3.1.24 \(\int \frac {A+B x}{(a+c x^2) \sqrt {d+f x^2}} \, dx\)

Optimal. Leaf size=101 \[ \frac {A \tan ^{-1}\left (\frac {x \sqrt {c d-a f}}{\sqrt {a} \sqrt {d+f x^2}}\right )}{\sqrt {a} \sqrt {c d-a f}}-\frac {B \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+f x^2}}{\sqrt {c d-a f}}\right )}{\sqrt {c} \sqrt {c d-a f}} \]

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Rubi [A]  time = 0.13, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1010, 377, 205, 444, 63, 208} \begin {gather*} \frac {A \tan ^{-1}\left (\frac {x \sqrt {c d-a f}}{\sqrt {a} \sqrt {d+f x^2}}\right )}{\sqrt {a} \sqrt {c d-a f}}-\frac {B \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+f x^2}}{\sqrt {c d-a f}}\right )}{\sqrt {c} \sqrt {c d-a f}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((a + c*x^2)*Sqrt[d + f*x^2]),x]

[Out]

(A*ArcTan[(Sqrt[c*d - a*f]*x)/(Sqrt[a]*Sqrt[d + f*x^2])])/(Sqrt[a]*Sqrt[c*d - a*f]) - (B*ArcTanh[(Sqrt[c]*Sqrt
[d + f*x^2])/Sqrt[c*d - a*f]])/(Sqrt[c]*Sqrt[c*d - a*f])

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 1010

Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Dist[g, Int[(a + c
*x^2)^p*(d + f*x^2)^q, x], x] + Dist[h, Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h,
p, q}, x]

Rubi steps

\begin {align*} \int \frac {A+B x}{\left (a+c x^2\right ) \sqrt {d+f x^2}} \, dx &=A \int \frac {1}{\left (a+c x^2\right ) \sqrt {d+f x^2}} \, dx+B \int \frac {x}{\left (a+c x^2\right ) \sqrt {d+f x^2}} \, dx\\ &=A \operatorname {Subst}\left (\int \frac {1}{a-(-c d+a f) x^2} \, dx,x,\frac {x}{\sqrt {d+f x^2}}\right )+\frac {1}{2} B \operatorname {Subst}\left (\int \frac {1}{(a+c x) \sqrt {d+f x}} \, dx,x,x^2\right )\\ &=\frac {A \tan ^{-1}\left (\frac {\sqrt {c d-a f} x}{\sqrt {a} \sqrt {d+f x^2}}\right )}{\sqrt {a} \sqrt {c d-a f}}+\frac {B \operatorname {Subst}\left (\int \frac {1}{a-\frac {c d}{f}+\frac {c x^2}{f}} \, dx,x,\sqrt {d+f x^2}\right )}{f}\\ &=\frac {A \tan ^{-1}\left (\frac {\sqrt {c d-a f} x}{\sqrt {a} \sqrt {d+f x^2}}\right )}{\sqrt {a} \sqrt {c d-a f}}-\frac {B \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+f x^2}}{\sqrt {c d-a f}}\right )}{\sqrt {c} \sqrt {c d-a f}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 154, normalized size = 1.52 \begin {gather*} \frac {\left (A \sqrt {c}-\sqrt {-a} B\right ) \tanh ^{-1}\left (\frac {\sqrt {c} d-\sqrt {-a} f x}{\sqrt {d+f x^2} \sqrt {c d-a f}}\right )-\left (\sqrt {-a} B+A \sqrt {c}\right ) \tanh ^{-1}\left (\frac {\sqrt {-a} f x+\sqrt {c} d}{\sqrt {d+f x^2} \sqrt {c d-a f}}\right )}{2 \sqrt {-a} \sqrt {c} \sqrt {c d-a f}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((a + c*x^2)*Sqrt[d + f*x^2]),x]

[Out]

((-(Sqrt[-a]*B) + A*Sqrt[c])*ArcTanh[(Sqrt[c]*d - Sqrt[-a]*f*x)/(Sqrt[c*d - a*f]*Sqrt[d + f*x^2])] - (Sqrt[-a]
*B + A*Sqrt[c])*ArcTanh[(Sqrt[c]*d + Sqrt[-a]*f*x)/(Sqrt[c*d - a*f]*Sqrt[d + f*x^2])])/(2*Sqrt[-a]*Sqrt[c]*Sqr
t[c*d - a*f])

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IntegrateAlgebraic [B]  time = 2.27, size = 563, normalized size = 5.57 \begin {gather*} \frac {A \tanh ^{-1}\left (\frac {c \sqrt {f} x^2}{\sqrt {a} \sqrt {a f-c d}}-\frac {c x \sqrt {d+f x^2}}{\sqrt {a} \sqrt {a f-c d}}+\frac {\sqrt {a} \sqrt {f}}{\sqrt {a f-c d}}\right )}{\sqrt {a} \sqrt {a f-c d}}+\frac {\left (B c d \sqrt {-2 \sqrt {a} \sqrt {f} \sqrt {a f-c d}+2 a f-c d}-a B f \sqrt {-2 \sqrt {a} \sqrt {f} \sqrt {a f-c d}+2 a f-c d}-\sqrt {a} B \sqrt {f} \sqrt {a f-c d} \sqrt {-2 \sqrt {a} \sqrt {f} \sqrt {a f-c d}+2 a f-c d}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \left (\sqrt {f} x-\sqrt {d+f x^2}\right )}{\sqrt {-2 \sqrt {a} \sqrt {f} \sqrt {a f-c d}+2 a f-c d}}\right )}{c^{3/2} d (c d-a f)}+\frac {\left (B c d \sqrt {2 \sqrt {a} \sqrt {f} \sqrt {a f-c d}+2 a f-c d}-a B f \sqrt {2 \sqrt {a} \sqrt {f} \sqrt {a f-c d}+2 a f-c d}+\sqrt {a} B \sqrt {f} \sqrt {a f-c d} \sqrt {2 \sqrt {a} \sqrt {f} \sqrt {a f-c d}+2 a f-c d}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \left (\sqrt {f} x-\sqrt {d+f x^2}\right )}{\sqrt {2 \sqrt {a} \sqrt {f} \sqrt {a f-c d}+2 a f-c d}}\right )}{c^{3/2} d (c d-a f)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/((a + c*x^2)*Sqrt[d + f*x^2]),x]

[Out]

((B*c*d*Sqrt[-(c*d) + 2*a*f - 2*Sqrt[a]*Sqrt[f]*Sqrt[-(c*d) + a*f]] - a*B*f*Sqrt[-(c*d) + 2*a*f - 2*Sqrt[a]*Sq
rt[f]*Sqrt[-(c*d) + a*f]] - Sqrt[a]*B*Sqrt[f]*Sqrt[-(c*d) + a*f]*Sqrt[-(c*d) + 2*a*f - 2*Sqrt[a]*Sqrt[f]*Sqrt[
-(c*d) + a*f]])*ArcTan[(Sqrt[c]*(Sqrt[f]*x - Sqrt[d + f*x^2]))/Sqrt[-(c*d) + 2*a*f - 2*Sqrt[a]*Sqrt[f]*Sqrt[-(
c*d) + a*f]]])/(c^(3/2)*d*(c*d - a*f)) + ((B*c*d*Sqrt[-(c*d) + 2*a*f + 2*Sqrt[a]*Sqrt[f]*Sqrt[-(c*d) + a*f]] -
a*B*f*Sqrt[-(c*d) + 2*a*f + 2*Sqrt[a]*Sqrt[f]*Sqrt[-(c*d) + a*f]] + Sqrt[a]*B*Sqrt[f]*Sqrt[-(c*d) + a*f]*Sqrt
[-(c*d) + 2*a*f + 2*Sqrt[a]*Sqrt[f]*Sqrt[-(c*d) + a*f]])*ArcTan[(Sqrt[c]*(Sqrt[f]*x - Sqrt[d + f*x^2]))/Sqrt[-
(c*d) + 2*a*f + 2*Sqrt[a]*Sqrt[f]*Sqrt[-(c*d) + a*f]]])/(c^(3/2)*d*(c*d - a*f)) + (A*ArcTanh[(Sqrt[a]*Sqrt[f])
/Sqrt[-(c*d) + a*f] + (c*Sqrt[f]*x^2)/(Sqrt[a]*Sqrt[-(c*d) + a*f]) - (c*x*Sqrt[d + f*x^2])/(Sqrt[a]*Sqrt[-(c*d
) + a*f])])/(Sqrt[a]*Sqrt[-(c*d) + a*f])

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fricas [B]  time = 0.54, size = 1515, normalized size = 15.00

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+a)/(f*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

-1/4*sqrt((B^2*a - A^2*c + 2*(a*c^2*d - a^2*c*f)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))/(a*c^
2*d - a^2*c*f))*log(((A*B^3*a + A^3*B*c)*f*x + (A^2*B*c^2*d - A^2*B*a*c*f + (B*a*c^3*d^2 - 2*B*a^2*c^2*d*f + B
*a^3*c*f^2)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))*sqrt(f*x^2 + d)*sqrt((B^2*a - A^2*c + 2*(a
*c^2*d - a^2*c*f)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))/(a*c^2*d - a^2*c*f)) + sqrt(-A^2*B^2
/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2))*((B^2*a*c^2 + A^2*c^3)*d^2 - (B^2*a^2*c + A^2*a*c^2)*d*f))/x) + 1/4*
sqrt((B^2*a - A^2*c + 2*(a*c^2*d - a^2*c*f)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))/(a*c^2*d -
a^2*c*f))*log(((A*B^3*a + A^3*B*c)*f*x - (A^2*B*c^2*d - A^2*B*a*c*f + (B*a*c^3*d^2 - 2*B*a^2*c^2*d*f + B*a^3*
c*f^2)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))*sqrt(f*x^2 + d)*sqrt((B^2*a - A^2*c + 2*(a*c^2*
d - a^2*c*f)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))/(a*c^2*d - a^2*c*f)) + sqrt(-A^2*B^2/(a*c
^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2))*((B^2*a*c^2 + A^2*c^3)*d^2 - (B^2*a^2*c + A^2*a*c^2)*d*f))/x) - 1/4*sqrt(
(B^2*a - A^2*c - 2*(a*c^2*d - a^2*c*f)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))/(a*c^2*d - a^2*
c*f))*log(((A*B^3*a + A^3*B*c)*f*x + (A^2*B*c^2*d - A^2*B*a*c*f - (B*a*c^3*d^2 - 2*B*a^2*c^2*d*f + B*a^3*c*f^2
)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))*sqrt(f*x^2 + d)*sqrt((B^2*a - A^2*c - 2*(a*c^2*d - a
^2*c*f)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))/(a*c^2*d - a^2*c*f)) - sqrt(-A^2*B^2/(a*c^3*d^
2 - 2*a^2*c^2*d*f + a^3*c*f^2))*((B^2*a*c^2 + A^2*c^3)*d^2 - (B^2*a^2*c + A^2*a*c^2)*d*f))/x) + 1/4*sqrt((B^2*
a - A^2*c - 2*(a*c^2*d - a^2*c*f)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))/(a*c^2*d - a^2*c*f))
*log(((A*B^3*a + A^3*B*c)*f*x - (A^2*B*c^2*d - A^2*B*a*c*f - (B*a*c^3*d^2 - 2*B*a^2*c^2*d*f + B*a^3*c*f^2)*sqr
t(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))*sqrt(f*x^2 + d)*sqrt((B^2*a - A^2*c - 2*(a*c^2*d - a^2*c*
f)*sqrt(-A^2*B^2/(a*c^3*d^2 - 2*a^2*c^2*d*f + a^3*c*f^2)))/(a*c^2*d - a^2*c*f)) - sqrt(-A^2*B^2/(a*c^3*d^2 - 2
*a^2*c^2*d*f + a^3*c*f^2))*((B^2*a*c^2 + A^2*c^3)*d^2 - (B^2*a^2*c + A^2*a*c^2)*d*f))/x)

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giac [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((B*x+A)/(c*x^2+a)/(f*x^2+d)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.02, size = 608, normalized size = 6.02 \begin {gather*} -\frac {A \ln \left (\frac {\frac {2 \sqrt {-a c}\, \left (x -\frac {\sqrt {-a c}}{c}\right ) f}{c}-\frac {2 \left (a f -c d \right )}{c}+2 \sqrt {-\frac {a f -c d}{c}}\, \sqrt {\left (x -\frac {\sqrt {-a c}}{c}\right )^{2} f +\frac {2 \sqrt {-a c}\, \left (x -\frac {\sqrt {-a c}}{c}\right ) f}{c}-\frac {a f -c d}{c}}}{x -\frac {\sqrt {-a c}}{c}}\right )}{2 \sqrt {-a c}\, \sqrt {-\frac {a f -c d}{c}}}+\frac {A \ln \left (\frac {-\frac {2 \sqrt {-a c}\, \left (x +\frac {\sqrt {-a c}}{c}\right ) f}{c}-\frac {2 \left (a f -c d \right )}{c}+2 \sqrt {-\frac {a f -c d}{c}}\, \sqrt {\left (x +\frac {\sqrt {-a c}}{c}\right )^{2} f -\frac {2 \sqrt {-a c}\, \left (x +\frac {\sqrt {-a c}}{c}\right ) f}{c}-\frac {a f -c d}{c}}}{x +\frac {\sqrt {-a c}}{c}}\right )}{2 \sqrt {-a c}\, \sqrt {-\frac {a f -c d}{c}}}-\frac {B \ln \left (\frac {\frac {2 \sqrt {-a c}\, \left (x -\frac {\sqrt {-a c}}{c}\right ) f}{c}-\frac {2 \left (a f -c d \right )}{c}+2 \sqrt {-\frac {a f -c d}{c}}\, \sqrt {\left (x -\frac {\sqrt {-a c}}{c}\right )^{2} f +\frac {2 \sqrt {-a c}\, \left (x -\frac {\sqrt {-a c}}{c}\right ) f}{c}-\frac {a f -c d}{c}}}{x -\frac {\sqrt {-a c}}{c}}\right )}{2 \sqrt {-\frac {a f -c d}{c}}\, c}-\frac {B \ln \left (\frac {-\frac {2 \sqrt {-a c}\, \left (x +\frac {\sqrt {-a c}}{c}\right ) f}{c}-\frac {2 \left (a f -c d \right )}{c}+2 \sqrt {-\frac {a f -c d}{c}}\, \sqrt {\left (x +\frac {\sqrt {-a c}}{c}\right )^{2} f -\frac {2 \sqrt {-a c}\, \left (x +\frac {\sqrt {-a c}}{c}\right ) f}{c}-\frac {a f -c d}{c}}}{x +\frac {\sqrt {-a c}}{c}}\right )}{2 \sqrt {-\frac {a f -c d}{c}}\, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(c*x^2+a)/(f*x^2+d)^(1/2),x)

[Out]

-1/2/(-a*c)^(1/2)/(-(a*f-c*d)/c)^(1/2)*ln((-2*(a*f-c*d)/c+2*f*(-a*c)^(1/2)/c*(x-(-a*c)^(1/2)/c)+2*(-(a*f-c*d)/
c)^(1/2)*((x-(-a*c)^(1/2)/c)^2*f+2*f*(-a*c)^(1/2)/c*(x-(-a*c)^(1/2)/c)-(a*f-c*d)/c)^(1/2))/(x-(-a*c)^(1/2)/c))
*A-1/2/c/(-(a*f-c*d)/c)^(1/2)*ln((-2*(a*f-c*d)/c+2*f*(-a*c)^(1/2)/c*(x-(-a*c)^(1/2)/c)+2*(-(a*f-c*d)/c)^(1/2)*
((x-(-a*c)^(1/2)/c)^2*f+2*f*(-a*c)^(1/2)/c*(x-(-a*c)^(1/2)/c)-(a*f-c*d)/c)^(1/2))/(x-(-a*c)^(1/2)/c))*B+1/2/(-
a*c)^(1/2)/(-(a*f-c*d)/c)^(1/2)*ln((-2*(a*f-c*d)/c-2*f*(-a*c)^(1/2)/c*(x+(-a*c)^(1/2)/c)+2*(-(a*f-c*d)/c)^(1/2
)*((x+(-a*c)^(1/2)/c)^2*f-2*f*(-a*c)^(1/2)/c*(x+(-a*c)^(1/2)/c)-(a*f-c*d)/c)^(1/2))/(x+(-a*c)^(1/2)/c))*A-1/2/
c/(-(a*f-c*d)/c)^(1/2)*ln((-2*(a*f-c*d)/c-2*f*(-a*c)^(1/2)/c*(x+(-a*c)^(1/2)/c)+2*(-(a*f-c*d)/c)^(1/2)*((x+(-a
*c)^(1/2)/c)^2*f-2*f*(-a*c)^(1/2)/c*(x+(-a*c)^(1/2)/c)-(a*f-c*d)/c)^(1/2))/(x+(-a*c)^(1/2)/c))*B

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{{\left (c x^{2} + a\right )} \sqrt {f x^{2} + d}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+a)/(f*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((c*x^2 + a)*sqrt(f*x^2 + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \left \{\begin {array}{cl} \frac {B\,\mathrm {atan}\left (\frac {c\,\sqrt {f\,x^2+d}}{\sqrt {a\,c\,f-c^2\,d}}\right )}{\sqrt {a\,c\,f-c^2\,d}}+\frac {A\,\mathrm {atan}\left (\frac {x\,\sqrt {c\,d-a\,f}}{\sqrt {a}\,\sqrt {f\,x^2+d}}\right )}{\sqrt {-a\,\left (a\,f-c\,d\right )}} & \text {\ if\ \ }0<c\,d-a\,f\\ \frac {A\,\ln \left (\frac {\sqrt {a\,\left (f\,x^2+d\right )}+x\,\sqrt {a\,f-c\,d}}{\sqrt {a\,\left (f\,x^2+d\right )}-x\,\sqrt {a\,f-c\,d}}\right )}{2\,\sqrt {a\,\left (a\,f-c\,d\right )}}+\frac {B\,\mathrm {atan}\left (\frac {c\,\sqrt {f\,x^2+d}}{\sqrt {a\,c\,f-c^2\,d}}\right )}{\sqrt {a\,c\,f-c^2\,d}} & \text {\ if\ \ }c\,d-a\,f<0\\ \int \frac {A+B\,x}{\left (c\,x^2+a\right )\,\sqrt {f\,x^2+d}} \,d x & \text {\ if\ \ }c\,d-a\,f\notin \mathbb {R}\vee a\,f=c\,d \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((a + c*x^2)*(d + f*x^2)^(1/2)),x)

[Out]

piecewise(0 < - a*f + c*d, (B*atan((c*(d + f*x^2)^(1/2))/(- c^2*d + a*c*f)^(1/2)))/(- c^2*d + a*c*f)^(1/2) + (
A*atan((x*(- a*f + c*d)^(1/2))/(a^(1/2)*(d + f*x^2)^(1/2))))/(-a*(a*f - c*d))^(1/2), - a*f + c*d < 0, (A*log((
(a*(d + f*x^2))^(1/2) + x*(a*f - c*d)^(1/2))/((a*(d + f*x^2))^(1/2) - x*(a*f - c*d)^(1/2))))/(2*(a*(a*f - c*d)
)^(1/2)) + (B*atan((c*(d + f*x^2)^(1/2))/(- c^2*d + a*c*f)^(1/2)))/(- c^2*d + a*c*f)^(1/2), ~in(- a*f + c*d, '
real') | a*f == c*d, int((A + B*x)/((a + c*x^2)*(d + f*x^2)^(1/2)), x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{\left (a + c x^{2}\right ) \sqrt {d + f x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x**2+a)/(f*x**2+d)**(1/2),x)

[Out]

Integral((A + B*x)/((a + c*x**2)*sqrt(d + f*x**2)), x)

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