3.4.10 \(\int \cos ^2(c+d x) (a+a \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [310]

Optimal. Leaf size=32 \[ a (B+C) x+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \sin (c+d x)}{d} \]

[Out]

a*(B+C)*x+a*C*arctanh(sin(d*x+c))/d+a*B*sin(d*x+c)/d

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Rubi [A]
time = 0.06, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {4157, 4081, 3855} \begin {gather*} \frac {a B \sin (c+d x)}{d}+a x (B+C)+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a*(B + C)*x + (a*C*ArcTanh[Sin[c + d*x]])/d + (a*B*Sin[c + d*x])/d

Rule 3855

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

Rule 4081

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 4157

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 46, normalized size = 1.44 \begin {gather*} a B x+a C x+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \cos (d x) \sin (c)}{d}+\frac {a B \cos (c) \sin (d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a*B*x + a*C*x + (a*C*ArcTanh[Sin[c + d*x]])/d + (a*B*Cos[d*x]*Sin[c])/d + (a*B*Cos[c]*Sin[d*x])/d

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Maple [A]
time = 0.72, size = 48, normalized size = 1.50

method	result	size
derivativedivides	\(\frac {B a \left (d x +c \right )+a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \sin \left (d x +c \right )+a C \left (d x +c \right )}{d}\)	\(48\)
default	\(\frac {B a \left (d x +c \right )+a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \sin \left (d x +c \right )+a C \left (d x +c \right )}{d}\)	\(48\)
risch	\(a B x +a x C -\frac {i B a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i B a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(83\)
norman	\(\frac {\left (B a +a C \right ) x +\left (-2 B a -2 a C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (B a +a C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 B a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 B a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 B a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 B a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(B*a*(d*x+c)+a*C*ln(sec(d*x+c)+tan(d*x+c))+B*a*sin(d*x+c)+a*C*(d*x+c))

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Maxima [A]
time = 0.28, size = 58, normalized size = 1.81 \begin {gather*} \frac {2 \, {\left (d x + c\right )} B a + 2 \, {\left (d x + c\right )} C a + C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a \sin \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.92, size = 51, normalized size = 1.59 \begin {gather*} \frac {2 \, {\left (B + C\right )} a d x + C a \log \left (\sin \left (d x + c\right ) + 1\right ) - C a \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a \sin \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

a*(Integral(B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(B*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(C*c
os(c + d*x)**2*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**3, x))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (32) = 64\).
time = 0.45, size = 79, normalized size = 2.47 \begin {gather*} \frac {C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (B a + C a\right )} {\left (d x + c\right )} + \frac {2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.02, size = 100, normalized size = 3.12 \begin {gather*} \frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {2\,B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x)),x)

[Out]

(B*a*sin(c + d*x))/d + (2*B*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a*atan(sin(c/2 + (d*x)/2)/
cos(c/2 + (d*x)/2)))/d + (2*C*a*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d

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