3.29.17 \(\int \frac {d+c x}{(-d+c x) \sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\)

Optimal. Leaf size=278 \[ \frac {4 d \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {-\sqrt {a^2 d^2+b^2 c^2}-a d}}\right )}{\sqrt {c} \sqrt {-\sqrt {a^2 d^2+b^2 c^2}-a d}}+\frac {4 d \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {\sqrt {a^2 d^2+b^2 c^2}-a d}}\right )}{\sqrt {c} \sqrt {\sqrt {a^2 d^2+b^2 c^2}-a d}}+\frac {2 \sqrt {a^2 x^2+b^2} (3 a c x+6 a d)+2 \left (3 a^2 c x^2+6 a^2 d x+b^2 c\right )}{3 a c \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}} \]

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Rubi [A]  time = 1.09, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {6742, 2117, 14, 2119, 1628, 828, 826, 1166, 208} \begin {gather*} -\frac {4 d \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {a d-\sqrt {a^2 d^2+b^2 c^2}}}\right )}{\sqrt {c} \sqrt {a d-\sqrt {a^2 d^2+b^2 c^2}}}-\frac {4 d \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {\sqrt {a^2 d^2+b^2 c^2}+a d}}\right )}{\sqrt {c} \sqrt {\sqrt {a^2 d^2+b^2 c^2}+a d}}+\frac {4 d}{c \sqrt {\sqrt {a^2 x^2+b^2}+a x}}-\frac {b^2}{3 a \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}}+\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*b^2/(a*(a*x + Sqrt[b^2 + a^2*x^2])^(3/2)) + (4*d)/(c*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]) + Sqrt[a*x + Sqrt[b
^2 + a^2*x^2]]/a - (4*d*ArcTanh[(Sqrt[c]*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]])/Sqrt[a*d - Sqrt[b^2*c^2 + a^2*d^2]]]
)/(Sqrt[c]*Sqrt[a*d - Sqrt[b^2*c^2 + a^2*d^2]]) - (4*d*ArcTanh[(Sqrt[c]*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]])/Sqrt[
a*d + Sqrt[b^2*c^2 + a^2*d^2]]])/(Sqrt[c]*Sqrt[a*d + Sqrt[b^2*c^2 + a^2*d^2]])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 828

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((
e*f - d*g)*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d
+ e*x)^(m + 1)*Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && FractionQ[m] && LtQ[m, -1]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rule 2119

Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dist[1/(2^(
m + 1)*e^(m + 1)), Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*(-(a*f^2*h) + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt
[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[m]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {d+c x}{(-d+c x) \sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx &=\int \left (\frac {1}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}}-\frac {2 d}{(d-c x) \sqrt {a x+\sqrt {b^2+a^2 x^2}}}\right ) \, dx\\ &=-\left ((2 d) \int \frac {1}{(d-c x) \sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\right )+\int \frac {1}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {b^2+x^2}{x^{5/2}} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{2 a}-(2 d) \operatorname {Subst}\left (\int \frac {b^2+x^2}{x^{3/2} \left (b^2 c+2 a d x-c x^2\right )} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {b^2}{x^{5/2}}+\frac {1}{\sqrt {x}}\right ) \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{2 a}-(2 d) \operatorname {Subst}\left (\int \left (-\frac {1}{c x^{3/2}}+\frac {2 \left (b^2 c+a d x\right )}{c x^{3/2} \left (b^2 c+2 a d x-c x^2\right )}\right ) \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )\\ &=-\frac {b^2}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}-\frac {4 d}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}-\frac {(4 d) \operatorname {Subst}\left (\int \frac {b^2 c+a d x}{x^{3/2} \left (b^2 c+2 a d x-c x^2\right )} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{c}\\ &=-\frac {b^2}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {4 d}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}-\frac {(4 d) \operatorname {Subst}\left (\int \frac {-a b^2 c d+b^2 c^2 x}{\sqrt {x} \left (b^2 c+2 a d x-c x^2\right )} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{b^2 c^2}\\ &=-\frac {b^2}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {4 d}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}-\frac {(8 d) \operatorname {Subst}\left (\int \frac {-a b^2 c d+b^2 c^2 x^2}{b^2 c+2 a d x^2-c x^4} \, dx,x,\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{b^2 c^2}\\ &=-\frac {b^2}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {4 d}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}-(4 d) \operatorname {Subst}\left (\int \frac {1}{a d-\sqrt {b^2 c^2+a^2 d^2}-c x^2} \, dx,x,\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )-(4 d) \operatorname {Subst}\left (\int \frac {1}{a d+\sqrt {b^2 c^2+a^2 d^2}-c x^2} \, dx,x,\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )\\ &=-\frac {b^2}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {4 d}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}-\frac {4 d \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {a d-\sqrt {b^2 c^2+a^2 d^2}}}\right )}{\sqrt {c} \sqrt {a d-\sqrt {b^2 c^2+a^2 d^2}}}-\frac {4 d \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {a d+\sqrt {b^2 c^2+a^2 d^2}}}\right )}{\sqrt {c} \sqrt {a d+\sqrt {b^2 c^2+a^2 d^2}}}\\ \end {align*}

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Mathematica [A]  time = 1.28, size = 374, normalized size = 1.35 \begin {gather*} \frac {\frac {12 a d \left (a d \left (a d-\sqrt {a^2 d^2+b^2 c^2}\right )+b^2 c^2\right ) \tan ^{-1}\left (\frac {b \sqrt {c}}{\sqrt {\sqrt {a^2 x^2+b^2}+a x} \sqrt {a d-\sqrt {a^2 d^2+b^2 c^2}}}\right )}{b \sqrt {a^2 d^2+b^2 c^2} \sqrt {a d-\sqrt {a^2 d^2+b^2 c^2}}}-\frac {12 a d \left (a d \left (\sqrt {a^2 d^2+b^2 c^2}+a d\right )+b^2 c^2\right ) \tan ^{-1}\left (\frac {b \sqrt {c}}{\sqrt {\sqrt {a^2 x^2+b^2}+a x} \sqrt {\sqrt {a^2 d^2+b^2 c^2}+a d}}\right )}{b \sqrt {a^2 d^2+b^2 c^2} \sqrt {\sqrt {a^2 d^2+b^2 c^2}+a d}}+\frac {\sqrt {c} \left (6 a \left (\sqrt {a^2 x^2+b^2}+a x\right ) (c x+2 d)+2 b^2 c\right )}{\left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}}}{3 a c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[c]*(2*b^2*c + 6*a*(2*d + c*x)*(a*x + Sqrt[b^2 + a^2*x^2])))/(a*x + Sqrt[b^2 + a^2*x^2])^(3/2) + (12*a*d
*(b^2*c^2 + a*d*(a*d - Sqrt[b^2*c^2 + a^2*d^2]))*ArcTan[(b*Sqrt[c])/(Sqrt[a*d - Sqrt[b^2*c^2 + a^2*d^2]]*Sqrt[
a*x + Sqrt[b^2 + a^2*x^2]])])/(b*Sqrt[b^2*c^2 + a^2*d^2]*Sqrt[a*d - Sqrt[b^2*c^2 + a^2*d^2]]) - (12*a*d*(b^2*c
^2 + a*d*(a*d + Sqrt[b^2*c^2 + a^2*d^2]))*ArcTan[(b*Sqrt[c])/(Sqrt[a*d + Sqrt[b^2*c^2 + a^2*d^2]]*Sqrt[a*x + S
qrt[b^2 + a^2*x^2]])])/(b*Sqrt[b^2*c^2 + a^2*d^2]*Sqrt[a*d + Sqrt[b^2*c^2 + a^2*d^2]]))/(3*a*c^(3/2))

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IntegrateAlgebraic [A]  time = 0.68, size = 278, normalized size = 1.00 \begin {gather*} \frac {2 (6 a d+3 a c x) \sqrt {b^2+a^2 x^2}+2 \left (b^2 c+6 a^2 d x+3 a^2 c x^2\right )}{3 a c \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {4 d \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {-a d-\sqrt {b^2 c^2+a^2 d^2}}}\right )}{\sqrt {c} \sqrt {-a d-\sqrt {b^2 c^2+a^2 d^2}}}+\frac {4 d \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {-a d+\sqrt {b^2 c^2+a^2 d^2}}}\right )}{\sqrt {c} \sqrt {-a d+\sqrt {b^2 c^2+a^2 d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(6*a*d + 3*a*c*x)*Sqrt[b^2 + a^2*x^2] + 2*(b^2*c + 6*a^2*d*x + 3*a^2*c*x^2))/(3*a*c*(a*x + Sqrt[b^2 + a^2*x
^2])^(3/2)) + (4*d*ArcTan[(Sqrt[c]*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]])/Sqrt[-(a*d) - Sqrt[b^2*c^2 + a^2*d^2]]])/(
Sqrt[c]*Sqrt[-(a*d) - Sqrt[b^2*c^2 + a^2*d^2]]) + (4*d*ArcTan[(Sqrt[c]*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]])/Sqrt[-
(a*d) + Sqrt[b^2*c^2 + a^2*d^2]]])/(Sqrt[c]*Sqrt[-(a*d) + Sqrt[b^2*c^2 + a^2*d^2]])

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fricas [B]  time = 0.54, size = 781, normalized size = 2.81 \begin {gather*} \frac {2 \, {\left (3 \, a b^{2} c \sqrt {-\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} + a d^{3}}{b^{2} c^{3}}} \log \left (32 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} d^{3} + 32 \, {\left (b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} - a d^{3}\right )} \sqrt {-\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} + a d^{3}}{b^{2} c^{3}}}\right ) - 3 \, a b^{2} c \sqrt {-\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} + a d^{3}}{b^{2} c^{3}}} \log \left (32 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} d^{3} - 32 \, {\left (b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} - a d^{3}\right )} \sqrt {-\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} + a d^{3}}{b^{2} c^{3}}}\right ) - 3 \, a b^{2} c \sqrt {\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} - a d^{3}}{b^{2} c^{3}}} \log \left (32 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} d^{3} + 32 \, {\left (b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} + a d^{3}\right )} \sqrt {\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} - a d^{3}}{b^{2} c^{3}}}\right ) + 3 \, a b^{2} c \sqrt {\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} - a d^{3}}{b^{2} c^{3}}} \log \left (32 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} d^{3} - 32 \, {\left (b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} + a d^{3}\right )} \sqrt {\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} d^{4} + a^{2} d^{6}}{b^{4} c^{6}}} - a d^{3}}{b^{2} c^{3}}}\right ) - {\left (a^{2} c x^{2} + 6 \, a^{2} d x - b^{2} c - \sqrt {a^{2} x^{2} + b^{2}} {\left (a c x + 6 \, a d\right )}\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}\right )}}{3 \, a b^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+d)/(c*x-d)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x, algorithm="fricas")

[Out]

2/3*(3*a*b^2*c*sqrt(-(b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*d^6)/(b^4*c^6)) + a*d^3)/(b^2*c^3))*log(32*sqrt(a*x + sq
rt(a^2*x^2 + b^2))*d^3 + 32*(b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*d^6)/(b^4*c^6)) - a*d^3)*sqrt(-(b^2*c^3*sqrt((b^2
*c^2*d^4 + a^2*d^6)/(b^4*c^6)) + a*d^3)/(b^2*c^3))) - 3*a*b^2*c*sqrt(-(b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*d^6)/(b
^4*c^6)) + a*d^3)/(b^2*c^3))*log(32*sqrt(a*x + sqrt(a^2*x^2 + b^2))*d^3 - 32*(b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*
d^6)/(b^4*c^6)) - a*d^3)*sqrt(-(b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*d^6)/(b^4*c^6)) + a*d^3)/(b^2*c^3))) - 3*a*b^2
*c*sqrt((b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*d^6)/(b^4*c^6)) - a*d^3)/(b^2*c^3))*log(32*sqrt(a*x + sqrt(a^2*x^2 +
b^2))*d^3 + 32*(b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*d^6)/(b^4*c^6)) + a*d^3)*sqrt((b^2*c^3*sqrt((b^2*c^2*d^4 + a^2
*d^6)/(b^4*c^6)) - a*d^3)/(b^2*c^3))) + 3*a*b^2*c*sqrt((b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*d^6)/(b^4*c^6)) - a*d^
3)/(b^2*c^3))*log(32*sqrt(a*x + sqrt(a^2*x^2 + b^2))*d^3 - 32*(b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*d^6)/(b^4*c^6))
+ a*d^3)*sqrt((b^2*c^3*sqrt((b^2*c^2*d^4 + a^2*d^6)/(b^4*c^6)) - a*d^3)/(b^2*c^3))) - (a^2*c*x^2 + 6*a^2*d*x
- b^2*c - sqrt(a^2*x^2 + b^2)*(a*c*x + 6*a*d))*sqrt(a*x + sqrt(a^2*x^2 + b^2)))/(a*b^2*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+d)/(c*x-d)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {c x +d}{\left (c x -d \right ) \sqrt {a x +\sqrt {a^{2} x^{2}+b^{2}}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x+d)/(c*x-d)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x)

[Out]

int((c*x+d)/(c*x-d)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+d)/(c*x-d)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {d+c\,x}{\sqrt {a\,x+\sqrt {a^2\,x^2+b^2}}\,\left (d-c\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(d + c*x)/((a*x + (b^2 + a^2*x^2)^(1/2))^(1/2)*(d - c*x)),x)

[Out]

int(-(d + c*x)/((a*x + (b^2 + a^2*x^2)^(1/2))^(1/2)*(d - c*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+d)/(c*x-d)/(a*x+(a**2*x**2+b**2)**(1/2))**(1/2),x)

[Out]

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

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