3.233 \(\int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=25 \[ \frac {b \sec (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4377, 12, 2606, 8, 3475} \[ \frac {b \sec (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 12

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4377

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Cos[c*(a +
b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[
Cos[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.00 \[ \frac {b \sec (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 34, normalized size = 1.36 \[ -\frac {a \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - b}{d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.29, size = 107, normalized size = 4.28 \[ \frac {a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {a + 2 \, b + \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 25, normalized size = 1.00 \[ \frac {b \sec \left (d x +c \right )}{d}+\frac {a \ln \left (\sec \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 32, normalized size = 1.28 \[ -\frac {a \log \left (-\sin \left (d x + c\right )^{2} + 1\right ) - \frac {2 \, b}{\cos \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.66, size = 40, normalized size = 1.60 \[ \frac {2\,a\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d}-\frac {2\,b}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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