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

Optimal. Leaf size=61 \[ \frac{(a C+b B) \tan (c+d x)}{d}+\frac{(a B+2 b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a B \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.196138, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {3029, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac{(a C+b B) \tan (c+d x)}{d}+\frac{(a B+2 b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a B \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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

Mathematica [A]  time = 0.0232534, size = 75, normalized size = 1.23 \[ \frac{a B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a B \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a C \tan (c+d x)}{d}+\frac{b B \tan (c+d x)}{d}+\frac{b C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 86, normalized size = 1.4 \begin{align*}{\frac{Cb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{bB\tan \left ( dx+c \right ) }{d}}+{\frac{aC\tan \left ( dx+c \right ) }{d}}+{\frac{Ba\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{Ba\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02818, size = 128, normalized size = 2.1 \begin{align*} -\frac{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 )} - 2 \, C b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, C a \tan \left (d x + c\right ) - 4 \, B b \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(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)) - 2*C*b*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 4*C*a*tan(d*x + c) - 4*B*b*tan(d*x + c))/d

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Fricas [A]  time = 1.3892, size = 247, normalized size = 4.05 \begin{align*} \frac{{\left (B a + 2 \, C b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (B a + 2 \, C b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (B a + 2 \,{\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((B*a + 2*C*b)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (B*a + 2*C*b)*cos(d*x + c)^2*log(-sin(d*x + c) + 1)
+ 2*(B*a + 2*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)

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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))*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.52171, size = 204, normalized size = 3.34 \begin{align*} \frac{{\left (B a + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (B a + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B b \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 ) + 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, 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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((B*a + 2*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (B*a + 2*C*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(
B*a*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*x + 1/2*c)^3 + B*a*tan(1/2*d*x + 1
/2*c) + 2*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d