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

Optimal. Leaf size=109 \[ -\frac{a}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{\left (a^2+b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)^2}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.0956814, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2721, 801} \[ -\frac{a}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{\left (a^2+b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)^2}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [A]  time = 0.286521, size = 162, normalized size = 1.49 \[ -\frac{a \left (-2 \left (\left (a^2+b^2\right ) \log (a+b \sin (c+d x))-a^2+b^2\right )+(a-b)^2 \log (1-\sin (c+d x))+(a+b)^2 \log (\sin (c+d x)+1)\right )+b \sin (c+d x) \left (-2 \left (a^2+b^2\right ) \log (a+b \sin (c+d x))+(a-b)^2 \log (1-\sin (c+d x))+(a+b)^2 \log (\sin (c+d x)+1)\right )}{2 d (a-b)^2 (a+b)^2 (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.078, size = 132, normalized size = 1.2 \begin{align*} -{\frac{a}{d \left ( a+b \right ) \left ( a-b \right ) \left ( a+b\sin \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ){a}^{2}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{2}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{2\,d \left ( a+b \right ) ^{2}}}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{2\, \left ( a-b \right ) ^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.7641, size = 167, normalized size = 1.53 \begin{align*} \frac{\frac{2 \,{\left (a^{2} + b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \, a}{a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )} - \frac{\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{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*(2*(a^2 + b^2)*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 + b^4) - 2*a/(a^3 - a*b^2 + (a^2*b - b^3)*sin(d*x
+ c)) - log(sin(d*x + c) + 1)/(a^2 - 2*a*b + b^2) - log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2))/d

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Fricas [A]  time = 1.69463, size = 463, normalized size = 4.25 \begin{align*} -\frac{2 \, a^{3} - 2 \, a b^{2} - 2 \,{\left (a^{3} + a b^{2} +{\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) +{\left (a^{3} + 2 \, a^{2} b + a b^{2} +{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{3} - 2 \, a^{2} b + a b^{2} +{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x + c\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \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*(2*a^3 - 2*a*b^2 - 2*(a^3 + a*b^2 + (a^2*b + b^3)*sin(d*x + c))*log(b*sin(d*x + c) + a) + (a^3 + 2*a^2*b
+ a*b^2 + (a^2*b + 2*a*b^2 + b^3)*sin(d*x + c))*log(sin(d*x + c) + 1) + (a^3 - 2*a^2*b + a*b^2 + (a^2*b - 2*a*
b^2 + b^3)*sin(d*x + c))*log(-sin(d*x + c) + 1))/((a^4*b - 2*a^2*b^3 + b^5)*d*sin(d*x + c) + (a^5 - 2*a^3*b^2
+ a*b^4)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{2}}\, 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 [A]  time = 1.28426, size = 211, normalized size = 1.94 \begin{align*} \frac{\frac{2 \,{\left (a^{2} b + b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{2 \,{\left (a^{2} b \sin \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) + 2 \, a^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}}}{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*(2*(a^2*b + b^3)*log(abs(b*sin(d*x + c) + a))/(a^4*b - 2*a^2*b^3 + b^5) - log(abs(sin(d*x + c) + 1))/(a^2
- 2*a*b + b^2) - log(abs(sin(d*x + c) - 1))/(a^2 + 2*a*b + b^2) - 2*(a^2*b*sin(d*x + c) + b^3*sin(d*x + c) + 2
*a^3)/((a^4 - 2*a^2*b^2 + b^4)*(b*sin(d*x + c) + a)))/d