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

Optimal. Leaf size=89 \[ -\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}-\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3572, 3532, 205, 3634, 63} \[ -\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}-\frac {\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[d]*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(a*f)) - (Sqrt[d]*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sq
rt[2]*Sqrt[d*Tan[e + f*x]])])/(Sqrt[2]*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 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 3572

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

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

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Mathematica [A]  time = 0.18, size = 98, normalized size = 1.10 \[ -\frac {\left (\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+2 \tan ^{-1}\left (\sqrt {\tan (e+f x)}\right )\right ) \sqrt {d \tan (e+f x)}}{2 a f \sqrt {\tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 211, normalized size = 2.37 \[ \left [\frac {\sqrt {2} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) - \sqrt {2}\right )} \sqrt {-d} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right )}{4 \, a f}, \frac {\sqrt {2} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) - \sqrt {2}\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) - 2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{2 \, a f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.98, size = 253, normalized size = 2.84 \[ \frac {\frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a f} + \frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a f} - \frac {8 \, d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a f} - \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a f}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.31, size = 359, normalized size = 4.03 \[ \frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{8 f a}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a}+\frac {d \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{8 f a \left (d^{2}\right )^{\frac {1}{4}}}+\frac {d \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a \left (d^{2}\right )^{\frac {1}{4}}}-\frac {d \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a \left (d^{2}\right )^{\frac {1}{4}}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right ) \sqrt {d}}{a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f/a*(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^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^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*d*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*d*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*d*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))*d^(1/2)/a/f

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maxima [A]  time = 0.50, size = 112, normalized size = 1.26 \[ \frac {\frac {d^{2} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{a} - \frac {2 \, d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a}}{2 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d^2*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*arc
tan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/a - 2*d^(3/2)*arctan(sqrt(d*tan(
f*x + e))/sqrt(d))/a)/(d*f)

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mupad [B]  time = 4.25, size = 103, normalized size = 1.16 \[ \frac {\sqrt {2}\,\sqrt {d}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{4\,a\,f}-\frac {\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*d^(1/2)*(2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2))) + 2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2
))/(2*d^(1/2)) + (2^(1/2)*(d*tan(e + f*x))^(3/2))/(2*d^(3/2)))))/(4*a*f) - (d^(1/2)*atan((d*tan(e + f*x))^(1/2
)/d^(1/2)))/(a*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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