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

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Rubi [A]  time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.16, 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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fricas [A]  time = 0.66, size = 62, normalized size = 1.13 \[ -\frac {4 \, a b \cos \left (d x + c\right ) - b^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + b^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \]

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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giac [A]  time = 4.94, size = 89, normalized size = 1.62 \[ \frac {b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{d} \]

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

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maple [A]  time = 1.86, size = 63, normalized size = 1.15 \[ -\frac {2 a b \cos \left (d x +c \right )}{d}+\frac {a^{2} \sin \left (d x +c \right )}{d}-\frac {b^{2} \sin \left (d x +c \right )}{d}+\frac {b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

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 = 0.33, size = 60, normalized size = 1.09 \[ \frac {b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} - 4 \, a b \cos \left (d x + c\right ) + 2 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \]

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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mupad [B]  time = 0.49, size = 66, normalized size = 1.20 \[ \frac {2\,b^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {4\,a\,b-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2-2\,b^2\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^2*atanh(tan(c/2 + (d*x)/2)))/d - (4*a*b - tan(c/2 + (d*x)/2)*(2*a^2 - 2*b^2))/(d*(tan(c/2 + (d*x)/2)^2 +
1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{2} \sec {\left (c + d x \right )}\, dx \]

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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