3.427 \(\int \cos (c+d x) (a+b \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=28 \[ \frac{(a-b) \sin (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(b*ArcTanh[Sin[c + d*x]])/d + ((a - b)*Sin[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.0327812, 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 = {3676, 388, 206} \[ \frac{(a-b) \sin (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTanh[Sin[c + d*x]])/d + ((a - b)*Sin[c + d*x])/d

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.025269, size = 47, normalized size = 1.68 \[ \frac{a \sin (c) \cos (d x)}{d}+\frac{a \cos (c) \sin (d x)}{d}-\frac{b \sin (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTanh[Sin[c + d*x]])/d + (a*Cos[d*x]*Sin[c])/d + (a*Cos[c]*Sin[d*x])/d - (b*Sin[c + d*x])/d

________________________________________________________________________________________

Maple [A]  time = 0.037, size = 44, normalized size = 1.6 \begin{align*}{\frac{\sin \left ( dx+c \right ) a}{d}}-{\frac{\sin \left ( dx+c \right ) b}{d}}+{\frac{b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.08571, size = 62, normalized size = 2.21 \begin{align*} \frac{b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 2 \, a \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.53931, size = 115, normalized size = 4.11 \begin{align*} \frac{b \log \left (\sin \left (d x + c\right ) + 1\right ) - b \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (a - b\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(c + d*x)**2)*cos(c + d*x), x)

________________________________________________________________________________________

Giac [A]  time = 1.68142, size = 65, normalized size = 2.32 \begin{align*} \frac{b{\left (\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 2 \, \sin \left (d x + c\right )\right )} + 2 \, a \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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