3.234 \(\int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=28 \[ \frac{a \sec (c+d x)}{d}+\frac{b \sec ^2(c+d x)}{2 d} \]

[Out]

(a*Sec[c + d*x])/d + (b*Sec[c + d*x]^2)/(2*d)

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Rubi [A]  time = 0.0501234, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {4377, 12, 2606, 30, 8} \[ \frac{a \sec (c+d x)}{d}+\frac{b \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[c + d*x])/d + (b*Sec[c + d*x]^2)/(2*d)

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])

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0197068, size = 28, normalized size = 1. \[ \frac{a \sec (c+d x)}{d}+\frac{b \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[c + d*x])/d + (b*Sec[c + d*x]^2)/(2*d)

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Maple [A]  time = 0.03, size = 25, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{b \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{2}}+a\sec \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02535, size = 36, normalized size = 1.29 \begin{align*} \frac{b \tan \left (d x + c\right )^{2} + \frac{2 \, a}{\cos \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.468432, size = 63, normalized size = 2.25 \begin{align*} \frac{2 \, a \cos \left (d x + c\right ) + b}{2 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sin{\left (c + d x \right )} + b \tan{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2, x)

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Giac [B]  time = 1.1582, size = 96, normalized size = 3.43 \begin{align*} \frac{2 \,{\left (a + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{d{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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