3.233 \(\int (a+a \cos (c+d x)) (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=86 \[ \frac {a (2 B+3 C) \tan (c+d x)}{3 d}+\frac {a (B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (B+C) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

1/2*a*(B+C)*arctanh(sin(d*x+c))/d+1/3*a*(2*B+3*C)*tan(d*x+c)/d+1/2*a*(B+C)*sec(d*x+c)*tan(d*x+c)/d+1/3*a*B*sec
(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.21, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3029, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {a (2 B+3 C) \tan (c+d x)}{3 d}+\frac {a (B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (B+C) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(a*(B + C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*B + 3*C)*Tan[c + d*x])/(3*d) + (a*(B + C)*Sec[c + d*x]*Tan[c +
d*x])/(2*d) + (a*B*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 8

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3029

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

Rule 3767

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]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin {align*} \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\int (a+a \cos (c+d x)) (B+C \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\int \left (a B+(a B+a C) \cos (c+d x)+a C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a B \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (3 a (B+C)+a (2 B+3 C) \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\frac {a B \sec ^2(c+d x) \tan (c+d x)}{3 d}+(a (B+C)) \int \sec ^3(c+d x) \, dx+\frac {1}{3} (a (2 B+3 C)) \int \sec ^2(c+d x) \, dx\\ &=\frac {a (B+C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a B \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} (a (B+C)) \int \sec (c+d x) \, dx-\frac {(a (2 B+3 C)) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {a (B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a (2 B+3 C) \tan (c+d x)}{3 d}+\frac {a (B+C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a B \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 56, normalized size = 0.65 \[ \frac {a \left (3 (B+C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 (B+C) \sec (c+d x)+2 B \tan ^2(c+d x)+6 (B+C)\right )\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(a*(3*(B + C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*(B + C) + 3*(B + C)*Sec[c + d*x] + 2*B*Tan[c + d*x]^2)))
/(6*d)

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fricas [A]  time = 0.48, size = 105, normalized size = 1.22 \[ \frac {3 \, {\left (B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + 2 \, B a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/12*(3*(B + C)*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(B + C)*a*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2
*(2*(2*B + 3*C)*a*cos(d*x + c)^2 + 3*(B + C)*a*cos(d*x + c) + 2*B*a)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.46, size = 154, normalized size = 1.79 \[ \frac {3 \, {\left (B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/6*(3*(B*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(B*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(
3*B*a*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*tan(1/2*d*x + 1/2*c)^5 - 4*B*a*tan(1/2*d*x + 1/2*c)^3 - 12*C*a*tan(1/2*d*
x + 1/2*c)^3 + 9*B*a*tan(1/2*d*x + 1/2*c) + 9*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

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maple [A]  time = 0.36, size = 128, normalized size = 1.49 \[ \frac {a C \tan \left (d x +c \right )}{d}+\frac {a B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a B \tan \left (d x +c \right )}{3 d}+\frac {a B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

1/d*a*C*tan(d*x+c)+1/2/d*a*B*sec(d*x+c)*tan(d*x+c)+1/2/d*a*B*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a*C*sec(d*x+c)*ta
n(d*x+c)+1/2/d*a*C*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*a*B*tan(d*x+c)+1/3/d*a*B*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.50, size = 127, normalized size = 1.48 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a - 3*B*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) - 3*C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d
*x + c) - 1)) + 12*C*a*tan(d*x + c))/d

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mupad [B]  time = 2.98, size = 126, normalized size = 1.47 \[ \frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B+C\right )}{d}-\frac {\left (B\,a+C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,B\,a}{3}-4\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,B\,a+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + a*cos(c + d*x)))/cos(c + d*x)^5,x)

[Out]

(a*atanh(tan(c/2 + (d*x)/2))*(B + C))/d - (tan(c/2 + (d*x)/2)*(3*B*a + 3*C*a) + tan(c/2 + (d*x)/2)^5*(B*a + C*
a) - tan(c/2 + (d*x)/2)^3*((4*B*a)/3 + 4*C*a))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 +
(d*x)/2)^6 - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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