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

Optimal. Leaf size=55 \[ \frac{a^2 \sin (c+d x)}{d}-\frac{2 a b \cos (c+d x)}{d}-\frac{b^2 \sin (c+d x)}{d}+\frac{b^2 \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(b^2*ArcTanh[Sin[c + d*x]])/d - (2*a*b*Cos[c + d*x])/d + (a^2*Sin[c + d*x])/d - (b^2*Sin[c + d*x])/d

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Rubi [A]  time = 0.07073, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3090, 2637, 2638, 2592, 321, 206} \[ \frac{a^2 \sin (c+d x)}{d}-\frac{2 a b \cos (c+d x)}{d}-\frac{b^2 \sin (c+d x)}{d}+\frac{b^2 \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*ArcTanh[Sin[c + d*x]])/d - (2*a*b*Cos[c + d*x])/d + (a^2*Sin[c + d*x])/d - (b^2*Sin[c + d*x])/d

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

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

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

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

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

Mathematica [A]  time = 0.141852, size = 84, normalized size = 1.53 \[ \frac{\left (a^2-b^2\right ) \sin (c+d x)-2 a b \cos (c+d x)+b^2 \left (\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b*Cos[c + d*x] + b^2*(-Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]]) + (a^2 - b^2)*Sin[c + d*x])/d

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Maple [A]  time = 0.085, size = 63, normalized size = 1.2 \begin{align*} -2\,{\frac{ab\cos \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{d}}-{\frac{{b}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}\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(sec(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.18698, size = 81, normalized size = 1.47 \begin{align*} \frac{b^{2}{\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 )} - 4 \, a b \cos \left (d x + c\right ) + 2 \, a^{2} \sin \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)*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2*(b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c)) - 4*a*b*cos(d*x + c) + 2*a^2*sin(d*x
+ c))/d

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Fricas [A]  time = 0.505515, size = 155, normalized size = 2.82 \begin{align*} -\frac{4 \, a b \cos \left (d x + c\right ) - b^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + b^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{2} - b^{2}\right )} \sin \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)*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(4*a*b*cos(d*x + c) - b^2*log(sin(d*x + c) + 1) + b^2*log(-sin(d*x + c) + 1) - 2*(a^2 - b^2)*sin(d*x + c)
)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17296, size = 120, normalized size = 2.18 \begin{align*} \frac{b^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - b^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a b\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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