3.6.11 \(\int \sec ^2(c+d x) (a+b \tan (c+d x)) \, dx\) [511]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 28, 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 = {3567, 3852, 8} \begin {gather*} \frac {a \tan (c+d x)}{d}+\frac {b \sec ^2(c+d x)}{2 d} \end {gather*}

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 8

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

Rule 3567

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 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*} \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 \text {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*}

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Mathematica [A]
time = 0.02, size = 28, normalized size = 1.00 \begin {gather*} \frac {b \sec ^2(c+d x)}{2 d}+\frac {a \tan (c+d x)}{d} \end {gather*}

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 = 0.17, size = 25, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {\frac {b}{2 \cos \left (d x +c \right )^{2}}+a \tan \left (d x +c \right )}{d}\)	\(25\)
default	\(\frac {\frac {b}{2 \cos \left (d x +c \right )^{2}}+a \tan \left (d x +c \right )}{d}\)	\(25\)
risch	\(\frac {2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i a}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 20, normalized size = 0.71 \begin {gather*} \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}{2 \, b d} \end {gather*}

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 = 0.37, size = 30, normalized size = 1.07 \begin {gather*} \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 {gather*}

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 = 1.34, size = 34, normalized size = 1.21 \begin {gather*} \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 {gather*}

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 = 0.52, size = 25, normalized size = 0.89 \begin {gather*} \frac {b \tan \left (d x + c\right )^{2} + 2 \, a \tan \left (d x + c\right )}{2 \, d} \end {gather*}

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

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Mupad [B]
time = 3.69, size = 23, normalized size = 0.82 \begin {gather*} \frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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