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

Optimal. Leaf size=96 \[ \frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a b^{3/2}}-\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {d \sqrt {c+d x^2}}{b} \]

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Rubi [A]  time = 0.11, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 84, 156, 63, 208} \begin {gather*} \frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a b^{3/2}}-\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {d \sqrt {c+d x^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.07, size = 102, normalized size = 1.06 \begin {gather*} \frac {a \sqrt {b} d \sqrt {c+d x^2}+(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )-b^{3/2} c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 106, normalized size = 1.10 \begin {gather*} \frac {(a d-b c)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{a b^{3/2}}-\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a}+\frac {d \sqrt {c+d x^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.42, size = 682, normalized size = 7.10 \begin {gather*} \left [\frac {2 \, b c^{\frac {3}{2}} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 4 \, \sqrt {d x^{2} + c} a d - {\left (b c - a d\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 \, a b}, \frac {4 \, b \sqrt {-c} c \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + 4 \, \sqrt {d x^{2} + c} a d - {\left (b c - a d\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 \, a b}, \frac {b c^{\frac {3}{2}} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {d x^{2} + c} a d + {\left (b c - a d\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 \, a b}, \frac {2 \, b \sqrt {-c} c \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + 2 \, \sqrt {d x^{2} + c} a d + {\left (b c - a d\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 \, a b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.37, size = 110, normalized size = 1.15 \begin {gather*} \frac {c^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{a \sqrt {-c}} + \frac {\sqrt {d x^{2} + c} d}{b} - \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(a*sqrt(-c)) + sqrt(d*x^2 + c)*d/b - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*arct
an(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a*b)

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maple [B]  time = 0.01, size = 1919, normalized size = 19.99

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)+1/4/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)^(1/2)*x+3/4/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+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-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-1/2/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))+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^2-1/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+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^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)-1/4/
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)^(1/2)*x-3/4/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+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-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+1/2/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))+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^2-1/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+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^2+1/3/a*(d*x^2+c)^(3/2)-1/a*c^(3/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)+1/a*
(d*x^2+c)^(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 {3}{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)^(3/2)/x/(b*x^2+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.81, size = 711, normalized size = 7.41 \begin {gather*} \frac {d\,\sqrt {d\,x^2+c}}{b}-\frac {\mathrm {atanh}\left (\frac {2\,a^3\,d^6\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{2\,a^3\,c^2\,d^6-8\,a^2\,b\,c^3\,d^5+12\,a\,b^2\,c^4\,d^4-6\,b^3\,c^5\,d^3}+\frac {8\,a^2\,c\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{8\,a^2\,c^3\,d^5+6\,b^2\,c^5\,d^3-\frac {2\,a^3\,c^2\,d^6}{b}-12\,a\,b\,c^4\,d^4}+\frac {6\,b^2\,c^3\,d^3\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{8\,a^2\,c^3\,d^5+6\,b^2\,c^5\,d^3-\frac {2\,a^3\,c^2\,d^6}{b}-12\,a\,b\,c^4\,d^4}-\frac {12\,a\,b\,c^2\,d^4\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{8\,a^2\,c^3\,d^5+6\,b^2\,c^5\,d^3-\frac {2\,a^3\,c^2\,d^6}{b}-12\,a\,b\,c^4\,d^4}\right )\,\sqrt {c^3}}{a}+\frac {\mathrm {atanh}\left (\frac {6\,c^3\,d^3\,\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}}{6\,b^3\,c^5\,d^3-10\,a^3\,c^2\,d^6-18\,a\,b^2\,c^4\,d^4+20\,a^2\,b\,c^3\,d^5+\frac {2\,a^4\,c\,d^7}{b}}-\frac {6\,a\,c^2\,d^4\,\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}}{2\,a^4\,c\,d^7-10\,a^3\,b\,c^2\,d^6+20\,a^2\,b^2\,c^3\,d^5-18\,a\,b^3\,c^4\,d^4+6\,b^4\,c^5\,d^3}+\frac {2\,a^2\,c\,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}}{2\,a^4\,b\,c\,d^7-10\,a^3\,b^2\,c^2\,d^6+20\,a^2\,b^3\,c^3\,d^5-18\,a\,b^4\,c^4\,d^4+6\,b^5\,c^5\,d^3}\right )\,\sqrt {-b^3\,{\left (a\,d-b\,c\right )}^3}}{a\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*(c + d*x^2)^(1/2))/b - (atanh((2*a^3*d^6*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(2*a^3*c^2*d^6 - 6*b^3*c^5*d^3 + 12
*a*b^2*c^4*d^4 - 8*a^2*b*c^3*d^5) + (8*a^2*c*d^5*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(8*a^2*c^3*d^5 + 6*b^2*c^5*d^3
- (2*a^3*c^2*d^6)/b - 12*a*b*c^4*d^4) + (6*b^2*c^3*d^3*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(8*a^2*c^3*d^5 + 6*b^2*
c^5*d^3 - (2*a^3*c^2*d^6)/b - 12*a*b*c^4*d^4) - (12*a*b*c^2*d^4*(c + d*x^2)^(1/2)*(c^3)^(1/2))/(8*a^2*c^3*d^5
+ 6*b^2*c^5*d^3 - (2*a^3*c^2*d^6)/b - 12*a*b*c^4*d^4))*(c^3)^(1/2))/a + (atanh((6*c^3*d^3*(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))/(6*b^3*c^5*d^3 - 10*a^3*c^2*d^6 - 18*a*b^2*c^4*
d^4 + 20*a^2*b*c^3*d^5 + (2*a^4*c*d^7)/b) - (6*a*c^2*d^4*(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))/(2*a^4*c*d^7 + 6*b^4*c^5*d^3 - 18*a*b^3*c^4*d^4 - 10*a^3*b*c^2*d^6 + 20*a^2*b^2*
c^3*d^5) + (2*a^2*c*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))/(6*
b^5*c^5*d^3 - 18*a*b^4*c^4*d^4 + 20*a^2*b^3*c^3*d^5 - 10*a^3*b^2*c^2*d^6 + 2*a^4*b*c*d^7))*(-b^3*(a*d - b*c)^3
)^(1/2))/(a*b^3)

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sympy [A]  time = 29.87, size = 92, normalized size = 0.96 \begin {gather*} \frac {d \sqrt {c + d x^{2}}}{b} + \frac {c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{a \sqrt {- c}} - \frac {\left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{a b^{2} \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)**(3/2)/x/(b*x**2+a),x)

[Out]

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

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