3.9 \(\int \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=23 \[ \frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2669, 3767, 8} \[ \frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*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 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
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 3767

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*} \int \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx &=\frac {a \sec (c+d x)}{d}+a \int \sec ^2(c+d x) \, dx\\ &=\frac {a \sec (c+d x)}{d}-\frac {a \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac {a \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \[ \frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 40, normalized size = 1.74 \[ \frac {a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.47, size = 19, normalized size = 0.83 \[ -\frac {2 \, a}{d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a/(d*(tan(1/2*d*x + 1/2*c) - 1))

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maple [A]  time = 0.14, size = 24, normalized size = 1.04 \[ \frac {\frac {a}{\cos \left (d x +c \right )}+a \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 23, normalized size = 1.00 \[ \frac {a \tan \left (d x + c\right ) + \frac {a}{\cos \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*tan(d*x + c) + a/cos(d*x + c))/d

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mupad [B]  time = 4.69, size = 19, normalized size = 0.83 \[ -\frac {2\,a}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a)/(d*(tan(c/2 + (d*x)/2) - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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