(a*d - b*c)^5)^(1/2)*((4*a^4*b^3*c*d^5 + 8*a^2*b^5*c^3*d^3 - 12*a^3*
b^4*c^2*d^4)/b^3 - ((4*a^3*b^5*d^3 - 8*a^2*b^6*c*d^2)*(-b^5*(a*d - b*c)^5)^(1/2)*(c + d*x^2)^(1/2))/(a*b^8)))/
(2*a*b^5)))/(2*a*b^5)))*(-b^5*(a*d - b*c)^5)^(1/2)*1i)/(a*b^5) - (d*(c + d*x^2)^(1/2)*(a*d - 2*b*c))/b^2

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sympy [A]  time = 67.14, size = 119, normalized size = 0.96 \begin {gather*} \frac {d \left (c + d x^{2}\right )^{\frac {3}{2}}}{3 b} + \frac {\sqrt {c + d x^{2}} \left (- a d^{2} + 2 b c d\right )}{b^{2}} + \frac {c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{a \sqrt {- c}} + \frac {\left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{a b^{3} \sqrt {\frac {a d - b c}{b}}} \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),x)

[Out]

d*(c + d*x**2)**(3/2)/(3*b) + sqrt(c + d*x**2)*(-a*d**2 + 2*b*c*d)/b**2 + c**3*atan(sqrt(c + d*x**2)/sqrt(-c))
/(a*sqrt(-c)) + (a*d - b*c)**3*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(a*b**3*sqrt((a*d - b*c)/b))

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