3.1.79 \(\int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [79]

Optimal. Leaf size=52 \[ -\frac {\cos (c+d x)}{a^2 d}+\frac {1}{d \left (a^2+a^2 \cos (c+d x)\right )}+\frac {2 \log (1+\cos (c+d x))}{a^2 d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3957, 2912, 12, 45} \begin {gather*} -\frac {\cos (c+d x)}{a^2 d}+\frac {1}{d \left (a^2 \cos (c+d x)+a^2\right )}+\frac {2 \log (\cos (c+d x)+1)}{a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sin (c+d x)}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac {\text {Subst}\left (\int \frac {x^2}{a^2 (-a+x)^2} \, dx,x,-a \cos (c+d x)\right )}{a d}\\ &=\frac {\text {Subst}\left (\int \frac {x^2}{(-a+x)^2} \, dx,x,-a \cos (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int \left (1+\frac {a^2}{(a-x)^2}-\frac {2 a}{a-x}\right ) \, dx,x,-a \cos (c+d x)\right )}{a^3 d}\\ &=-\frac {\cos (c+d x)}{a^2 d}+\frac {1}{d \left (a^2+a^2 \cos (c+d x)\right )}+\frac {2 \log (1+\cos (c+d x))}{a^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 64, normalized size = 1.23 \begin {gather*} -\frac {\left (-3+\cos (2 (c+d x))-8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 \cos (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{4 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((-3 + Cos[2*(c + d*x)] - 8*Log[Cos[(c + d*x)/2]] - 8*Cos[c + d*x]*Log[Cos[(c + d*x)/2]])*Sec[(c + d*x)/2
]^2)/(a^2*d)

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Maple [A]
time = 0.04, size = 51, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {-\frac {1}{\sec \left (d x +c \right )}-2 \ln \left (\sec \left (d x +c \right )\right )-\frac {1}{1+\sec \left (d x +c \right )}+2 \ln \left (1+\sec \left (d x +c \right )\right )}{a^{2} d}\)	\(51\)
default	\(\frac {-\frac {1}{\sec \left (d x +c \right )}-2 \ln \left (\sec \left (d x +c \right )\right )-\frac {1}{1+\sec \left (d x +c \right )}+2 \ln \left (1+\sec \left (d x +c \right )\right )}{a^{2} d}\)	\(51\)
norman	\(\frac {\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {5}{2 a d}}{a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}\)	\(71\)
risch	\(-\frac {2 i x}{a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}-\frac {4 i c}{a^{2} d}+\frac {2 \,{\mathrm e}^{i \left (d x +c \right )}}{a^{2} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{2}}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 46, normalized size = 0.88 \begin {gather*} \frac {\frac {1}{a^{2} \cos \left (d x + c\right ) + a^{2}} - \frac {\cos \left (d x + c\right )}{a^{2}} + \frac {2 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.07, size = 58, normalized size = 1.12 \begin {gather*} -\frac {\cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + \cos \left (d x + c\right ) - 1}{a^{2} d \cos \left (d x + c\right ) + a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sin {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(sin(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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Giac [A]
time = 0.48, size = 52, normalized size = 1.00 \begin {gather*} -\frac {\cos \left (d x + c\right )}{a^{2} d} + \frac {2 \, \log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{a^{2} d} + \frac {1}{a^{2} d {\left (\cos \left (d x + c\right ) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.92, size = 46, normalized size = 0.88 \begin {gather*} \frac {2\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{a^2\,d}-\frac {{\cos \left (c+d\,x\right )}^2-2}{a^2\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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