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

Optimal. Leaf size=86 \[ \frac{a (2 A+3 C) \tan (c+d x)}{3 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac{b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A b \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

(b*(A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*A + 3*C)*Tan[c + d*x])/(3*d) + (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 = 0.172846, antiderivative size = 86, normalized size of antiderivative = 1., 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 = {3032, 3021, 2748, 3767, 8, 3770} \[ \frac{a (2 A+3 C) \tan (c+d x)}{3 d}+\frac{a A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac{b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A b \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*A + 3*C)*Tan[c + d*x])/(3*d) + (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 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 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 (A+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 b+a (2 A+3 C) \cos (c+d x)+3 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{a A \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{6} \int (2 a (2 A+3 C)+3 b (A+2 C) \cos (c+d x)) \sec ^2(c+d x) \, dx\\ &=\frac{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} (b (A+2 C)) \int \sec (c+d x) \, dx+\frac{1}{3} (a (2 A+3 C)) \int \sec ^2(c+d x) \, dx\\ &=\frac{b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{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 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (2 A+3 C) \tan (c+d x)}{3 d}+\frac{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 = 0.323874, size = 59, normalized size = 0.69 \[ \frac{\tan (c+d x) \left (2 a A \tan ^2(c+d x)+6 a (A+C)+3 A b \sec (c+d x)\right )+3 b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.048, size = 108, normalized size = 1.3 \begin{align*}{\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{Cb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\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{aC\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+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

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

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Maxima [A]  time = 1.01852, size = 144, normalized size = 1.67 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a - 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 )} + 6 \, C b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\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+b*cos(d*x+c))*(A+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*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*C*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*a*tan(d*x + c))/d

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Fricas [A]  time = 1.51209, size = 285, normalized size = 3.31 \begin{align*} \frac{3 \,{\left (A + 2 \, C\right )} b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (A + 2 \, C\right )} b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (2 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, A b \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+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/12*(3*(A + 2*C)*b*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(A + 2*C)*b*cos(d*x + c)^3*log(-sin(d*x + c) + 1)
+ 2*(2*(2*A + 3*C)*a*cos(d*x + c)^2 + 3*A*b*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+b*cos(d*x+c))*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.58034, size = 248, normalized size = 2.88 \begin{align*} \frac{3 \,{\left (A b + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (A b + 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 (6 \, A 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} - 3 \, A 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 \, C a \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 ) + 6 \, C 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 )\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+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

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