3.236 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=124 \[ \frac{a \left (a^2 A+6 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 (a B+2 A b) \tan (c+d x)}{d}-\frac{b^2 (a A-2 b B) \sin (c+d x)}{2 d}+b^2 x (3 a B+A b)+\frac{a A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.339024, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2989, 3031, 3023, 2735, 3770} \[ \frac{a \left (a^2 A+6 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 (a B+2 A b) \tan (c+d x)}{d}-\frac{b^2 (a A-2 b B) \sin (c+d x)}{2 d}+b^2 x (3 a B+A b)+\frac{a A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx &=\frac{a A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \int (a+b \cos (c+d x)) \left (2 a (2 A b+a B)+\left (a^2 A+2 A b^2+4 a b B\right ) \cos (c+d x)-b (a A-2 b B) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (2 A b+a B) \tan (c+d x)}{d}+\frac{a A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-a \left (a^2 A+6 A b^2+6 a b B\right )-2 b^2 (A b+3 a B) \cos (c+d x)+b^2 (a A-2 b B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac{b^2 (a A-2 b B) \sin (c+d x)}{2 d}+\frac{a^2 (2 A b+a B) \tan (c+d x)}{d}+\frac{a A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{1}{2} \int \left (-a \left (a^2 A+6 A b^2+6 a b B\right )-2 b^2 (A b+3 a B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^2 (A b+3 a B) x-\frac{b^2 (a A-2 b B) \sin (c+d x)}{2 d}+\frac{a^2 (2 A b+a B) \tan (c+d x)}{d}+\frac{a A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a \left (a^2 A+6 A b^2+6 a b B\right )\right ) \int \sec (c+d x) \, dx\\ &=b^2 (A b+3 a B) x+\frac{a \left (a^2 A+6 A b^2+6 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{b^2 (a A-2 b B) \sin (c+d x)}{2 d}+\frac{a^2 (2 A b+a B) \tan (c+d x)}{d}+\frac{a A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 2.03464, size = 277, normalized size = 2.23 \[ \frac{-2 a \left (a^2 A+6 a b B+6 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a \left (a^2 A+6 a b B+6 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{4 a^2 (a B+3 A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 a^2 (a B+3 A b) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{a^3 A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^3 A}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+4 b^2 (c+d x) (3 a B+A b)+4 b^3 B \sin (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^2*(A*b + 3*a*B)*(c + d*x) - 2*a*(a^2*A + 6*A*b^2 + 6*a*b*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*
a*(a^2*A + 6*A*b^2 + 6*a*b*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^3*A)/(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])^2 + (4*a^2*(3*A*b + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (a^3*A)/(Cos[(c +
d*x)/2] + Sin[(c + d*x)/2])^2 + (4*a^2*(3*A*b + a*B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) +
4*b^3*B*Sin[c + d*x])/(4*d)

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Maple [A]  time = 0.085, size = 172, normalized size = 1.4 \begin{align*}{\frac{A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{3}B\tan \left ( dx+c \right ) }{d}}+3\,{\frac{A{a}^{2}b\tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}bB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{Aa{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,Ba{b}^{2}x+3\,{\frac{Ba{b}^{2}c}{d}}+A{b}^{3}x+{\frac{A{b}^{3}c}{d}}+{\frac{B{b}^{3}\sin \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))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x)

[Out]

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

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Maxima [A]  time = 1.1441, size = 228, normalized size = 1.84 \begin{align*} \frac{12 \,{\left (d x + c\right )} B a b^{2} + 4 \,{\left (d x + c\right )} A b^{3} - A 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 )} + 6 \, B a^{2} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A 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 )} + 4 \, B b^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right ) + 12 \, A a^{2} 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))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.46356, size = 401, normalized size = 3.23 \begin{align*} \frac{4 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d x \cos \left (d x + c\right )^{2} +{\left (A a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, B b^{3} \cos \left (d x + c\right )^{2} + A a^{3} + 2 \,{\left (B a^{3} + 3 \, A a^{2} 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))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/4*(4*(3*B*a*b^2 + A*b^3)*d*x*cos(d*x + c)^2 + (A*a^3 + 6*B*a^2*b + 6*A*a*b^2)*cos(d*x + c)^2*log(sin(d*x + c
) + 1) - (A*a^3 + 6*B*a^2*b + 6*A*a*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*B*b^3*cos(d*x + c)^2 + A
*a^3 + 2*(B*a^3 + 3*A*a^2*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))**3*(A+B*cos(d*x+c))*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.70623, size = 323, normalized size = 2.6 \begin{align*} \frac{\frac{4 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 2 \,{\left (3 \, B a b^{2} + A b^{3}\right )}{\left (d x + c\right )} +{\left (A a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (A a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A a^{2} 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))^3*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="giac")

[Out]

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