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

Optimal. Leaf size=91 \[ \frac{a (2 A+3 (B+C)) \tan (c+d x)}{3 d}+\frac{a (A+B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

(a*(A + B + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*A + 3*(B + C))*Tan[c + d*x])/(3*d) + (a*(A + B)*Sec[c +
d*x]*Tan[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 = 0.211266, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3031, 3021, 2748, 3767, 8, 3770} \[ \frac{a (2 A+3 (B+C)) \tan (c+d x)}{3 d}+\frac{a (A+B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (A+B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + 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)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (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, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -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 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+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \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 \left (-3 a (A+B)-a (2 A+3 (B+C)) \cos (c+d x)-3 a C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a (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}{6} \int (-2 a (2 A+3 (B+C))-3 a (A+B+2 C) \cos (c+d x)) \sec ^2(c+d x) \, dx\\ &=\frac{a (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 (A+B+2 C)) \int \sec (c+d x) \, dx+\frac{1}{3} (a (2 A+3 (B+C))) \int \sec ^2(c+d x) \, dx\\ &=\frac{a (A+B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (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{(a (2 A+3 (B+C))) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{a (A+B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (2 A+3 (B+C)) \tan (c+d x)}{3 d}+\frac{a (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 = 0.367292, size = 60, normalized size = 0.66 \[ \frac{a \left (3 (A+B+2 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 (A+B) \sec (c+d x)+6 (A+B+C)+2 A \tan ^2(c+d x)\right )\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.068, size = 160, normalized size = 1.8 \begin{align*}{\frac{2\,aA\tan \left ( dx+c \right ) }{3\,d}}+{\frac{aA \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,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}}+{\frac{aC\tan \left ( dx+c \right ) }{d}}+{\frac{aA\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{aA\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{Ba\tan \left ( dx+c \right ) }{d}}+{\frac{aC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*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*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))+1/d*a*C*tan(d*x+c)+1/2*a*A*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a*A*ln(sec(d*x+c)+tan(d*x+c))+a*B*tan(d*
x+c)/d+1/d*a*C*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.01358, size = 219, normalized size = 2.41 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a - 3 \, A 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 \, 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 )} + 6 \, C a{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a \tan \left (d x + c\right ) + 12 \, C a \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.98741, size = 312, normalized size = 3.43 \begin{align*} \frac{3 \,{\left (A + B + 2 \, C\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (A + B + 2 \, 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 \, A + 3 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \, A a\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+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/12*(3*(A + B + 2*C)*a*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(A + B + 2*C)*a*cos(d*x + c)^3*log(-sin(d*x +
c) + 1) + 2*(2*(2*A + 3*B + 3*C)*a*cos(d*x + c)^2 + 3*(A + B)*a*cos(d*x + c) + 2*A*a)*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+a*cos(d*x+c))*(A+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 = 1.2607, size = 277, normalized size = 3.04 \begin{align*} \frac{3 \,{\left (A a + B a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (A a + B a + 2 \, 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 \, 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} + 6 \, C a \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 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 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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