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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 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])

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
]

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 = 0.04, 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])

________________________________________________________________________________________

fricas [A]  time = 0.89, size = 91, normalized size = 3.03 \[ \left [\frac {\sqrt {-a \cos \left (d x + c\right )^{2} + a} \log \left (-\frac {\sin \left (d x + c\right ) + 1}{\sin \left (d x + c\right ) - 1}\right )}{2 \, a d \sin \left (d x + c\right )}, -\frac {\sqrt {-a} \arctan \left (\frac {\sqrt {-a \cos \left (d x + c\right )^{2} + a} \sqrt {-a}}{a}\right )}{a d}\right ] \]

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)]

________________________________________________________________________________________

giac [B]  time = 0.35, size = 61, normalized size = 2.03 \[ \frac {\frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\sqrt {a} d} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.15, size = 30, normalized size = 1.00 \[ \frac {\sin \left (d x +c \right ) \arctanh \left (\sin \left (d x +c \right )\right )}{\sqrt {a \left (\sin ^{2}\left (d x +c \right )\right )}\, d} \]

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

________________________________________________________________________________________

maxima [B]  time = 0.43, size = 76, normalized size = 2.53 \[ \frac {\frac {\left (-1\right )^{2 \, a \sin \left (d x + c\right )} \log \left (-\frac {a \sin \left (d x + c\right )}{\sin \left (d x + c\right ) + 1}\right )}{\sqrt {a}} + \frac {\left (-1\right )^{2 \, a \sin \left (d x + c\right )} \log \left (-\frac {a \sin \left (d x + c\right )}{\sin \left (d x + c\right ) - 1}\right )}{\sqrt {a}}}{2 \, d} \]

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]

1/2*((-1)^(2*a*sin(d*x + c))*log(-a*sin(d*x + c)/(sin(d*x + c) + 1))/sqrt(a) + (-1)^(2*a*sin(d*x + c))*log(-a*
sin(d*x + c)/(sin(d*x + c) - 1))/sqrt(a))/d

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {tan}\left (c+d\,x\right )}{\sqrt {a\,{\sin \left (c+d\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________