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

Optimal. Leaf size=93 \[ \frac{(2 a A+3 b B) \tan (c+d x)}{3 d}+\frac{(a B+A b) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.163321, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{(2 a A+3 b B) \tan (c+d x)}{3 d}+\frac{(a B+A b) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(a B+A b) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.263559, size = 67, normalized size = 0.72 \[ \frac{3 (a B+A b) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 (a B+A b) \sec (c+d x)+2 a A \tan ^2(c+d x)+6 a A+6 b B\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.073, size = 128, normalized size = 1.4 \begin{align*}{\frac{2\,A\tan \left ( dx+c \right ) a}{3\,d}}+{\frac{aA \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{aB\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{aB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{Ab\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{Ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{Bb\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03363, size = 171, normalized size = 1.84 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A 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 \, 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 )} + 12 \, B b \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(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 - 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*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)) + 12*B*b*tan(d*x + c))/d

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Fricas [A]  time = 1.40996, size = 298, normalized size = 3.2 \begin{align*} \frac{3 \,{\left (B a + A b\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (B a + A b\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (2 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )^{2} + 2 \, A a + 3 \,{\left (B 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.35696, size = 284, normalized size = 3.05 \begin{align*} \frac{3 \,{\left (B a + A b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (B a + A b\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 \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B 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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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