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

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

[Out]

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

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Rubi [A]  time = 0.0284686, antiderivative size = 28, normalized size of antiderivative = 1., 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 = {3486, 3767, 8} \[ \frac{a \tan (c+d x)}{d}+\frac{b \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || 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]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0133805, size = 28, normalized size = 1. \[ \frac{a \tan (c+d x)}{d}+\frac{b \sec ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 25, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{b}{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+a\tan \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+b*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.5035, size = 27, normalized size = 0.96 \begin{align*} \frac{{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}{2 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*tan(d*x + c) + a)^2/(b*d)

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Fricas [A]  time = 1.76126, size = 81, normalized size = 2.89 \begin{align*} \frac{2 \, a \cos \left (d x + c\right ) \sin \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+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.19389, size = 34, normalized size = 1.21 \begin{align*} \begin{cases} \frac{a \tan{\left (c + d x \right )} + \frac{b \tan ^{2}{\left (c + d x \right )}}{2}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan{\left (c \right )}\right ) \sec ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((a*tan(c + d*x) + b*tan(c + d*x)**2/2)/d, Ne(d, 0)), (x*(a + b*tan(c))*sec(c)**2, True))

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Giac [A]  time = 1.2868, size = 34, normalized size = 1.21 \begin{align*} \frac{b \tan \left (d x + c\right )^{2} + 2 \, a \tan \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+b*tan(d*x+c)),x, algorithm="giac")

[Out]

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