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

Optimal. Leaf size=43 \[ \frac{\log \left (a+b \sin ^2(c+d x)\right )}{2 d (a+b)}-\frac{\log (\cos (c+d x))}{d (a+b)} \]

[Out]

-(Log[Cos[c + d*x]]/((a + b)*d)) + Log[a + b*Sin[c + d*x]^2]/(2*(a + b)*d)

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Rubi [A]  time = 0.0385448, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3194, 36, 31} \[ \frac{\log \left (a+b \sin ^2(c+d x)\right )}{2 d (a+b)}-\frac{\log (\cos (c+d x))}{d (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(a + b*Sin[c + d*x]^2),x]

[Out]

-(Log[Cos[c + d*x]]/((a + b)*d)) + Log[a + b*Sin[c + d*x]^2]/(2*(a + b)*d)

Rule 3194

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0323504, size = 37, normalized size = 0.86 \[ \frac{\log \left (a-b \cos ^2(c+d x)+b\right )-2 \log (\cos (c+d x))}{2 a d+2 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/(a + b*Sin[c + d*x]^2),x]

[Out]

(-2*Log[Cos[c + d*x]] + Log[a + b - b*Cos[c + d*x]^2])/(2*a*d + 2*b*d)

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Maple [A]  time = 0.065, size = 47, normalized size = 1.1 \begin{align*} -{\frac{\ln \left ( \cos \left ( dx+c \right ) \right ) }{ \left ( a+b \right ) d}}+{\frac{\ln \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}-a-b \right ) }{2\, \left ( a+b \right ) d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(cos(d*x+c))/(a+b)/d+1/2/d/(a+b)*ln(b*cos(d*x+c)^2-a-b)

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Maxima [A]  time = 0.983951, size = 58, normalized size = 1.35 \begin{align*} \frac{\frac{\log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a + b} - \frac{\log \left (\sin \left (d x + c\right )^{2} - 1\right )}{a + b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.87446, size = 99, normalized size = 2.3 \begin{align*} \frac{\log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 2 \, \log \left (-\cos \left (d x + c\right )\right )}{2 \,{\left (a + b\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(-b*cos(d*x + c)^2 + a + b) - 2*log(-cos(d*x + c)))/((a + b)*d)

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

[Out]

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

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Giac [B]  time = 1.1399, size = 149, normalized size = 3.47 \begin{align*} \frac{\frac{\log \left (a - \frac{2 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{4 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a + b} - \frac{2 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right )}{a + b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(a - 2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a*(cos(d*
x + c) - 1)^2/(cos(d*x + c) + 1)^2)/(a + b) - 2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1))/(a + b))/
d