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

Optimal. Leaf size=112 \[ \frac {\left (a^2 (2 A+3 C)+2 A b^2\right ) \tan (c+d x)}{3 d}+\frac {a b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a A b \tan (c+d x) \sec (c+d x)}{3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^2 C x \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3048, 3031, 3021, 2735, 3770} \[ \frac {\left (a^2 (2 A+3 C)+2 A b^2\right ) \tan (c+d x)}{3 d}+\frac {a b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a A b \tan (c+d x) \sec (c+d x)}{3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^2}{3 d}+b^2 C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^2*C*x + (a*b*(A + 2*C)*ArcTanh[Sin[c + d*x]])/d + ((2*A*b^2 + a^2*(2*A + 3*C))*Tan[c + d*x])/(3*d) + (a*A*b*
Sec[c + d*x]*Tan[c + d*x])/(3*d) + (A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 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 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 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(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
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*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] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -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))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac {A (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 (2 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 A b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-2 \left (2 A b^2+a^2 (2 A+3 C)\right )-6 a b (A+2 C) \cos (c+d x)-6 b^2 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {\left (2 A b^2+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac {a A b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-6 a b (A+2 C)-6 b^2 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=b^2 C x+\frac {\left (2 A b^2+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac {a A b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+(a b (A+2 C)) \int \sec (c+d x) \, dx\\ &=b^2 C x+\frac {a b (A+2 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {\left (2 A b^2+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}+\frac {a A b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 76, normalized size = 0.68 \[ \frac {3 \tan (c+d x) \left (a^2 (A+C)+a A b \sec (c+d x)+A b^2\right )+a^2 A \tan ^3(c+d x)+3 a b (A+2 C) \tanh ^{-1}(\sin (c+d x))+3 b^2 C d x}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.96, size = 136, normalized size = 1.21 \[ \frac {6 \, C b^{2} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (A + 2 \, C\right )} a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + 2 \, C\right )} a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, A a b \cos \left (d x + c\right ) + A a^{2} + {\left ({\left (2 \, A + 3 \, C\right )} a^{2} + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.48, size = 262, normalized size = 2.34 \[ \frac {3 \, {\left (d x + c\right )} C b^{2} + 3 \, {\left (A a b + 2 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A a b + 2 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.33, size = 145, normalized size = 1.29 \[ \frac {2 a^{2} A \tan \left (d x +c \right )}{3 d}+\frac {a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a^{2} C \tan \left (d x +c \right )}{d}+\frac {a A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{2} \tan \left (d x +c \right )}{d}+b^{2} C x +\frac {b^{2} C c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 136, normalized size = 1.21 \[ \frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 6 \, {\left (d x + c\right )} C b^{2} - 3 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{2} \tan \left (d x + c\right ) + 6 \, A b^{2} \tan \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.47, size = 209, normalized size = 1.87 \[ \frac {2\,C\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a\,b\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}-\frac {A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^2)/cos(c + d*x)^4,x)

[Out]

(2*C*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*A*a^2*sin(c + d*x))/(3*d*cos(c + d*x)) + (A*a^2*s
in(c + d*x))/(3*d*cos(c + d*x)^3) + (A*b^2*sin(c + d*x))/(d*cos(c + d*x)) + (C*a^2*sin(c + d*x))/(d*cos(c + d*
x)) - (A*a*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d - (C*a*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(
c/2 + (d*x)/2))*4i)/d + (A*a*b*sin(c + d*x))/(d*cos(c + d*x)^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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