3.530 \(\int (a+b \cos (c+d x)) (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=140 \[ \frac {a (4 A+5 C) \tan ^3(c+d x)}{15 d}+\frac {a (4 A+5 C) \tan (c+d x)}{5 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b (3 A+4 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {A b \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[Out]

1/8*b*(3*A+4*C)*arctanh(sin(d*x+c))/d+1/5*a*(4*A+5*C)*tan(d*x+c)/d+1/8*b*(3*A+4*C)*sec(d*x+c)*tan(d*x+c)/d+1/4
*A*b*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a*A*sec(d*x+c)^4*tan(d*x+c)/d+1/15*a*(4*A+5*C)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.20, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3032, 3021, 2748, 3767, 3768, 3770} \[ \frac {a (4 A+5 C) \tan ^3(c+d x)}{15 d}+\frac {a (4 A+5 C) \tan (c+d x)}{5 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b (3 A+4 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {A b \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(b*(3*A + 4*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(4*A + 5*C)*Tan[c + d*x])/(5*d) + (b*(3*A + 4*C)*Sec[c + d*x]
*Tan[c + d*x])/(8*d) + (A*b*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*A*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (a*
(4*A + 5*C)*Tan[c + d*x]^3)/(15*d)

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 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 3032

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

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+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \left (5 A b+a (4 A+5 C) \cos (c+d x)+5 b C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {A b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int (4 a (4 A+5 C)+5 b (3 A+4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac {A b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (b (3 A+4 C)) \int \sec ^3(c+d x) \, dx+\frac {1}{5} (a (4 A+5 C)) \int \sec ^4(c+d x) \, dx\\ &=\frac {b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} (b (3 A+4 C)) \int \sec (c+d x) \, dx-\frac {(a (4 A+5 C)) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (4 A+5 C) \tan (c+d x)}{5 d}+\frac {b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {A b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (4 A+5 C) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.80, size = 96, normalized size = 0.69 \[ \frac {\tan (c+d x) \left (8 a \left (5 (2 A+C) \tan ^2(c+d x)+3 A \tan ^4(c+d x)+15 (A+C)\right )+15 b (3 A+4 C) \sec (c+d x)+30 A b \sec ^3(c+d x)\right )+15 b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(15*b*(3*A + 4*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*b*(3*A + 4*C)*Sec[c + d*x] + 30*A*b*Sec[c + d*x]^3
+ 8*a*(15*(A + C) + 5*(2*A + C)*Tan[c + d*x]^2 + 3*A*Tan[c + d*x]^4)))/(120*d)

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fricas [A]  time = 0.92, size = 147, normalized size = 1.05 \[ \frac {15 \, {\left (3 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (4 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{4} + 15 \, {\left (3 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{2} + 30 \, A b \cos \left (d x + c\right ) + 24 \, A a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(15*(3*A + 4*C)*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*A + 4*C)*b*cos(d*x + c)^5*log(-sin(d*x +
c) + 1) + 2*(16*(4*A + 5*C)*a*cos(d*x + c)^4 + 15*(3*A + 4*C)*b*cos(d*x + c)^3 + 8*(4*A + 5*C)*a*cos(d*x + c)^
2 + 30*A*b*cos(d*x + c) + 24*A*a)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [B]  time = 0.39, size = 334, normalized size = 2.39 \[ \frac {15 \, {\left (3 \, A b + 4 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, A b + 4 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(15*(3*A*b + 4*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*A*b + 4*C*b)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) - 2*(120*A*a*tan(1/2*d*x + 1/2*c)^9 + 120*C*a*tan(1/2*d*x + 1/2*c)^9 - 75*A*b*tan(1/2*d*x + 1/2*c)^9 -
60*C*b*tan(1/2*d*x + 1/2*c)^9 - 160*A*a*tan(1/2*d*x + 1/2*c)^7 - 320*C*a*tan(1/2*d*x + 1/2*c)^7 + 30*A*b*tan(1
/2*d*x + 1/2*c)^7 + 120*C*b*tan(1/2*d*x + 1/2*c)^7 + 464*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*C*a*tan(1/2*d*x + 1/
2*c)^5 - 160*A*a*tan(1/2*d*x + 1/2*c)^3 - 320*C*a*tan(1/2*d*x + 1/2*c)^3 - 30*A*b*tan(1/2*d*x + 1/2*c)^3 - 120
*C*b*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*tan(1/2*d*x + 1/2*c) + 120*C*a*tan(1/2*d*x + 1/2*c) + 75*A*b*tan(1/2*d*x
+ 1/2*c) + 60*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 0.37, size = 192, normalized size = 1.37 \[ \frac {8 a A \tan \left (d x +c \right )}{15 d}+\frac {a A \left (\sec ^{4}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {4 a A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d}+\frac {2 a C \tan \left (d x +c \right )}{3 d}+\frac {a C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {A b \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {C b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

8/15*a*A*tan(d*x+c)/d+1/5*a*A*sec(d*x+c)^4*tan(d*x+c)/d+4/15*a*A*sec(d*x+c)^2*tan(d*x+c)/d+2/3/d*a*C*tan(d*x+c
)+1/3/d*a*C*tan(d*x+c)*sec(d*x+c)^2+1/4*A*b*sec(d*x+c)^3*tan(d*x+c)/d+3/8*A*b*sec(d*x+c)*tan(d*x+c)/d+3/8/d*A*
b*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*C*b*tan(d*x+c)*sec(d*x+c)+1/2/d*C*b*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.77, size = 175, normalized size = 1.25 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 15 \, A b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*
C*a - 15*A*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x +
c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*C*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log
(sin(d*x + c) - 1)))/d

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mupad [B]  time = 4.83, size = 233, normalized size = 1.66 \[ \frac {b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A+4\,C\right )}{4\,d}-\frac {\left (2\,A\,a-\frac {5\,A\,b}{4}+2\,C\,a-C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {A\,b}{2}-\frac {8\,A\,a}{3}-\frac {16\,C\,a}{3}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a}{3}-\frac {A\,b}{2}-\frac {16\,C\,a}{3}-2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+\frac {5\,A\,b}{4}+2\,C\,a+C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x)))/cos(c + d*x)^6,x)

[Out]

(b*atanh(tan(c/2 + (d*x)/2))*(3*A + 4*C))/(4*d) - (tan(c/2 + (d*x)/2)*(2*A*a + (5*A*b)/4 + 2*C*a + C*b) + tan(
c/2 + (d*x)/2)^5*((116*A*a)/15 + (20*C*a)/3) + tan(c/2 + (d*x)/2)^9*(2*A*a - (5*A*b)/4 + 2*C*a - C*b) - tan(c/
2 + (d*x)/2)^3*((8*A*a)/3 + (A*b)/2 + (16*C*a)/3 + 2*C*b) - tan(c/2 + (d*x)/2)^7*((8*A*a)/3 - (A*b)/2 + (16*C*
a)/3 - 2*C*b))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d
*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 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+b*cos(d*x+c))*(A+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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