3.361 \(\int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx\)

Optimal. Leaf size=81 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a \sqrt {d} f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {d} (\tan (e+f x)+1)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a \sqrt {d} f} \]

[Out]

arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a/f/d^(1/2)+1/2*arctanh(1/2*d^(1/2)*(1+tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))
^(1/2))/a/f*2^(1/2)/d^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3574, 3532, 208, 3634, 63, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a \sqrt {d} f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {d} (\tan (e+f x)+1)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a \sqrt {d} f} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])),x]

[Out]

ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]]/(a*Sqrt[d]*f) + ArcTanh[(Sqrt[d]*(1 + Tan[e + f*x]))/(Sqrt[2]*Sqrt[d*Tan[
e + f*x]])]/(Sqrt[2]*a*Sqrt[d]*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 3532

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*d^2)/f,
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 3574

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

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rubi steps

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

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Mathematica [A]  time = 0.51, size = 107, normalized size = 1.32 \[ \frac {\sqrt {\tan (e+f x)} \left (4 \tan ^{-1}\left (\sqrt {\tan (e+f x)}\right )+\sqrt {2} \left (\log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )-\log \left (-\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}-1\right )\right )\right )}{4 a f \sqrt {d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])),x]

[Out]

((4*ArcTan[Sqrt[Tan[e + f*x]]] + Sqrt[2]*(-Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] + Log[1 + Sqrt[
2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]))*Sqrt[Tan[e + f*x]])/(4*a*f*Sqrt[d*Tan[e + f*x]])

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fricas [A]  time = 0.53, size = 209, normalized size = 2.58 \[ \left [-\frac {\sqrt {2} \sqrt {-d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) + \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right )}{2 \, a d f}, \frac {\sqrt {2} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{4 \, a d f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e)),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(2)*sqrt(-d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d*tan(f*x + e)))
+ sqrt(-d)*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)))/(a*d*f), 1/4*(sqrt(
2)*sqrt(d)*log((d*tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d)*(tan(f*x + e) + 1) + 4*d*tan(f*x + e
) + d)/(tan(f*x + e)^2 + 1)) + 4*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d)))/(a*d*f)]

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giac [B]  time = 0.93, size = 260, normalized size = 3.21 \[ \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{4 \, a d^{2} f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{4 \, a d^{2} f} + \frac {\arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a \sqrt {d} f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{8 \, a d^{2} f} - \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{8 \, a d^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e)),x, algorithm="giac")

[Out]

1/4*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x + e)))
/sqrt(abs(d)))/(a*d^2*f) + 1/4*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d
)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a*d^2*f) + arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a*sqrt(d)*f) + 1/
8*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs
(d))/(a*d^2*f) - 1/8*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))
*sqrt(abs(d)) + abs(d))/(a*d^2*f)

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maple [B]  time = 0.31, size = 364, normalized size = 4.49 \[ \frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{8 f a d}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a d}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a d}-\frac {\sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{8 f a \left (d^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a \left (d^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a \left (d^{2}\right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{a f \sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e)),x)

[Out]

1/8/f/a/d*(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*
x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/4/f/a/d*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)
^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/4/f/a/d*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+
1)-1/8/f/a*2^(1/2)/(d^2)^(1/4)*ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f
*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/4/f/a*2^(1/2)/(d^2)^(1/4)*arctan(2^(1/2)/(d^2)^
(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/4/f/a*2^(1/2)/(d^2)^(1/4)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+
arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a/f/d^(1/2)

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maxima [A]  time = 0.74, size = 110, normalized size = 1.36 \[ \frac {\frac {d {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a} + \frac {4 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a}}{4 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e)),x, algorithm="maxima")

[Out]

1/4*(d*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - sqrt(2)*log(d*tan(f*x
+ e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/a + 4*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/
a)/(d*f)

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mupad [B]  time = 4.34, size = 78, normalized size = 0.96 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,\sqrt {d}\,f}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,d^{9/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{12\,d^5\,\mathrm {tan}\left (e+f\,x\right )+12\,d^5}\right )}{2\,a\,\sqrt {d}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d*tan(e + f*x))^(1/2)*(a + a*tan(e + f*x))),x)

[Out]

atan((d*tan(e + f*x))^(1/2)/d^(1/2))/(a*d^(1/2)*f) + (2^(1/2)*atanh((12*2^(1/2)*d^(9/2)*(d*tan(e + f*x))^(1/2)
)/(12*d^5*tan(e + f*x) + 12*d^5)))/(2*a*d^(1/2)*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\sqrt {d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + \sqrt {d \tan {\left (e + f x \right )}}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))**(1/2)/(a+a*tan(f*x+e)),x)

[Out]

Integral(1/(sqrt(d*tan(e + f*x))*tan(e + f*x) + sqrt(d*tan(e + f*x))), x)/a

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