3.1.47 \(\int \frac {\tan (c+d x)}{a+a \sin (c+d x)} \, dx\) [47]

Optimal. Leaf size=37 \[ \frac {\tanh ^{-1}(\sin (c+d x))}{2 a d}+\frac {1}{2 d (a+a \sin (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.57, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2785, 2686, 30, 2691, 3855} \begin {gather*} \frac {\sec ^2(c+d x)}{2 a d}+\frac {\tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac {\tan (c+d x) \sec (c+d x)}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(a + a*Sin[c + d*x]),x]

[Out]

ArcTanh[Sin[c + d*x]]/(2*a*d) + Sec[c + d*x]^2/(2*a*d) - (Sec[c + d*x]*Tan[c + d*x])/(2*a*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2785

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 28, normalized size = 0.76 \begin {gather*} \frac {\tanh ^{-1}(\sin (c+d x))+\frac {1}{1+\sin (c+d x)}}{2 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/(a + a*Sin[c + d*x]),x]

[Out]

(ArcTanh[Sin[c + d*x]] + (1 + Sin[c + d*x])^(-1))/(2*a*d)

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Maple [A]
time = 0.16, size = 43, normalized size = 1.16

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{4}+\frac {1}{2+2 \sin \left (d x +c \right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{4}}{d a}\)	\(43\)
default	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{4}+\frac {1}{2+2 \sin \left (d x +c \right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{4}}{d a}\)	\(43\)
risch	\(\frac {i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-1/4*ln(sin(d*x+c)-1)+1/2/(1+sin(d*x+c))+1/4*ln(1+sin(d*x+c)))

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Maxima [A]
time = 0.31, size = 47, normalized size = 1.27 \begin {gather*} \frac {\frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a} + \frac {2}{a \sin \left (d x + c\right ) + a}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(sin(d*x + c) + 1)/a - log(sin(d*x + c) - 1)/a + 2/(a*sin(d*x + c) + a))/d

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Fricas [A]
time = 0.37, size = 58, normalized size = 1.57 \begin {gather*} \frac {{\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2}{4 \, {\left (a d \sin \left (d x + c\right ) + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((sin(d*x + c) + 1)*log(sin(d*x + c) + 1) - (sin(d*x + c) + 1)*log(-sin(d*x + c) + 1) + 2)/(a*d*sin(d*x +
c) + a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\tan {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 9.67, size = 58, normalized size = 1.57 \begin {gather*} \frac {\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac {\sin \left (d x + c\right ) - 1}{a {\left (\sin \left (d x + c\right ) + 1\right )}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(abs(sin(d*x + c) + 1))/a - log(abs(sin(d*x + c) - 1))/a - (sin(d*x + c) - 1)/(a*(sin(d*x + c) + 1)))/
d

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Mupad [B]
time = 6.66, size = 61, normalized size = 1.65 \begin {gather*} \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(tan(c/2 + (d*x)/2))/(a*d) - tan(c/2 + (d*x)/2)/(d*(a + 2*a*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2))

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