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

Optimal. Leaf size=170 \[ -\frac {\sin ^4(c+d x) \left (a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)+b \left (5 a^4-10 a^2 b^2+b^4\right )\right )}{4 d}+\frac {\sin ^2(c+d x) \left (4 b \left (5 a^4-b^4\right )+5 a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \cot (c+d x)\right )}{8 d}+\frac {1}{8} a x \left (3 a^4+10 a^2 b^2+15 b^4\right )-\frac {b^5 \log (\sin (c+d x))}{d}+\frac {b^5 \log (\tan (c+d x))}{d} \]

[Out]

1/8*a*(3*a^4+10*a^2*b^2+15*b^4)*x-b^5*ln(sin(d*x+c))/d+b^5*ln(tan(d*x+c))/d+1/8*(4*b*(5*a^4-b^4)+5*a*(a^2-3*b^
2)*(a^2+b^2)*cot(d*x+c))*sin(d*x+c)^2/d-1/4*(b*(5*a^4-10*a^2*b^2+b^4)+a*(a^4-10*a^2*b^2+5*b^4)*cot(d*x+c))*sin
(d*x+c)^4/d

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Rubi [A]  time = 0.22, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3088, 1805, 801, 635, 203, 260} \[ -\frac {\sin ^4(c+d x) \left (a \left (-10 a^2 b^2+a^4+5 b^4\right ) \cot (c+d x)+b \left (-10 a^2 b^2+5 a^4+b^4\right )\right )}{4 d}+\frac {\sin ^2(c+d x) \left (5 a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \cot (c+d x)+4 b \left (5 a^4-b^4\right )\right )}{8 d}+\frac {1}{8} a x \left (10 a^2 b^2+3 a^4+15 b^4\right )-\frac {b^5 \log (\sin (c+d x))}{d}+\frac {b^5 \log (\tan (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*a^4 + 10*a^2*b^2 + 15*b^4)*x)/8 - (b^5*Log[Sin[c + d*x]])/d + (b^5*Log[Tan[c + d*x]])/d + ((4*b*(5*a^4 -
b^4) + 5*a*(a^2 - 3*b^2)*(a^2 + b^2)*Cot[c + d*x])*Sin[c + d*x]^2)/(8*d) - ((b*(5*a^4 - 10*a^2*b^2 + b^4) + a
*(a^4 - 10*a^2*b^2 + 5*b^4)*Cot[c + d*x])*Sin[c + d*x]^4)/(4*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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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 \sec (c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(b+a x)^5}{x \left (1+x^2\right )^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^4(c+d x)}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {-4 b^5+a \left (a^4-10 a^2 b^2-15 b^4\right ) x-20 a^4 b x^2-4 a^5 x^3}{x \left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{4 d}\\ &=\frac {\left (4 b \left (5 a^4-b^4\right )+5 a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{8 d}-\frac {\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^4(c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {8 b^5+a \left (3 a^4+10 a^2 b^2+15 b^4\right ) x}{x \left (1+x^2\right )} \, dx,x,\cot (c+d x)\right )}{8 d}\\ &=\frac {\left (4 b \left (5 a^4-b^4\right )+5 a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{8 d}-\frac {\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^4(c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \left (\frac {8 b^5}{x}+\frac {3 a^5+10 a^3 b^2+15 a b^4-8 b^5 x}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{8 d}\\ &=\frac {b^5 \log (\tan (c+d x))}{d}+\frac {\left (4 b \left (5 a^4-b^4\right )+5 a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{8 d}-\frac {\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^4(c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {3 a^5+10 a^3 b^2+15 a b^4-8 b^5 x}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 d}\\ &=\frac {b^5 \log (\tan (c+d x))}{d}+\frac {\left (4 b \left (5 a^4-b^4\right )+5 a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{8 d}-\frac {\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^4(c+d x)}{4 d}+\frac {b^5 \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}-\frac {\left (a \left (3 a^4+10 a^2 b^2+15 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 d}\\ &=\frac {1}{8} a \left (3 a^4+10 a^2 b^2+15 b^4\right ) x-\frac {b^5 \log (\sin (c+d x))}{d}+\frac {b^5 \log (\tan (c+d x))}{d}+\frac {\left (4 b \left (5 a^4-b^4\right )+5 a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{8 d}-\frac {\left (b \left (5 a^4-10 a^2 b^2+b^4\right )+a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)\right ) \sin ^4(c+d x)}{4 d}\\ \end {align*}

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Mathematica [B]  time = 6.44, size = 711, normalized size = 4.18 \[ \frac {b^5 \left (\frac {\cos ^4(c+d x) (a+b \tan (c+d x))^6 \left (a b \tan (c+d x)+b^2\right )}{4 b^6 \left (a^2+b^2\right )}-\frac {\frac {\cos ^2(c+d x) (a+b \tan (c+d x))^6 \left (b \left (a \left (2 b^2-3 a^2\right )+3 a b^2\right ) \tan (c+d x)-3 a^2 b^2+b^2 \left (2 b^2-3 a^2\right )\right )}{2 b^4 \left (a^2+b^2\right )}-\frac {\left (3 a^4-29 a^2 b^2+5 a^2 \left (3 a^2-5 b^2\right )+8 b^4\right ) \left (\frac {1}{2} b^2 \left (10 a^2-b^2\right ) \tan ^2(c+d x)+5 a b \left (2 a^2-b^2\right ) \tan (c+d x)+\frac {1}{2} \left (5 a^4-10 a^2 b^2+\frac {a^5-10 a^3 b^2+5 a b^4}{\sqrt {-b^2}}+b^4\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\frac {1}{2} \left (5 a^4-10 a^2 b^2-\frac {a^5-10 a^3 b^2+5 a b^4}{\sqrt {-b^2}}+b^4\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {5}{3} a b^3 \tan ^3(c+d x)+\frac {1}{4} b^4 \tan ^4(c+d x)\right )-5 a \left (3 a^2-5 b^2\right ) \left (a b^2 \left (10 a^2-3 b^2\right ) \tan ^2(c+d x)+\frac {1}{3} b^3 \left (15 a^2-b^2\right ) \tan ^3(c+d x)+b \left (15 a^4-15 a^2 b^2+b^4\right ) \tan (c+d x)+\frac {1}{2} \left (6 a^5-20 a^3 b^2+\frac {a^6-15 a^4 b^2+15 a^2 b^4-b^6}{\sqrt {-b^2}}+6 a b^4\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\frac {1}{2} \left (6 a^5-20 a^3 b^2-\frac {a^6-15 a^4 b^2+15 a^2 b^4-b^6}{\sqrt {-b^2}}+6 a b^4\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {3}{2} a b^4 \tan ^4(c+d x)+\frac {1}{5} b^5 \tan ^5(c+d x)\right )}{2 b^2 \left (a^2+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^5*((Cos[c + d*x]^4*(a + b*Tan[c + d*x])^6*(b^2 + a*b*Tan[c + d*x]))/(4*b^6*(a^2 + b^2)) - ((Cos[c + d*x]^2*
(a + b*Tan[c + d*x])^6*(-3*a^2*b^2 + b^2*(-3*a^2 + 2*b^2) + b*(3*a*b^2 + a*(-3*a^2 + 2*b^2))*Tan[c + d*x]))/(2
*b^4*(a^2 + b^2)) - ((3*a^4 - 29*a^2*b^2 + 8*b^4 + 5*a^2*(3*a^2 - 5*b^2))*(((5*a^4 - 10*a^2*b^2 + b^4 + (a^5 -
10*a^3*b^2 + 5*a*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]])/2 + ((5*a^4 - 10*a^2*b^2 + b^4 - (a^5 - 1
0*a^3*b^2 + 5*a*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]])/2 + 5*a*b*(2*a^2 - b^2)*Tan[c + d*x] + (b^2
*(10*a^2 - b^2)*Tan[c + d*x]^2)/2 + (5*a*b^3*Tan[c + d*x]^3)/3 + (b^4*Tan[c + d*x]^4)/4) - 5*a*(3*a^2 - 5*b^2)
*(((6*a^5 - 20*a^3*b^2 + 6*a*b^4 + (a^6 - 15*a^4*b^2 + 15*a^2*b^4 - b^6)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c
+ d*x]])/2 + ((6*a^5 - 20*a^3*b^2 + 6*a*b^4 - (a^6 - 15*a^4*b^2 + 15*a^2*b^4 - b^6)/Sqrt[-b^2])*Log[Sqrt[-b^2]
+ b*Tan[c + d*x]])/2 + b*(15*a^4 - 15*a^2*b^2 + b^4)*Tan[c + d*x] + a*b^2*(10*a^2 - 3*b^2)*Tan[c + d*x]^2 + (
b^3*(15*a^2 - b^2)*Tan[c + d*x]^3)/3 + (3*a*b^4*Tan[c + d*x]^4)/2 + (b^5*Tan[c + d*x]^5)/5))/(2*b^2*(a^2 + b^2
)))/(4*b^2*(a^2 + b^2))))/d

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fricas [A]  time = 0.53, size = 160, normalized size = 0.94 \[ -\frac {8 \, b^{5} \log \left (-\cos \left (d x + c\right )\right ) + 2 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x + 8 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left (2 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{5} + 10 \, a^{3} b^{2} - 25 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*b^5*log(-cos(d*x + c)) + 2*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4 - (3*a^5 + 10*a^3*b^2 + 15*a*b^
4)*d*x + 8*(5*a^2*b^3 - b^5)*cos(d*x + c)^2 - (2*(a^5 - 10*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^3 + (3*a^5 + 10*a^3
*b^2 - 25*a*b^4)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 4.85, size = 199, normalized size = 1.17 \[ \frac {4 \, b^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} {\left (d x + c\right )} - \frac {6 \, b^{5} \tan \left (d x + c\right )^{4} - 3 \, a^{5} \tan \left (d x + c\right )^{3} - 10 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} + 25 \, a b^{4} \tan \left (d x + c\right )^{3} + 40 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} + 4 \, b^{5} \tan \left (d x + c\right )^{2} - 5 \, a^{5} \tan \left (d x + c\right ) + 10 \, a^{3} b^{2} \tan \left (d x + c\right ) + 15 \, a b^{4} \tan \left (d x + c\right ) + 10 \, a^{4} b + 20 \, a^{2} b^{3}}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*b^5*log(tan(d*x + c)^2 + 1) + (3*a^5 + 10*a^3*b^2 + 15*a*b^4)*(d*x + c) - (6*b^5*tan(d*x + c)^4 - 3*a^5
*tan(d*x + c)^3 - 10*a^3*b^2*tan(d*x + c)^3 + 25*a*b^4*tan(d*x + c)^3 + 40*a^2*b^3*tan(d*x + c)^2 + 4*b^5*tan(
d*x + c)^2 - 5*a^5*tan(d*x + c) + 10*a^3*b^2*tan(d*x + c) + 15*a*b^4*tan(d*x + c) + 10*a^4*b + 20*a^2*b^3)/(ta
n(d*x + c)^2 + 1)^2)/d

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maple [A]  time = 0.24, size = 272, normalized size = 1.60 \[ \frac {a^{5} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}+\frac {3 a^{5} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {3 a^{5} x}{8}+\frac {3 a^{5} c}{8 d}-\frac {5 a^{4} \left (\cos ^{4}\left (d x +c \right )\right ) b}{4 d}-\frac {5 a^{3} b^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{2 d}+\frac {5 a^{3} b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{4 d}+\frac {5 a^{3} b^{2} x}{4}+\frac {5 a^{3} b^{2} c}{4 d}+\frac {5 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{2 d}-\frac {5 a \,b^{4} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{4 d}-\frac {15 a \,b^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {15 a \,b^{4} x}{8}+\frac {15 a \,b^{4} c}{8 d}-\frac {b^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b^{5}}{2 d}-\frac {b^{5} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^5*cos(d*x+c)^3*sin(d*x+c)/d+3/8*a^5*cos(d*x+c)*sin(d*x+c)/d+3/8*a^5*x+3/8/d*a^5*c-5/4/d*a^4*cos(d*x+c)^4
*b-5/2*a^3*b^2*cos(d*x+c)^3*sin(d*x+c)/d+5/4*a^3*b^2*cos(d*x+c)*sin(d*x+c)/d+5/4*a^3*b^2*x+5/4/d*a^3*b^2*c+5/2
/d*a^2*b^3*sin(d*x+c)^4-5/4/d*a*b^4*cos(d*x+c)*sin(d*x+c)^3-15/8*a*b^4*cos(d*x+c)*sin(d*x+c)/d+15/8*a*b^4*x+15
/8/d*a*b^4*c-1/4/d*b^5*sin(d*x+c)^4-1/2/d*sin(d*x+c)^2*b^5-1/d*b^5*ln(cos(d*x+c))

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maxima [A]  time = 0.33, size = 170, normalized size = 1.00 \[ \frac {80 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} - 40 \, {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2} a^{4} b + {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{5} + 10 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} b^{2} + 5 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{4} - 8 \, {\left (\sin \left (d x + c\right )^{4} + 2 \, \sin \left (d x + c\right )^{2} + 2 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} b^{5}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(80*a^2*b^3*sin(d*x + c)^4 - 40*(sin(d*x + c)^2 - 1)^2*a^4*b + (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(
2*d*x + 2*c))*a^5 + 10*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^3*b^2 + 5*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(
2*d*x + 2*c))*a*b^4 - 8*(sin(d*x + c)^4 + 2*sin(d*x + c)^2 + 2*log(sin(d*x + c)^2 - 1))*b^5)/d

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mupad [B]  time = 2.65, size = 297, normalized size = 1.75 \[ \frac {4\,b^5\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )-4\,b^5\,\ln \left (\frac {\cos \left (c+d\,x\right )}{\cos \left (c+d\,x\right )+1}\right )+3\,a^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {3\,b^5\,\cos \left (2\,c+2\,d\,x\right )}{2}-\frac {b^5\,\cos \left (4\,c+4\,d\,x\right )}{8}+a^5\,\sin \left (2\,c+2\,d\,x\right )+\frac {a^5\,\sin \left (4\,c+4\,d\,x\right )}{8}+15\,a\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {5\,a^4\,b\,\cos \left (2\,c+2\,d\,x\right )}{2}-\frac {5\,a^4\,b\,\cos \left (4\,c+4\,d\,x\right )}{8}-5\,a\,b^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {5\,a\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{8}+10\,a^3\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-5\,a^2\,b^3\,\cos \left (2\,c+2\,d\,x\right )+\frac {5\,a^2\,b^3\,\cos \left (4\,c+4\,d\,x\right )}{4}-\frac {5\,a^3\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{4}}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*b^5*log(1/cos(c/2 + (d*x)/2)^2) - 4*b^5*log(cos(c + d*x)/(cos(c + d*x) + 1)) + 3*a^5*atan(sin(c/2 + (d*x)/2
)/cos(c/2 + (d*x)/2)) + (3*b^5*cos(2*c + 2*d*x))/2 - (b^5*cos(4*c + 4*d*x))/8 + a^5*sin(2*c + 2*d*x) + (a^5*si
n(4*c + 4*d*x))/8 + 15*a*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (5*a^4*b*cos(2*c + 2*d*x))/2 - (5*a
^4*b*cos(4*c + 4*d*x))/8 - 5*a*b^4*sin(2*c + 2*d*x) + (5*a*b^4*sin(4*c + 4*d*x))/8 + 10*a^3*b^2*atan(sin(c/2 +
(d*x)/2)/cos(c/2 + (d*x)/2)) - 5*a^2*b^3*cos(2*c + 2*d*x) + (5*a^2*b^3*cos(4*c + 4*d*x))/4 - (5*a^3*b^2*sin(4
*c + 4*d*x))/4)/(4*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(sec(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Timed out

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