3.96 \(\int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

Optimal. Leaf size=126 \[ \frac {5 a \left (a^2+b^2\right ) \sin ^2(c+d x) (a \cot (c+d x)+b) (a-b \cot (c+d x))}{16 d}+\frac {5}{16} a x \left (a^2+b^2\right )^2+\frac {\sin ^6(c+d x) (a \cot (c+d x)+b)^5}{6 d}+\frac {5 a \sin ^4(c+d x) (a \cot (c+d x)+b)^3 (a-b \cot (c+d x))}{24 d} \]

[Out]

5/16*a*(a^2+b^2)^2*x+5/16*a*(a^2+b^2)*(b+a*cot(d*x+c))*(a-b*cot(d*x+c))*sin(d*x+c)^2/d+5/24*a*(b+a*cot(d*x+c))
^3*(a-b*cot(d*x+c))*sin(d*x+c)^4/d+1/6*(b+a*cot(d*x+c))^5*sin(d*x+c)^6/d

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Rubi [A]  time = 0.09, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3088, 805, 723, 203} \[ \frac {5 a \left (a^2+b^2\right ) \sin ^2(c+d x) (a \cot (c+d x)+b) (a-b \cot (c+d x))}{16 d}+\frac {5}{16} a x \left (a^2+b^2\right )^2+\frac {\sin ^6(c+d x) (a \cot (c+d x)+b)^5}{6 d}+\frac {5 a \sin ^4(c+d x) (a \cot (c+d x)+b)^3 (a-b \cot (c+d x))}{24 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(5*a*(a^2 + b^2)^2*x)/16 + (5*a*(a^2 + b^2)*(b + a*Cot[c + d*x])*(a - b*Cot[c + d*x])*Sin[c + d*x]^2)/(16*d) +
(5*a*(b + a*Cot[c + d*x])^3*(a - b*Cot[c + d*x])*Sin[c + d*x]^4)/(24*d) + ((b + a*Cot[c + d*x])^5*Sin[c + d*x
]^6)/(6*d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 723

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 805

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[(m*(c*d*f + a*e*g))/(2*a*c*(p + 1)), Int[(d + e*
x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[Simplif
y[m + 2*p + 3], 0] && LtQ[p, -1]

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps

\begin {align*} \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x (b+a x)^5}{\left (1+x^2\right )^4} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {(b+a \cot (c+d x))^5 \sin ^6(c+d x)}{6 d}-\frac {(5 a) \operatorname {Subst}\left (\int \frac {(b+a x)^4}{\left (1+x^2\right )^3} \, dx,x,\cot (c+d x)\right )}{6 d}\\ &=\frac {5 a (b+a \cot (c+d x))^3 (a-b \cot (c+d x)) \sin ^4(c+d x)}{24 d}+\frac {(b+a \cot (c+d x))^5 \sin ^6(c+d x)}{6 d}-\frac {\left (5 a \left (a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {(b+a x)^2}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{8 d}\\ &=\frac {5 a \left (a^2+b^2\right ) (b+a \cot (c+d x)) (a-b \cot (c+d x)) \sin ^2(c+d x)}{16 d}+\frac {5 a (b+a \cot (c+d x))^3 (a-b \cot (c+d x)) \sin ^4(c+d x)}{24 d}+\frac {(b+a \cot (c+d x))^5 \sin ^6(c+d x)}{6 d}-\frac {\left (5 a \left (a^2+b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{16 d}\\ &=\frac {5}{16} a \left (a^2+b^2\right )^2 x+\frac {5 a \left (a^2+b^2\right ) (b+a \cot (c+d x)) (a-b \cot (c+d x)) \sin ^2(c+d x)}{16 d}+\frac {5 a (b+a \cot (c+d x))^3 (a-b \cot (c+d x)) \sin ^4(c+d x)}{24 d}+\frac {(b+a \cot (c+d x))^5 \sin ^6(c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 188, normalized size = 1.49 \[ \frac {6 b \left (b^4-5 a^4\right ) \cos (4 (c+d x))+60 a \left (a^2+b^2\right )^2 (c+d x)+15 a \left (3 a^4+2 a^2 b^2-b^4\right ) \sin (2 (c+d x))+3 a \left (3 a^4-10 a^2 b^2-5 b^4\right ) \sin (4 (c+d x))+a \left (a^4-10 a^2 b^2+5 b^4\right ) \sin (6 (c+d x))-15 b \left (5 a^4+6 a^2 b^2+b^4\right ) \cos (2 (c+d x))-b \left (5 a^4-10 a^2 b^2+b^4\right ) \cos (6 (c+d x))}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(60*a*(a^2 + b^2)^2*(c + d*x) - 15*b*(5*a^4 + 6*a^2*b^2 + b^4)*Cos[2*(c + d*x)] + 6*b*(-5*a^4 + b^4)*Cos[4*(c
+ d*x)] - b*(5*a^4 - 10*a^2*b^2 + b^4)*Cos[6*(c + d*x)] + 15*a*(3*a^4 + 2*a^2*b^2 - b^4)*Sin[2*(c + d*x)] + 3*
a*(3*a^4 - 10*a^2*b^2 - 5*b^4)*Sin[4*(c + d*x)] + a*(a^4 - 10*a^2*b^2 + 5*b^4)*Sin[6*(c + d*x)])/(192*d)

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fricas [A]  time = 0.62, size = 182, normalized size = 1.44 \[ -\frac {24 \, b^{5} \cos \left (d x + c\right )^{2} + 8 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{6} + 24 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} - 15 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x - {\left (8 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 10 \, {\left (a^{5} + 2 \, a^{3} b^{2} - 7 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

-1/48*(24*b^5*cos(d*x + c)^2 + 8*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^6 + 24*(5*a^2*b^3 - b^5)*cos(d*x +
c)^4 - 15*(a^5 + 2*a^3*b^2 + a*b^4)*d*x - (8*(a^5 - 10*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^5 + 10*(a^5 + 2*a^3*b^2
- 7*a*b^4)*cos(d*x + c)^3 + 15*(a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.53, size = 211, normalized size = 1.67 \[ \frac {5}{16} \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} x - \frac {{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {{\left (5 \, a^{4} b - b^{5}\right )} \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {5 \, {\left (5 \, a^{4} b + 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {{\left (3 \, a^{5} - 10 \, a^{3} b^{2} - 5 \, a b^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {5 \, {\left (3 \, a^{5} + 2 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

5/16*(a^5 + 2*a^3*b^2 + a*b^4)*x - 1/192*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(6*d*x + 6*c)/d - 1/32*(5*a^4*b - b^5
)*cos(4*d*x + 4*c)/d - 5/64*(5*a^4*b + 6*a^2*b^3 + b^5)*cos(2*d*x + 2*c)/d + 1/192*(a^5 - 10*a^3*b^2 + 5*a*b^4
)*sin(6*d*x + 6*c)/d + 1/64*(3*a^5 - 10*a^3*b^2 - 5*a*b^4)*sin(4*d*x + 4*c)/d + 5/64*(3*a^5 + 2*a^3*b^2 - a*b^
4)*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.12, size = 236, normalized size = 1.87 \[ \frac {\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{6}+5 a \,b^{4} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{8}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+10 a^{2} b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )+10 a^{3} b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {5 a^{4} b \left (\cos ^{6}\left (d x +c \right )\right )}{6}+a^{5} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^5,x)

[Out]

1/d*(1/6*b^5*sin(d*x+c)^6+5*a*b^4*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*sin(d*x+c)*cos(d*x+c)^3+1/16*cos(d*x+c)*
sin(d*x+c)+1/16*d*x+1/16*c)+10*a^2*b^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+10*a^3*b^2*(-1/6*sin
(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-5/6*a^4*b*cos(d*x+c)^6+a^5
*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c))

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maxima [A]  time = 0.33, size = 187, normalized size = 1.48 \[ -\frac {160 \, a^{4} b \cos \left (d x + c\right )^{6} - 32 \, b^{5} \sin \left (d x + c\right )^{6} + {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{5} - 10 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} b^{2} + 160 \, {\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} a^{2} b^{3} + 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{4}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/192*(160*a^4*b*cos(d*x + c)^6 - 32*b^5*sin(d*x + c)^6 + (4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x
+ 4*c) - 48*sin(2*d*x + 2*c))*a^5 - 10*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^3*b^2 +
160*(2*sin(d*x + c)^6 - 3*sin(d*x + c)^4)*a^2*b^3 + 5*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x + 4*
c))*a*b^4)/d

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mupad [B]  time = 2.32, size = 472, normalized size = 3.75 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (-\frac {11\,a^5}{8}+\frac {5\,a^3\,b^2}{4}+\frac {5\,a\,b^4}{8}\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-\frac {11\,a^5}{8}+\frac {5\,a^3\,b^2}{4}+\frac {5\,a\,b^4}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {15\,a^5}{4}-\frac {65\,a^3\,b^2}{2}+\frac {95\,a\,b^4}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {15\,a^5}{4}-\frac {65\,a^3\,b^2}{2}+\frac {95\,a\,b^4}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {5\,a^5}{24}-\frac {235\,a^3\,b^2}{12}+\frac {85\,a\,b^4}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {5\,a^5}{24}-\frac {235\,a^3\,b^2}{12}+\frac {85\,a\,b^4}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {100\,a^4\,b}{3}-\frac {80\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+40\,a^2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+40\,a^2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {5\,a\,\mathrm {atan}\left (\frac {5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2+b^2\right )}^2}{8\,\left (\frac {5\,a^5}{8}+\frac {5\,a^3\,b^2}{4}+\frac {5\,a\,b^4}{8}\right )}\right )\,{\left (a^2+b^2\right )}^2}{8\,d}-\frac {5\,a\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,{\left (a^2+b^2\right )}^2}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)*(a*cos(c + d*x) + b*sin(c + d*x))^5,x)

[Out]

(tan(c/2 + (d*x)/2)^11*((5*a*b^4)/8 - (11*a^5)/8 + (5*a^3*b^2)/4) - tan(c/2 + (d*x)/2)*((5*a*b^4)/8 - (11*a^5)
/8 + (5*a^3*b^2)/4) + tan(c/2 + (d*x)/2)^5*((95*a*b^4)/4 + (15*a^5)/4 - (65*a^3*b^2)/2) - tan(c/2 + (d*x)/2)^7
*((95*a*b^4)/4 + (15*a^5)/4 - (65*a^3*b^2)/2) - tan(c/2 + (d*x)/2)^3*((85*a*b^4)/24 + (5*a^5)/24 - (235*a^3*b^
2)/12) + tan(c/2 + (d*x)/2)^9*((85*a*b^4)/24 + (5*a^5)/24 - (235*a^3*b^2)/12) + tan(c/2 + (d*x)/2)^6*((100*a^4
*b)/3 + (32*b^5)/3 - (80*a^2*b^3)/3) + 40*a^2*b^3*tan(c/2 + (d*x)/2)^4 + 40*a^2*b^3*tan(c/2 + (d*x)/2)^8 + 10*
a^4*b*tan(c/2 + (d*x)/2)^2 + 10*a^4*b*tan(c/2 + (d*x)/2)^10)/(d*(6*tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2
)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)
) + (5*a*atan((5*a*tan(c/2 + (d*x)/2)*(a^2 + b^2)^2)/(8*((5*a*b^4)/8 + (5*a^5)/8 + (5*a^3*b^2)/4)))*(a^2 + b^2
)^2)/(8*d) - (5*a*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(a^2 + b^2)^2)/(8*d)

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sympy [A]  time = 4.17, size = 609, normalized size = 4.83 \[ \begin {cases} \frac {5 a^{5} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a^{5} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a^{5} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a^{5} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a^{5} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 a^{5} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 a^{5} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {5 a^{4} b \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac {5 a^{3} b^{2} x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {15 a^{3} b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {15 a^{3} b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {5 a^{3} b^{2} x \cos ^{6}{\left (c + d x \right )}}{8} + \frac {5 a^{3} b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a^{3} b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {5 a^{3} b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} + \frac {5 a^{2} b^{3} \sin ^{6}{\left (c + d x \right )}}{6 d} + \frac {5 a^{2} b^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac {5 a b^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a b^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 a b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 a b^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 a b^{4} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {5 a b^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {5 a b^{4} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {b^{5} \sin ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{5} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Piecewise((5*a**5*x*sin(c + d*x)**6/16 + 15*a**5*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a**5*x*sin(c + d*x)
**2*cos(c + d*x)**4/16 + 5*a**5*x*cos(c + d*x)**6/16 + 5*a**5*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*a**5*sin
(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*a**5*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 5*a**4*b*cos(c + d*x)**6/(6
*d) + 5*a**3*b**2*x*sin(c + d*x)**6/8 + 15*a**3*b**2*x*sin(c + d*x)**4*cos(c + d*x)**2/8 + 15*a**3*b**2*x*sin(
c + d*x)**2*cos(c + d*x)**4/8 + 5*a**3*b**2*x*cos(c + d*x)**6/8 + 5*a**3*b**2*sin(c + d*x)**5*cos(c + d*x)/(8*
d) + 5*a**3*b**2*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) - 5*a**3*b**2*sin(c + d*x)*cos(c + d*x)**5/(8*d) + 5*a*
*2*b**3*sin(c + d*x)**6/(6*d) + 5*a**2*b**3*sin(c + d*x)**4*cos(c + d*x)**2/(2*d) + 5*a*b**4*x*sin(c + d*x)**6
/16 + 15*a*b**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a*b**4*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*a*b*
*4*x*cos(c + d*x)**6/16 + 5*a*b**4*sin(c + d*x)**5*cos(c + d*x)/(16*d) - 5*a*b**4*sin(c + d*x)**3*cos(c + d*x)
**3/(6*d) - 5*a*b**4*sin(c + d*x)*cos(c + d*x)**5/(16*d) + b**5*sin(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a*cos(c)
+ b*sin(c))**5*cos(c), True))

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