3.850 \(\int \frac{\tan (c+d x)}{\sqrt{a \sin ^2(c+d x)}} \, dx\)

Optimal. Leaf size=30 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a \sin ^2(c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d} \]

[Out]

ArcTanh[Sqrt[a*Sin[c + d*x]^2]/Sqrt[a]]/(Sqrt[a]*d)

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Rubi [A]  time = 0.0351015, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3205, 63, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a \sin ^2(c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/Sqrt[a*Sin[c + d*x]^2],x]

[Out]

ArcTanh[Sqrt[a*Sin[c + d*x]^2]/Sqrt[a]]/(Sqrt[a]*d)

Rule 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0385442, size = 31, normalized size = 1.03 \[ \frac{\sin (c+d x) \tanh ^{-1}(\sin (c+d x))}{d \sqrt{a \sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/Sqrt[a*Sin[c + d*x]^2],x]

[Out]

(ArcTanh[Sin[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a*Sin[c + d*x]^2])

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Maple [A]  time = 0.043, size = 30, normalized size = 1. \begin{align*}{\frac{\sin \left ( dx+c \right ){\it Artanh} \left ( \sin \left ( dx+c \right ) \right ) }{d}{\frac{1}{\sqrt{a \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a*sin(d*x+c)^2)^(1/2)*sin(d*x+c)*arctanh(sin(d*x+c))/d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.32307, size = 221, normalized size = 7.37 \begin{align*} \left [\frac{\sqrt{-a \cos \left (d x + c\right )^{2} + a} \log \left (-\frac{\sin \left (d x + c\right ) + 1}{\sin \left (d x + c\right ) - 1}\right )}{2 \, a d \sin \left (d x + c\right )}, -\frac{\sqrt{-a} \arctan \left (\frac{\sqrt{-a \cos \left (d x + c\right )^{2} + a} \sqrt{-a}}{a}\right )}{a d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(-a*cos(d*x + c)^2 + a)*log(-(sin(d*x + c) + 1)/(sin(d*x + c) - 1))/(a*d*sin(d*x + c)), -sqrt(-a)*arc
tan(sqrt(-a*cos(d*x + c)^2 + a)*sqrt(-a)/a)/(a*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (c + d x \right )}}{\sqrt{a \sin ^{2}{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a*sin(d*x+c)**2)**(1/2),x)

[Out]

Integral(tan(c + d*x)/sqrt(a*sin(c + d*x)**2), x)

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Giac [B]  time = 1.23317, size = 82, normalized size = 2.73 \begin{align*} \frac{\frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{\sqrt{a} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(abs(tan(1/2*d*x + 1/2*c) + 1))/sgn(tan(1/2*d*x + 1/2*c)) - log(abs(tan(1/2*d*x + 1/2*c) - 1))/sgn(tan(1/2
*d*x + 1/2*c)))/(sqrt(a)*d)