3.194 \(\int \frac{\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=149 \[ -\frac{a}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{a^2+b^2}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{a \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)^3}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)^3} \]

[Out]

-Log[1 - Sin[c + d*x]]/(2*(a + b)^3*d) - Log[1 + Sin[c + d*x]]/(2*(a - b)^3*d) + (a*(a^2 + 3*b^2)*Log[a + b*Si
n[c + d*x]])/((a^2 - b^2)^3*d) - a/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - (a^2 + b^2)/((a^2 - b^2)^2*d*(a
+ b*Sin[c + d*x]))

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Rubi [A]  time = 0.132163, antiderivative size = 149, 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} \[ -\frac{a}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{a^2+b^2}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{a \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)^3}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/(a + b*Sin[c + d*x])^3,x]

[Out]

-Log[1 - Sin[c + d*x]]/(2*(a + b)^3*d) - Log[1 + Sin[c + d*x]]/(2*(a - b)^3*d) + (a*(a^2 + 3*b^2)*Log[a + b*Si
n[c + d*x]])/((a^2 - b^2)^3*d) - a/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - (a^2 + b^2)/((a^2 - b^2)^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))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{(a+x)^3 \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)^3 (b-x)}+\frac{a}{(a-b) (a+b) (a+x)^3}+\frac{a^2+b^2}{(a-b)^2 (a+b)^2 (a+x)^2}+\frac{a^3+3 a b^2}{(a-b)^3 (a+b)^3 (a+x)}-\frac{1}{2 (a-b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\log (1-\sin (c+d x))}{2 (a+b)^3 d}-\frac{\log (1+\sin (c+d x))}{2 (a-b)^3 d}+\frac{a \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac{a}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac{a^2+b^2}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.02021, size = 213, normalized size = 1.43 \[ \frac{\frac{2 b}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{4 a b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+a \left (\frac{b \left (\frac{\left (a^2-b^2\right ) \left (-5 a^2-4 a b \sin (c+d x)+b^2\right )}{(a+b \sin (c+d x))^2}+2 \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))\right )}{\left (a^2-b^2\right )^3}+\frac{\log (1-\sin (c+d x))}{(a+b)^3}-\frac{\log (\sin (c+d x)+1)}{(a-b)^3}\right )-\frac{\log (1-\sin (c+d x))}{(a+b)^2}+\frac{\log (\sin (c+d x)+1)}{(a-b)^2}}{2 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/(a + b*Sin[c + d*x])^3,x]

[Out]

(-(Log[1 - Sin[c + d*x]]/(a + b)^2) + Log[1 + Sin[c + d*x]]/(a - b)^2 - (4*a*b*Log[a + b*Sin[c + d*x]])/(a^2 -
b^2)^2 + (2*b)/((a^2 - b^2)*(a + b*Sin[c + d*x])) + a*(Log[1 - Sin[c + d*x]]/(a + b)^3 - Log[1 + Sin[c + d*x]
]/(a - b)^3 + (b*(2*(3*a^2 + b^2)*Log[a + b*Sin[c + d*x]] + ((a^2 - b^2)*(-5*a^2 + b^2 - 4*a*b*Sin[c + d*x]))/
(a + b*Sin[c + d*x])^2))/(a^2 - b^2)^3))/(2*b*d)

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Maple [A]  time = 0.1, size = 198, normalized size = 1.3 \begin{align*} -{\frac{a}{2\,d \left ( a+b \right ) \left ( a-b \right ) \left ( a+b\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-{\frac{{b}^{2}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( a+b\sin \left ( dx+c \right ) \right ) }}+{\frac{{a}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}+3\,{\frac{a\ln \left ( a+b\sin \left ( dx+c \right ) \right ){b}^{2}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{2\,d \left ( a+b \right ) ^{3}}}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{2\, \left ( a-b \right ) ^{3}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))^3,x)

[Out]

-1/2/d*a/(a+b)/(a-b)/(a+b*sin(d*x+c))^2-1/d/(a+b)^2/(a-b)^2/(a+b*sin(d*x+c))*a^2-1/d/(a+b)^2/(a-b)^2/(a+b*sin(
d*x+c))*b^2+1/d*a^3/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))+3/d*a/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))*b^2-1/2/d/(a+b
)^3*ln(sin(d*x+c)-1)-1/2*ln(1+sin(d*x+c))/(a-b)^3/d

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Maxima [A]  time = 1.91543, size = 308, normalized size = 2.07 \begin{align*} \frac{\frac{2 \,{\left (a^{3} + 3 \, a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac{3 \, a^{3} + a b^{2} + 2 \,{\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} +{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )} - \frac{\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{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))^3,x, algorithm="maxima")

[Out]

1/2*(2*(a^3 + 3*a*b^2)*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - (3*a^3 + a*b^2 + 2*(a^2*b
+ b^3)*sin(d*x + c))/(a^6 - 2*a^4*b^2 + a^2*b^4 + (a^4*b^2 - 2*a^2*b^4 + b^6)*sin(d*x + c)^2 + 2*(a^5*b - 2*a
^3*b^3 + a*b^5)*sin(d*x + c)) - log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - log(sin(d*x + c) - 1)/
(a^3 + 3*a^2*b + 3*a*b^2 + b^3))/d

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Fricas [B]  time = 2.6488, size = 995, normalized size = 6.68 \begin{align*} \frac{3 \, a^{5} - 2 \, a^{3} b^{2} - a b^{4} - 2 \,{\left (a^{5} + 4 \, a^{3} b^{2} + 3 \, a b^{4} -{\left (a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b + 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) +{\left (a^{5} + 3 \, a^{4} b + 4 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5} -{\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{5} - 3 \, a^{4} b + 4 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5} -{\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (a^{4} b - b^{5}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \sin \left (d x + c\right ) -{\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\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))^3,x, algorithm="fricas")

[Out]

1/2*(3*a^5 - 2*a^3*b^2 - a*b^4 - 2*(a^5 + 4*a^3*b^2 + 3*a*b^4 - (a^3*b^2 + 3*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b
+ 3*a^2*b^3)*sin(d*x + c))*log(b*sin(d*x + c) + a) + (a^5 + 3*a^4*b + 4*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4 + b^5 -
(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cos(d*x + c)^2 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*sin(d*x + c))
*log(sin(d*x + c) + 1) + (a^5 - 3*a^4*b + 4*a^3*b^2 - 4*a^2*b^3 + 3*a*b^4 - b^5 - (a^3*b^2 - 3*a^2*b^3 + 3*a*b
^4 - b^5)*cos(d*x + c)^2 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 2*
(a^4*b - b^5)*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(a^7*b - 3*a^5*b^3 +
3*a^3*b^5 - a*b^7)*d*sin(d*x + c) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*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 )^{3}}\, 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))**3,x)

[Out]

Integral(tan(c + d*x)/(a + b*sin(c + d*x))**3, x)

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Giac [A]  time = 1.58231, size = 347, normalized size = 2.33 \begin{align*} \frac{\frac{2 \,{\left (a^{3} b + 3 \, a b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{3 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 9 \, a b^{4} \sin \left (d x + c\right )^{2} + 8 \, a^{4} b \sin \left (d x + c\right ) + 18 \, a^{2} b^{3} \sin \left (d x + c\right ) - 2 \, b^{5} \sin \left (d x + c\right ) + 6 \, a^{5} + 7 \, a^{3} b^{2} - a b^{4}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{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))^3,x, algorithm="giac")

[Out]

1/2*(2*(a^3*b + 3*a*b^3)*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - log(abs(sin(d*x
+ c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (3*a
^3*b^2*sin(d*x + c)^2 + 9*a*b^4*sin(d*x + c)^2 + 8*a^4*b*sin(d*x + c) + 18*a^2*b^3*sin(d*x + c) - 2*b^5*sin(d*
x + c) + 6*a^5 + 7*a^3*b^2 - a*b^4)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(b*sin(d*x + c) + a)^2))/d