3.26.56 \(\int \frac {1}{(d+c x) \sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\)

Optimal. Leaf size=215 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {a d-\sqrt {a^2 d^2+b^2 c^2}}}\right )}{\sqrt {c} \sqrt {a d-\sqrt {a^2 d^2+b^2 c^2}}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {\sqrt {a^2 d^2+b^2 c^2}+a d}}\right )}{\sqrt {c} \sqrt {\sqrt {a^2 d^2+b^2 c^2}+a d}}+\frac {2}{c \sqrt {\sqrt {a^2 x^2+b^2}+a x}} \]

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Rubi [A]  time = 0.50, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2119, 1628, 828, 826, 1166, 205} \begin {gather*} \frac {2 \tan ^{-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 {2 \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}{c \sqrt {\sqrt {a^2 x^2+b^2}+a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + c*x)*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]),x]

[Out]

2/(c*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]) + (2*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]]) + (2*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]])

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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{(d+c x) \sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx &=\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 )\\ &=\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 {2}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {2 \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 {2}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}-\frac {2 \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 {2}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}-\frac {4 \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 {2}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+2 \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 )+2 \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 {2}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {2 \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 {2 \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 {align*}

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Mathematica [A]  time = 0.64, size = 333, normalized size = 1.55 \begin {gather*} -\frac {2 \left (-\frac {\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 {\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 {\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}}{\sqrt {\sqrt {a^2 x^2+b^2}+a x}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + c*x)*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.66, size = 215, normalized size = 1.00 \begin {gather*} \frac {2}{c \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {2 \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 {2 \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[1/((d + c*x)*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]),x]

[Out]

2/(c*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]) + (2*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]]) + (2*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 = 0.65, size = 673, normalized size = 3.13 \begin {gather*} -\frac {b^{2} c \sqrt {\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} + a d}{b^{2} c^{3}}} \log \left (4 \, {\left (b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} - a d\right )} \sqrt {\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} + a d}{b^{2} c^{3}}} + 4 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}\right ) - b^{2} c \sqrt {\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} + a d}{b^{2} c^{3}}} \log \left (-4 \, {\left (b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} - a d\right )} \sqrt {\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} + a d}{b^{2} c^{3}}} + 4 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}\right ) - b^{2} c \sqrt {-\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} - a d}{b^{2} c^{3}}} \log \left (4 \, {\left (b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} + a d\right )} \sqrt {-\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} - a d}{b^{2} c^{3}}} + 4 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}\right ) + b^{2} c \sqrt {-\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} - a d}{b^{2} c^{3}}} \log \left (-4 \, {\left (b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} + a d\right )} \sqrt {-\frac {b^{2} c^{3} \sqrt {\frac {b^{2} c^{2} + a^{2} d^{2}}{b^{4} c^{6}}} - a d}{b^{2} c^{3}}} + 4 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}\right ) + 2 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} {\left (a x - \sqrt {a^{2} x^{2} + b^{2}}\right )}}{b^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*c*sqrt((b^2*c^3*sqrt((b^2*c^2 + a^2*d^2)/(b^4*c^6)) + a*d)/(b^2*c^3))*log(4*(b^2*c^3*sqrt((b^2*c^2 + a^2
*d^2)/(b^4*c^6)) - a*d)*sqrt((b^2*c^3*sqrt((b^2*c^2 + a^2*d^2)/(b^4*c^6)) + a*d)/(b^2*c^3)) + 4*sqrt(a*x + sqr
t(a^2*x^2 + b^2))) - b^2*c*sqrt((b^2*c^3*sqrt((b^2*c^2 + a^2*d^2)/(b^4*c^6)) + a*d)/(b^2*c^3))*log(-4*(b^2*c^3
*sqrt((b^2*c^2 + a^2*d^2)/(b^4*c^6)) - a*d)*sqrt((b^2*c^3*sqrt((b^2*c^2 + a^2*d^2)/(b^4*c^6)) + a*d)/(b^2*c^3)
) + 4*sqrt(a*x + sqrt(a^2*x^2 + b^2))) - b^2*c*sqrt(-(b^2*c^3*sqrt((b^2*c^2 + a^2*d^2)/(b^4*c^6)) - a*d)/(b^2*
c^3))*log(4*(b^2*c^3*sqrt((b^2*c^2 + a^2*d^2)/(b^4*c^6)) + a*d)*sqrt(-(b^2*c^3*sqrt((b^2*c^2 + a^2*d^2)/(b^4*c
^6)) - a*d)/(b^2*c^3)) + 4*sqrt(a*x + sqrt(a^2*x^2 + b^2))) + b^2*c*sqrt(-(b^2*c^3*sqrt((b^2*c^2 + a^2*d^2)/(b
^4*c^6)) - a*d)/(b^2*c^3))*log(-4*(b^2*c^3*sqrt((b^2*c^2 + a^2*d^2)/(b^4*c^6)) + a*d)*sqrt(-(b^2*c^3*sqrt((b^2
*c^2 + a^2*d^2)/(b^4*c^6)) - a*d)/(b^2*c^3)) + 4*sqrt(a*x + sqrt(a^2*x^2 + b^2))) + 2*sqrt(a*x + sqrt(a^2*x^2
+ b^2))*(a*x - sqrt(a^2*x^2 + b^2)))/(b^2*c)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\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(1/(c*x+d)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(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 \[\int \frac {1}{\left (c x +d \right ) \sqrt {a x +\sqrt {a^{2} x^{2}+b^{2}}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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 {1}{\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(1/(c*x+d)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(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 {1}{\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(1/((a*x + (b^2 + a^2*x^2)^(1/2))^(1/2)*(d + c*x)),x)

[Out]

int(1/((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 {1}{\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(1/(c*x+d)/(a*x+(a**2*x**2+b**2)**(1/2))**(1/2),x)

[Out]

Integral(1/(sqrt(a*x + sqrt(a**2*x**2 + b**2))*(c*x + d)), x)

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