\(\int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx\) [384]

Optimal result

Integrand size = 19, antiderivative size = 23 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \]

[Out]

b*sec(d*x+c)/d+a*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2748, 3852, 8} \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \tan (c+d x)}{d}+\frac {b \sec (c+d x)}{d} \]

[In]

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

[Out]

(b*Sec[c + d*x])/d + (a*Tan[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \]

[In]

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

[Out]

(b*Sec[c + d*x])/d + (a*Tan[c + d*x])/d

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right ) a +\frac {b}{\cos \left (d x +c \right )}}{d}\)	\(24\)
default	\(\frac {\tan \left (d x +c \right ) a +\frac {b}{\cos \left (d x +c \right )}}{d}\)	\(24\)
risch	\(\frac {2 i a +2 b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )}\)	\(35\)
parallelrisch	\(\frac {-2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(36\)
norman	\(\frac {-\frac {2 b}{d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(88\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \sin \left (d x + c\right ) + b}{d \cos \left (d x + c\right )} \]

[In]

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

[Out]

(a*sin(d*x + c) + b)/(d*cos(d*x + c))

Sympy [F]

\[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \tan \left (d x + c\right ) + \frac {b}{\cos \left (d x + c\right )}}{d} \]

[In]

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

[Out]

(a*tan(d*x + c) + b/cos(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.69 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b+a\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

(b + a*sin(c + d*x))/(d*cos(c + d*x))