3.229 \(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=116 \[ \frac {\left (2 a^2 A+6 a b B+3 A b^2\right ) \tan (c+d x)}{3 d}+\frac {\left (a^2 B+2 a A b+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {a (a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

1/2*(2*A*a*b+B*a^2+2*B*b^2)*arctanh(sin(d*x+c))/d+1/3*(2*A*a^2+3*A*b^2+6*B*a*b)*tan(d*x+c)/d+1/2*a*(2*A*b+B*a)
*sec(d*x+c)*tan(d*x+c)/d+1/3*a^2*A*sec(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.27, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2988, 3021, 2748, 3767, 8, 3770} \[ \frac {\left (2 a^2 A+6 a b B+3 A b^2\right ) \tan (c+d x)}{3 d}+\frac {\left (a^2 B+2 a A b+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {a (a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]

[Out]

((2*a*A*b + a^2*B + 2*b^2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + ((2*a^2*A + 3*A*b^2 + 6*a*b*B)*Tan[c + d*x])/(3*d)
+ (a*(2*A*b + a*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^2*A*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 2988

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

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

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Mathematica [A]  time = 0.47, size = 92, normalized size = 0.79 \[ \frac {3 \left (a^2 B+2 a A b+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (2 \left (a^2 A \tan ^2(c+d x)+3 a^2 A+6 a b B+3 A b^2\right )+3 a (a B+2 A b) \sec (c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]

[Out]

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

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fricas [A]  time = 1.06, size = 150, normalized size = 1.29 \[ \frac {3 \, {\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{2} + 2 \, {\left (2 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/12*(3*(B*a^2 + 2*A*a*b + 2*B*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(B*a^2 + 2*A*a*b + 2*B*b^2)*cos(d
*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*A*a^2 + 2*(2*A*a^2 + 6*B*a*b + 3*A*b^2)*cos(d*x + c)^2 + 3*(B*a^2 + 2*
A*a*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [B]  time = 0.82, size = 294, normalized size = 2.53 \[ \frac {3 \, {\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/6*(3*(B*a^2 + 2*A*a*b + 2*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(B*a^2 + 2*A*a*b + 2*B*b^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*A*a*b*tan(
1/2*d*x + 1/2*c)^5 + 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^2*tan(1/2*d*x +
1/2*c)^3 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) +
3*B*a^2*tan(1/2*d*x + 1/2*c) + 6*A*a*b*tan(1/2*d*x + 1/2*c) + 12*B*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*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 = 0.13, size = 174, normalized size = 1.50 \[ \frac {2 a^{2} A \tan \left (d x +c \right )}{3 d}+\frac {a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {B \,a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {A a b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 B a b \tan \left (d x +c \right )}{d}+\frac {A \,b^{2} \tan \left (d x +c \right )}{d}+\frac {b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^4,x)

[Out]

2/3*a^2*A*tan(d*x+c)/d+1/3*a^2*A*sec(d*x+c)^2*tan(d*x+c)/d+1/2/d*B*a^2*sec(d*x+c)*tan(d*x+c)+1/2/d*B*a^2*ln(se
c(d*x+c)+tan(d*x+c))+1/d*A*a*b*sec(d*x+c)*tan(d*x+c)+1/d*A*a*b*ln(sec(d*x+c)+tan(d*x+c))+2/d*B*a*b*tan(d*x+c)+
1/d*A*b^2*tan(d*x+c)+1/d*b^2*B*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.44, size = 172, normalized size = 1.48 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} - 3 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, B a b \tan \left (d x + c\right ) + 12 \, A b^{2} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 - 3*B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x +
c) + 1) + log(sin(d*x + c) - 1)) - 6*A*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log
(sin(d*x + c) - 1)) + 6*B*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*B*a*b*tan(d*x + c) + 12*A*b
^2*tan(d*x + c))/d

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mupad [B]  time = 3.66, size = 227, normalized size = 1.96 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^2}{2}+A\,a\,b+B\,b^2\right )}{2\,B\,a^2+4\,A\,a\,b+4\,B\,b^2}\right )\,\left (B\,a^2+2\,A\,a\,b+2\,B\,b^2\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-B\,a^2-2\,A\,a\,b+4\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A\,a^2}{3}-8\,B\,a\,b-4\,A\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+B\,a^2+2\,A\,a\,b+4\,B\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^2)/cos(c + d*x)^4,x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((B*a^2)/2 + B*b^2 + A*a*b))/(2*B*a^2 + 4*B*b^2 + 4*A*a*b))*(B*a^2 + 2*B*b^2 + 2*
A*a*b))/d - (tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + B*a^2 + 2*A*a*b + 4*B*a*b) - tan(c/2 + (d*x)/2)^3*((4*A*a
^2)/3 + 4*A*b^2 + 8*B*a*b) + tan(c/2 + (d*x)/2)^5*(2*A*a^2 + 2*A*b^2 - B*a^2 - 2*A*a*b + 4*B*a*b))/(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+b*cos(d*x+c))**2*(A+B*cos(d*x+c))*sec(d*x+c)**4,x)

[Out]

Timed out

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