3.2.83 \(\int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [183]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 109, normalized size of antiderivative = 1.00, 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 = {2800, 815} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Log[1 - Sin[c + d*x]]/((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 815

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]

Rule 2800

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]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 162, normalized size = 1.49 \begin {gather*} -\frac {a \left ((a-b)^2 \log (1-\sin (c+d x))+(a+b)^2 \log (1+\sin (c+d x))-2 \left (-a^2+b^2+\left (a^2+b^2\right ) \log (a+b \sin (c+d x))\right )\right )+b \left ((a-b)^2 \log (1-\sin (c+d x))+(a+b)^2 \log (1+\sin (c+d x))-2 \left (a^2+b^2\right ) \log (a+b \sin (c+d x))\right ) \sin (c+d x)}{2 (a-b)^2 (a+b)^2 d (a+b \sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.36, size = 98, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {-\frac {a}{\left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(98\)
default	\(\frac {-\frac {a}{\left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}}{d}\)	\(98\)
risch	\(\frac {i x}{a^{2}+2 a b +b^{2}}+\frac {i c}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {i x}{a^{2}-2 a b +b^{2}}+\frac {i c}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 i a^{2} x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {2 i a^{2} c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i b^{2} x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {2 i b^{2} c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\)	\(401\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 124, normalized size = 1.14 \begin {gather*} \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 {gather*}

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 = 0.40, size = 195, normalized size = 1.79 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

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 = 5.15, size = 156, normalized size = 1.43 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 6.76, size = 158, normalized size = 1.45 \begin {gather*} \frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^2+b^2\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,{\left (a-b\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,{\left (a+b\right )}^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2-b^2\right )\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\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 + b*sin(c + d*x))^2,x)

[Out]

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

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