3.32 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=150 \[ \frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac {a^4 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {(3 A-8 B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (13 A+12 B)-\frac {(3 A-B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]

[Out]

1/2*a^4*(13*A+12*B)*x+a^4*(4*A+B)*arctanh(sin(d*x+c))/d+5/2*a^4*(A+2*B)*sin(d*x+c)/d-1/3*(3*A-B)*(a^2+a^2*cos(
d*x+c))^2*sin(d*x+c)/d-1/6*(3*A-8*B)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d+a*A*(a+a*cos(d*x+c))^3*tan(d*x+c)/d

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Rubi [A]  time = 0.45, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2975, 2976, 2968, 3023, 2735, 3770} \[ \frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac {a^4 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {(3 A-B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}-\frac {(3 A-8 B) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (13 A+12 B)+\frac {a A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(13*A + 12*B)*x)/2 + (a^4*(4*A + B)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(A + 2*B)*Sin[c + d*x])/(2*d) - ((3
*A - B)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) - ((3*A - 8*B)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(
6*d) + (a*A*(a + a*Cos[c + d*x])^3*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 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2975

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^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2976

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

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 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+a \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx &=\frac {a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\int (a+a \cos (c+d x))^3 (a (4 A+B)-a (3 A-B) \cos (c+d x)) \sec (c+d x) \, dx\\ &=-\frac {(3 A-B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 d}+\frac {a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{3} \int (a+a \cos (c+d x))^2 \left (3 a^2 (4 A+B)-a^2 (3 A-8 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {(3 A-B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 d}-\frac {(3 A-8 B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{6} \int (a+a \cos (c+d x)) \left (6 a^3 (4 A+B)+15 a^3 (A+2 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {(3 A-B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 d}-\frac {(3 A-8 B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{6} \int \left (6 a^4 (4 A+B)+\left (6 a^4 (4 A+B)+15 a^4 (A+2 B)\right ) \cos (c+d x)+15 a^4 (A+2 B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}-\frac {(3 A-B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 d}-\frac {(3 A-8 B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{6} \int \left (6 a^4 (4 A+B)+3 a^4 (13 A+12 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} a^4 (13 A+12 B) x+\frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}-\frac {(3 A-B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 d}-\frac {(3 A-8 B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\left (a^4 (4 A+B)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^4 (13 A+12 B) x+\frac {a^4 (4 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}-\frac {(3 A-B) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 d}-\frac {(3 A-8 B) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 1.71, size = 312, normalized size = 2.08 \[ \frac {1}{192} a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (\frac {3 (16 A+27 B) \sin (c) \cos (d x)}{d}+\frac {3 (A+4 B) \sin (2 c) \cos (2 d x)}{d}+\frac {3 (16 A+27 B) \cos (c) \sin (d x)}{d}+\frac {3 (A+4 B) \cos (2 c) \sin (2 d x)}{d}-\frac {12 (4 A+B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {12 (4 A+B) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {12 A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {12 A \sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+78 A x+\frac {B \sin (3 c) \cos (3 d x)}{d}+\frac {B \cos (3 c) \sin (3 d x)}{d}+72 B x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(78*A*x + 72*B*x - (12*(4*A + B)*Log[Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]])/d + (12*(4*A + B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (3*(16*A + 27*B)*Cos[d*x]*Sin[c])/d
+ (3*(A + 4*B)*Cos[2*d*x]*Sin[2*c])/d + (B*Cos[3*d*x]*Sin[3*c])/d + (3*(16*A + 27*B)*Cos[c]*Sin[d*x])/d + (3*(
A + 4*B)*Cos[2*c]*Sin[2*d*x])/d + (B*Cos[3*c]*Sin[3*d*x])/d + (12*A*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Co
s[(c + d*x)/2] - Sin[(c + d*x)/2])) + (12*A*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]))))/192

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fricas [A]  time = 0.95, size = 150, normalized size = 1.00 \[ \frac {3 \, {\left (13 \, A + 12 \, B\right )} a^{4} d x \cos \left (d x + c\right ) + 3 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, B a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, A + 5 \, B\right )} a^{4} \cos \left (d x + c\right ) + 6 \, A a^{4}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(13*A + 12*B)*a^4*d*x*cos(d*x + c) + 3*(4*A + B)*a^4*cos(d*x + c)*log(sin(d*x + c) + 1) - 3*(4*A + B)*a
^4*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*B*a^4*cos(d*x + c)^3 + 3*(A + 4*B)*a^4*cos(d*x + c)^2 + 8*(3*A + 5
*B)*a^4*cos(d*x + c) + 6*A*a^4)*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 0.89, size = 226, normalized size = 1.51 \[ -\frac {\frac {12 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (13 \, A a^{4} + 12 \, B a^{4}\right )} {\left (d x + c\right )} - 6 \, {\left (4 \, A a^{4} + B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 6 \, {\left (4 \, A a^{4} + B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 76 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 54 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(13*A*a^4 + 12*B*a^4)*(d*x + c) - 6*(4*A*
a^4 + B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(4*A*a^4 + B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(
21*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 30*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 48*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 76*B*a^4
*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1/2*c) + 54*B*a^4*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 = 0.16, size = 190, normalized size = 1.27 \[ \frac {A \,a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {13 A \,a^{4} x}{2}+\frac {13 A \,a^{4} c}{2 d}+\frac {B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{4}}{3 d}+\frac {20 a^{4} B \sin \left (d x +c \right )}{3 d}+\frac {4 A \,a^{4} \sin \left (d x +c \right )}{d}+\frac {2 a^{4} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+6 a^{4} B x +\frac {6 a^{4} B c}{d}+\frac {4 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right )}{d}+\frac {a^{4} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*A*a^4*cos(d*x+c)*sin(d*x+c)+13/2*A*a^4*x+13/2/d*A*a^4*c+1/3/d*B*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3/d*a^4*B
*sin(d*x+c)+4/d*A*a^4*sin(d*x+c)+2/d*a^4*B*cos(d*x+c)*sin(d*x+c)+6*a^4*B*x+6/d*a^4*B*c+4/d*A*a^4*ln(sec(d*x+c)
+tan(d*x+c))+1/d*A*a^4*tan(d*x+c)+1/d*a^4*B*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.40, size = 187, normalized size = 1.25 \[ \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 72 \, {\left (d x + c\right )} A a^{4} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 48 \, {\left (d x + c\right )} B a^{4} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{4} \sin \left (d x + c\right ) + 72 \, B a^{4} \sin \left (d x + c\right ) + 12 \, A a^{4} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 72*(d*x + c)*A*a^4 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^
4 + 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 + 48*(d*x + c)*B*a^4 + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin
(d*x + c) - 1)) + 6*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*A*a^4*sin(d*x + c) + 72*B*a^4*s
in(d*x + c) + 12*A*a^4*tan(d*x + c))/d

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mupad [B]  time = 0.42, size = 242, normalized size = 1.61 \[ \frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {20\,B\,a^4\,\sin \left (c+d\,x\right )}{3\,d}+\frac {13\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {A\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {2\,B\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*A*a^4*sin(c + d*x))/d + (20*B*a^4*sin(c + d*x))/(3*d) + (13*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
)))/d + (8*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2)))/d + (2*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^4*sin(c + d*x))/(d*cos(c + d*x
)) + (B*a^4*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (A*a^4*cos(c + d*x)*sin(c + d*x))/(2*d) + (2*B*a^4*cos(c + d*
x)*sin(c + d*x))/d

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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+a*cos(d*x+c))**4*(A+B*cos(d*x+c))*sec(d*x+c)**2,x)

[Out]

Timed out

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