3.235 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=131 \[ -\frac {b \left (2 a^2 A-3 a b B-A b^2\right ) \sin (c+d x)}{d}+\frac {1}{2} b x \left (6 a^2 B+6 a A b+b^2 B\right )+\frac {a^2 (a B+3 A b) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b^2 (2 a A-b B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 131, normalized size of antiderivative = 1.00, 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, 3033, 3023, 2735, 3770} \[ -\frac {b \left (2 a^2 A-3 a b B-A b^2\right ) \sin (c+d x)}{d}+\frac {1}{2} b x \left (6 a^2 B+6 a A b+b^2 B\right )+\frac {a^2 (a B+3 A b) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b^2 (2 a A-b B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a A \tan (c+d x) (a+b \cos (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

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

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Mathematica [A]  time = 0.68, size = 217, normalized size = 1.66 \[ \frac {\frac {4 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+2 b (c+d x) \left (6 a^2 B+6 a A b+b^2 B\right )-4 a^2 (a B+3 A b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 (a B+3 A b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 (3 a B+A b) \sin (c+d x)+b^3 B \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.97, size = 152, normalized size = 1.16 \[ \frac {{\left (6 \, B a^{2} b + 6 \, A a b^{2} + B b^{3}\right )} d x \cos \left (d x + c\right ) + {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (B b^{3} \cos \left (d x + c\right )^{2} + 2 \, A a^{3} + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

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)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.29, size = 234, normalized size = 1.79 \[ -\frac {\frac {4 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (6 \, B a^{2} b + 6 \, A a b^{2} + B b^{3}\right )} {\left (d x + c\right )} - 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

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)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.13, size = 168, normalized size = 1.28 \[ \frac {A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+3 a^{2} b B x +\frac {3 B \,a^{2} b c}{d}+3 A \,b^{2} a x +\frac {3 A a \,b^{2} c}{d}+\frac {3 B \,b^{2} a \sin \left (d x +c \right )}{d}+\frac {A \,b^{3} \sin \left (d x +c \right )}{d}+\frac {b^{3} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {b^{3} B x}{2}+\frac {b^{3} B c}{2 d} \]

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)^2,x)

[Out]

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

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maxima [A]  time = 0.32, size = 144, normalized size = 1.10 \[ \frac {12 \, {\left (d x + c\right )} B a^{2} b + 12 \, {\left (d x + c\right )} A a b^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} + 2 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a b^{2} \sin \left (d x + c\right ) + 4 \, A b^{3} \sin \left (d x + c\right ) + 4 \, A a^{3} \tan \left (d x + c\right )}{4 \, d} \]

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)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.35, size = 236, normalized size = 1.80 \[ \frac {B\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,B\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}-A\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+A\,a^3\,\sin \left (c+d\,x\right )+\frac {B\,b^3\,\sin \left (c+d\,x\right )}{8}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^3)/cos(c + d*x)^2,x)

[Out]

(B*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - B*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i
+ 6*A*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - A*a^2*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x
)/2))*6i + 6*B*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + ((A*b^3*sin(2*c + 2*d*x))/2 + (B*b^3*sin
(3*c + 3*d*x))/8 + A*a^3*sin(c + d*x) + (B*b^3*sin(c + d*x))/8 + (3*B*a*b^2*sin(2*c + 2*d*x))/2)/(d*cos(c + d*
x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{3} \sec ^{2}{\left (c + d x \right )}\, dx \]

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)**2,x)

[Out]

Integral((A + B*cos(c + d*x))*(a + b*cos(c + d*x))**3*sec(c + d*x)**2, x)

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