3.791 \(\int (a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=145 \[ \frac{a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \tan (c+d x)}{3 d}+\frac{\left (3 a^2 b B+a^3 C+6 a b^2 C+2 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^3 C x \]

[Out]

b^3*C*x + ((3*a^2*b*B + 2*b^3*B + a^3*C + 6*a*b^2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*a^2*B + 8*b^2*B + 9*
a*b*C)*Tan[c + d*x])/(3*d) + (a^2*(5*b*B + 3*a*C)*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (a*B*(a + b*Cos[c + d*x])
^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.429343, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3029, 2989, 3031, 3021, 2735, 3770} \[ \frac{a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \tan (c+d x)}{3 d}+\frac{\left (3 a^2 b B+a^3 C+6 a b^2 C+2 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 (3 a C+5 b B) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{a B \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^3 C x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

b^3*C*x + ((3*a^2*b*B + 2*b^3*B + a^3*C + 6*a*b^2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*a^2*B + 8*b^2*B + 9*
a*b*C)*Tan[c + d*x])/(3*d) + (a^2*(5*b*B + 3*a*C)*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (a*B*(a + b*Cos[c + d*x])
^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 2989

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

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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\int (a+b \cos (c+d x))^3 (B+C \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac{a B (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int (a+b \cos (c+d x)) \left (a (5 b B+3 a C)+\left (2 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)+3 b^2 C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a^2 (5 b B+3 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-2 a \left (2 a^2 B+8 b^2 B+9 a b C\right )-3 \left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right ) \cos (c+d x)-6 b^3 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a \left (2 a^2 B+8 b^2 B+9 a b C\right ) \tan (c+d x)}{3 d}+\frac{a^2 (5 b B+3 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-3 \left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right )-6 b^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^3 C x+\frac{a \left (2 a^2 B+8 b^2 B+9 a b C\right ) \tan (c+d x)}{3 d}+\frac{a^2 (5 b B+3 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{2} \left (-3 a^2 b B-2 b^3 B-a^3 C-6 a b^2 C\right ) \int \sec (c+d x) \, dx\\ &=b^3 C x+\frac{\left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a \left (2 a^2 B+8 b^2 B+9 a b C\right ) \tan (c+d x)}{3 d}+\frac{a^2 (5 b B+3 a C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{a B (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.56333, size = 108, normalized size = 0.74 \[ \frac{3 \left (3 a^2 b B+a^3 C+6 a b^2 C+2 b^3 B\right ) \tanh ^{-1}(\sin (c+d x))+3 a \tan (c+d x) \left (2 a^2 B+a (a C+3 b B) \sec (c+d x)+6 a b C+6 b^2 B\right )+2 a^3 B \tan ^3(c+d x)+6 b^3 C d x}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

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

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Maple [A]  time = 0.06, size = 223, normalized size = 1.5 \begin{align*}{b}^{3}Cx+{\frac{C{b}^{3}c}{d}}+{\frac{{b}^{3}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{a{b}^{2}B\tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}bC\tan \left ( dx+c \right ) }{d}}+{\frac{3\,{a}^{2}bB\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}bB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,{a}^{3}B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

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

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Maxima [A]  time = 1.03589, size = 292, normalized size = 2.01 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 12 \,{\left (d x + c\right )} C b^{3} - 3 \, C a^{3}{\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 )} - 9 \, B a^{2} 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 )} + 18 \, C a b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, C a^{2} b \tan \left (d x + c\right ) + 36 \, B a b^{2} \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))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 12*(d*x + c)*C*b^3 - 3*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2
- 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 9*B*a^2*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(
sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 18*C*a*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*
b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*C*a^2*b*tan(d*x + c) + 36*B*a*b^2*tan(d*x + c))/d

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Fricas [A]  time = 1.54638, size = 458, normalized size = 3.16 \begin{align*} \frac{12 \, C b^{3} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, B a^{3} + 2 \,{\left (2 \, B a^{3} + 9 \, C a^{2} b + 9 \, B a b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (C a^{3} + 3 \, B a^{2} 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))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/12*(12*C*b^3*d*x*cos(d*x + c)^3 + 3*(C*a^3 + 3*B*a^2*b + 6*C*a*b^2 + 2*B*b^3)*cos(d*x + c)^3*log(sin(d*x + c
) + 1) - 3*(C*a^3 + 3*B*a^2*b + 6*C*a*b^2 + 2*B*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*B*a^3 + 2*(2
*B*a^3 + 9*C*a^2*b + 9*B*a*b^2)*cos(d*x + c)^2 + 3*(C*a^3 + 3*B*a^2*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))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.54403, size = 454, normalized size = 3.13 \begin{align*} \frac{6 \,{\left (d x + c\right )} C b^{3} + 3 \,{\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, B 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*C*b^3 + 3*(C*a^3 + 3*B*a^2*b + 6*C*a*b^2 + 2*B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(C
*a^3 + 3*B*a^2*b + 6*C*a*b^2 + 2*B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*B*a^3*tan(1/2*d*x + 1/2*c)^5
- 3*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*
B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*B*a*b
^2*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^3*tan(1/2*d*x + 1/2*c) + 3*C*a^3*tan(1/2*d*x + 1/2*c) + 9*B*a^2*b*tan(1/2*d*
x + 1/2*c) + 18*C*a^2*b*tan(1/2*d*x + 1/2*c) + 18*B*a*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3
)/d