3.100 \(\int \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

Optimal. Leaf size=169 \[ -\frac {2 b^3 \left (5 a^2-b^2\right ) \log (\sin (c+d x))}{d}+\frac {2 b^3 \left (5 a^2-b^2\right ) \log (\tan (c+d x))}{d}+\frac {\sin ^2(c+d x) \left (a \left (a^4-10 a^2 b^2+5 b^4\right ) \cot (c+d x)+b \left (5 a^4-10 a^2 b^2+b^4\right )\right )}{2 d}+\frac {1}{2} a x \left (a^4+10 a^2 b^2-15 b^4\right )+\frac {5 a b^4 \tan (c+d x)}{d}+\frac {b^5 \tan ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3088, 1805, 1802, 635, 203, 260} \[ -\frac {2 b^3 \left (5 a^2-b^2\right ) \log (\sin (c+d x))}{d}+\frac {2 b^3 \left (5 a^2-b^2\right ) \log (\tan (c+d x))}{d}+\frac {\sin ^2(c+d x) \left (a \left (-10 a^2 b^2+a^4+5 b^4\right ) \cot (c+d x)+b \left (-10 a^2 b^2+5 a^4+b^4\right )\right )}{2 d}+\frac {1}{2} a x \left (10 a^2 b^2+a^4-15 b^4\right )+\frac {5 a b^4 \tan (c+d x)}{d}+\frac {b^5 \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(a*(a^4 + 10*a^2*b^2 - 15*b^4)*x)/2 - (2*b^3*(5*a^2 - b^2)*Log[Sin[c + d*x]])/d + (2*b^3*(5*a^2 - b^2)*Log[Tan
[c + d*x]])/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]^2)/(2
*d) + (5*a*b^4*Tan[c + d*x])/d + (b^5*Tan[c + d*x]^2)/(2*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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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

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Mathematica [B]  time = 6.36, size = 571, normalized size = 3.38 \[ \frac {b^3 \left (\frac {\cos ^2(c+d x) (a+b \tan (c+d x))^6 \left (a b \tan (c+d x)+b^2\right )}{2 b^4 \left (a^2+b^2\right )}-\frac {\left (4 b^2-6 a^2\right ) \left (\frac {1}{2} b^2 \left (10 a^2-b^2\right ) \tan ^2(c+d x)+5 a b \left (2 a^2-b^2\right ) \tan (c+d x)+\frac {1}{2} \left (5 a^4-10 a^2 b^2+\frac {a^5-10 a^3 b^2+5 a b^4}{\sqrt {-b^2}}+b^4\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\frac {1}{2} \left (5 a^4-10 a^2 b^2-\frac {a^5-10 a^3 b^2+5 a b^4}{\sqrt {-b^2}}+b^4\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {5}{3} a b^3 \tan ^3(c+d x)+\frac {1}{4} b^4 \tan ^4(c+d x)\right )+5 a \left (a b^2 \left (10 a^2-3 b^2\right ) \tan ^2(c+d x)+\frac {1}{3} b^3 \left (15 a^2-b^2\right ) \tan ^3(c+d x)+b \left (15 a^4-15 a^2 b^2+b^4\right ) \tan (c+d x)+\frac {1}{2} \left (6 a^5-20 a^3 b^2+\frac {a^6-15 a^4 b^2+15 a^2 b^4-b^6}{\sqrt {-b^2}}+6 a b^4\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\frac {1}{2} \left (6 a^5-20 a^3 b^2-\frac {a^6-15 a^4 b^2+15 a^2 b^4-b^6}{\sqrt {-b^2}}+6 a b^4\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {3}{2} a b^4 \tan ^4(c+d x)+\frac {1}{5} b^5 \tan ^5(c+d x)\right )}{2 b^2 \left (a^2+b^2\right )}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(b^3*((Cos[c + d*x]^2*(a + b*Tan[c + d*x])^6*(b^2 + a*b*Tan[c + d*x]))/(2*b^4*(a^2 + b^2)) - ((-6*a^2 + 4*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 - 10*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*(((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))))/d

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fricas [A]  time = 0.46, size = 177, normalized size = 1.05 \[ \frac {2 \, b^{5} - 2 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} - 8 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5} + 2 \, {\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} d x\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (10 \, a b^{4} \cos \left (d x + c\right ) + {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.58, size = 173, normalized size = 1.02 \[ \frac {b^{5} \tan \left (d x + c\right )^{2} + 10 \, a b^{4} \tan \left (d x + c\right ) + {\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} {\left (d x + c\right )} + 2 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac {10 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - 2 \, b^{5} \tan \left (d x + c\right )^{2} - a^{5} \tan \left (d x + c\right ) + 10 \, a^{3} b^{2} \tan \left (d x + c\right ) - 5 \, a b^{4} \tan \left (d x + c\right ) + 5 \, a^{4} b - b^{5}}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.24, size = 291, normalized size = 1.72 \[ \frac {a^{5} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a^{5} x}{2}+\frac {a^{5} c}{2 d}-\frac {5 a^{4} b \left (\cos ^{2}\left (d x +c \right )\right )}{2 d}-\frac {5 a^{3} b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+5 a^{3} b^{2} x +\frac {5 a^{3} b^{2} c}{d}-\frac {5 a^{2} b^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{d}-\frac {10 a^{2} b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {5 a \,b^{4} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{d}+\frac {15 a \,b^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {15 a \,b^{4} x}{2}-\frac {15 a \,b^{4} c}{2 d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {b^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{2 d}+\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b^{5}}{d}+\frac {2 b^{5} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^5,x)

[Out]

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

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maxima [A]  time = 0.43, size = 179, normalized size = 1.06 \[ \frac {10 \, a^{4} b \sin \left (d x + c\right )^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{5} + 10 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} b^{2} - 20 \, {\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} a^{2} b^{3} - 10 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a b^{4} + 2 \, {\left (\sin \left (d x + c\right )^{2} - \frac {1}{\sin \left (d x + c\right )^{2} - 1} + 2 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} b^{5}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

1/4*(10*a^4*b*sin(d*x + c)^2 + (2*d*x + 2*c + sin(2*d*x + 2*c))*a^5 + 10*(2*d*x + 2*c - sin(2*d*x + 2*c))*a^3*
b^2 - 20*(sin(d*x + c)^2 + log(sin(d*x + c)^2 - 1))*a^2*b^3 - 10*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 +
1) - 2*tan(d*x + c))*a*b^4 + 2*(sin(d*x + c)^2 - 1/(sin(d*x + c)^2 - 1) + 2*log(sin(d*x + c)^2 - 1))*b^5)/d

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mupad [B]  time = 2.53, size = 354, normalized size = 2.09 \[ \frac {2\,\left (b^5\,\ln \left (\frac {\cos \left (c+d\,x\right )}{\cos \left (c+d\,x\right )+1}\right )-b^5\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )+\frac {a^5\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}-5\,a^2\,b^3\,\ln \left (\frac {\cos \left (c+d\,x\right )}{\cos \left (c+d\,x\right )+1}\right )-\frac {15\,a\,b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+5\,a^2\,b^3\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )+5\,a^3\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {5\,a^4\,b}{16}+\frac {9\,b^5}{16}-\frac {5\,a^2\,b^3}{8}-\frac {b^5\,\cos \left (4\,c+4\,d\,x\right )}{16}+\frac {a^5\,\sin \left (2\,c+2\,d\,x\right )}{8}+\frac {a^5\,\sin \left (4\,c+4\,d\,x\right )}{16}-\frac {5\,a^4\,b\,\cos \left (4\,c+4\,d\,x\right )}{16}+\frac {25\,a\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{8}+\frac {5\,a\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{16}+\frac {5\,a^2\,b^3\,\cos \left (4\,c+4\,d\,x\right )}{8}-\frac {5\,a^3\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4}-\frac {5\,a^3\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{8}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

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)^3,x)

[Out]

(2*(b^5*log(cos(c + d*x)/(cos(c + d*x) + 1)) - b^5*log(1/cos(c/2 + (d*x)/2)^2) + (a^5*atan(sin(c/2 + (d*x)/2)/
cos(c/2 + (d*x)/2)))/2 - 5*a^2*b^3*log(cos(c + d*x)/(cos(c + d*x) + 1)) - (15*a*b^4*atan(sin(c/2 + (d*x)/2)/co
s(c/2 + (d*x)/2)))/2 + 5*a^2*b^3*log(1/cos(c/2 + (d*x)/2)^2) + 5*a^3*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2))))/d + ((5*a^4*b)/16 + (9*b^5)/16 - (5*a^2*b^3)/8 - (b^5*cos(4*c + 4*d*x))/16 + (a^5*sin(2*c + 2*d*x))/8
+ (a^5*sin(4*c + 4*d*x))/16 - (5*a^4*b*cos(4*c + 4*d*x))/16 + (25*a*b^4*sin(2*c + 2*d*x))/8 + (5*a*b^4*sin(4*
c + 4*d*x))/16 + (5*a^2*b^3*cos(4*c + 4*d*x))/8 - (5*a^3*b^2*sin(2*c + 2*d*x))/4 - (5*a^3*b^2*sin(4*c + 4*d*x)
)/8)/(d*(cos(2*c + 2*d*x)/2 + 1/2))

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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(sec(d*x+c)**3*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Timed out

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