\(\int (a+a \sec (c+d x)) \sin (c+d x) \, dx\) [5]

Optimal result

Integrand size = 17, antiderivative size = 26 \[ \int (a+a \sec (c+d x)) \sin (c+d x) \, dx=-\frac {a \cos (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3957, 2786, 45} \[ \int (a+a \sec (c+d x)) \sin (c+d x) \, dx=-\frac {a \cos (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

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

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 2786

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 - (p + 1)/2)/(a - x)^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

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*} \text {integral}& = -\int (-a-a \cos (c+d x)) \tan (c+d x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {-a+x}{x} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (1-\frac {a}{x}\right ) \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = -\frac {a \cos (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int (a+a \sec (c+d x)) \sin (c+d x) \, dx=-\frac {a \cos (c) \cos (d x)}{d}-\frac {a \log (\cos (c+d x))}{d}+\frac {a \sin (c) \sin (d x)}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {a \left (-\frac {1}{\sec \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\)	\(24\)
default	\(\frac {a \left (-\frac {1}{\sec \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )\right )\right )}{d}\)	\(24\)
parts	\(-\frac {a \cos \left (d x +c \right )}{d}+\frac {a \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(26\)
risch	\(i a x +\frac {2 i a c}{d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {a \cos \left (d x +c \right )}{d}\)	\(45\)
parallelrisch	\(-\frac {a \left (-\ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+1+\cos \left (d x +c \right )\right )}{d}\)	\(53\)
norman	\(-\frac {2 a}{d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {a \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(78\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int (a+a \sec (c+d x)) \sin (c+d x) \, dx=-\frac {a \cos \left (d x + c\right ) + a \log \left (-\cos \left (d x + c\right )\right )}{d} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (a+a \sec (c+d x)) \sin (c+d x) \, dx=a \left (\int \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int (a+a \sec (c+d x)) \sin (c+d x) \, dx=-\frac {a \cos \left (d x + c\right ) + a \log \left (\cos \left (d x + c\right )\right )}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int (a+a \sec (c+d x)) \sin (c+d x) \, dx=-\frac {a \cos \left (d x + c\right )}{d} - \frac {a \log \left (\frac {{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int (a+a \sec (c+d x)) \sin (c+d x) \, dx=-\frac {a\,\left (\cos \left (c+d\,x\right )+\ln \left (\cos \left (c+d\,x\right )\right )\right )}{d} \]

[In]

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

[Out]

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