\(\int \sec (c+d x) (a+b \sin (c+d x)) \, dx\) [379]

Optimal result

Integrand size = 17, antiderivative size = 43 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {(a+b) \log (1-\sin (c+d x))}{2 d}+\frac {(a-b) \log (1+\sin (c+d x))}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 43, 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 = {2747, 647, 31} \[ \int \sec (c+d x) (a+b \sin (c+d x)) \, dx=\frac {(a-b) \log (\sin (c+d x)+1)}{2 d}-\frac {(a+b) \log (1-\sin (c+d x))}{2 d} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.60 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {b \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-b \ln \left (\cos \left (d x +c \right )\right )}{d}\)	\(32\)
default	\(\frac {a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-b \ln \left (\cos \left (d x +c \right )\right )}{d}\)	\(32\)
parallelrisch	\(\frac {b \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (a -b \right )}{d}\)	\(58\)
norman	\(\frac {b \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {\left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(62\)
risch	\(i b x +\frac {2 i b c}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d}-\frac {a \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) b}{d}\)	\(90\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (a - b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a + b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (a - b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a + b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \, dx=\frac {{\left (a - b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (a + b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \]

[In]

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

[Out]

1/2*((a - b)*log(abs(sin(d*x + c) + 1)) - (a + b)*log(abs(sin(d*x + c) - 1)))/d

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.26 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \, dx=-\frac {\frac {a\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{2}-\frac {a\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{2}+\frac {b\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{2}+\frac {b\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{2}}{d} \]

[In]

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

[Out]

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