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

Optimal. Leaf size=147 \[ \frac {b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac {4 a b^2 \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac {b \left (5 a^4-10 a^2 b^2+b^4\right ) \log (\cos (c+d x))}{d}+a x \left (a^4-10 a^2 b^2+5 b^4\right )+\frac {b (a+b \tan (c+d x))^4}{4 d}+\frac {2 a b (a+b \tan (c+d x))^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3086, 3482, 3528, 3525, 3475} \[ \frac {b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac {4 a b^2 \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac {b \left (-10 a^2 b^2+5 a^4+b^4\right ) \log (\cos (c+d x))}{d}+a x \left (-10 a^2 b^2+a^4+5 b^4\right )+\frac {b (a+b \tan (c+d x))^4}{4 d}+\frac {2 a b (a+b \tan (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3086

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int (a+b \tan (c+d x))^5 \, dx\\ &=\frac {b (a+b \tan (c+d x))^4}{4 d}+\int (a+b \tan (c+d x))^3 \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx\\ &=\frac {2 a b (a+b \tan (c+d x))^3}{3 d}+\frac {b (a+b \tan (c+d x))^4}{4 d}+\int (a+b \tan (c+d x))^2 \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\frac {b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac {2 a b (a+b \tan (c+d x))^3}{3 d}+\frac {b (a+b \tan (c+d x))^4}{4 d}+\int (a+b \tan (c+d x)) \left (a^4-6 a^2 b^2+b^4+4 a b \left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=a \left (a^4-10 a^2 b^2+5 b^4\right ) x+\frac {4 a b^2 \left (a^2-b^2\right ) \tan (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac {2 a b (a+b \tan (c+d x))^3}{3 d}+\frac {b (a+b \tan (c+d x))^4}{4 d}+\left (b \left (5 a^4-10 a^2 b^2+b^4\right )\right ) \int \tan (c+d x) \, dx\\ &=a \left (a^4-10 a^2 b^2+5 b^4\right ) x-\frac {b \left (5 a^4-10 a^2 b^2+b^4\right ) \log (\cos (c+d x))}{d}+\frac {4 a b^2 \left (a^2-b^2\right ) \tan (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) (a+b \tan (c+d x))^2}{2 d}+\frac {2 a b (a+b \tan (c+d x))^3}{3 d}+\frac {b (a+b \tan (c+d x))^4}{4 d}\\ \end {align*}

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Mathematica [C]  time = 0.74, size = 126, normalized size = 0.86 \[ \frac {60 a b^2 \left (2 a^2-b^2\right ) \tan (c+d x)-6 b^3 \left (b^2-10 a^2\right ) \tan ^2(c+d x)+20 a b^4 \tan ^3(c+d x)+6 (b-i a)^5 \log (-\tan (c+d x)+i)+6 (b+i a)^5 \log (\tan (c+d x)+i)+3 b^5 \tan ^4(c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*((-I)*a + b)^5*Log[I - Tan[c + d*x]] + 6*(I*a + b)^5*Log[I + Tan[c + d*x]] + 60*a*b^2*(2*a^2 - b^2)*Tan[c +
d*x] - 6*b^3*(-10*a^2 + b^2)*Tan[c + d*x]^2 + 20*a*b^4*Tan[c + d*x]^3 + 3*b^5*Tan[c + d*x]^4)/(12*d)

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fricas [A]  time = 0.64, size = 155, normalized size = 1.05 \[ \frac {12 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} d x \cos \left (d x + c\right )^{4} - 12 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) + 3 \, b^{5} + 12 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 20 \, {\left (a b^{4} \cos \left (d x + c\right ) + 2 \, {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(a^5 - 10*a^3*b^2 + 5*a*b^4)*d*x*cos(d*x + c)^4 - 12*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4*log(
-cos(d*x + c)) + 3*b^5 + 12*(5*a^2*b^3 - b^5)*cos(d*x + c)^2 + 20*(a*b^4*cos(d*x + c) + 2*(3*a^3*b^2 - 2*a*b^4
)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [A]  time = 0.62, size = 144, normalized size = 0.98 \[ \frac {3 \, b^{5} \tan \left (d x + c\right )^{4} + 20 \, a b^{4} \tan \left (d x + c\right )^{3} + 60 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - 6 \, b^{5} \tan \left (d x + c\right )^{2} + 120 \, a^{3} b^{2} \tan \left (d x + c\right ) - 60 \, a b^{4} \tan \left (d x + c\right ) + 12 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} {\left (d x + c\right )} + 6 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^5*tan(d*x + c)^4 + 20*a*b^4*tan(d*x + c)^3 + 60*a^2*b^3*tan(d*x + c)^2 - 6*b^5*tan(d*x + c)^2 + 120*
a^3*b^2*tan(d*x + c) - 60*a*b^4*tan(d*x + c) + 12*(a^5 - 10*a^3*b^2 + 5*a*b^4)*(d*x + c) + 6*(5*a^4*b - 10*a^2
*b^3 + b^5)*log(tan(d*x + c)^2 + 1))/d

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maple [A]  time = 0.23, size = 202, normalized size = 1.37 \[ a^{5} x +\frac {a^{5} c}{d}-\frac {5 a^{4} b \ln \left (\cos \left (d x +c \right )\right )}{d}-10 a^{3} b^{2} x +\frac {10 \tan \left (d x +c \right ) a^{3} b^{2}}{d}-\frac {10 a^{3} b^{2} c}{d}+\frac {5 a^{2} b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {10 a^{2} b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {5 a \,b^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {5 a \,b^{4} \tan \left (d x +c \right )}{d}+5 a \,b^{4} x +\frac {5 a \,b^{4} c}{d}+\frac {b^{5} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {b^{5} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b^{5} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 174, normalized size = 1.18 \[ \frac {12 \, {\left (d x + c\right )} a^{5} - 120 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} b^{2} + 20 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a b^{4} + 3 \, b^{5} {\left (\frac {4 \, \sin \left (d x + c\right )^{2} - 3}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 2 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 60 \, a^{2} b^{3} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 30 \, a^{4} b \log \left (-\sin \left (d x + c\right )^{2} + 1\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(d*x + c)*a^5 - 120*(d*x + c - tan(d*x + c))*a^3*b^2 + 20*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x +
c))*a*b^4 + 3*b^5*((4*sin(d*x + c)^2 - 3)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 2*log(sin(d*x + c)^2 - 1)
) - 60*a^2*b^3*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1)) - 30*a^4*b*log(-sin(d*x + c)^2 + 1))/d

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mupad [B]  time = 3.45, size = 971, normalized size = 6.61 \[ \text {result too large to display} \]

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)^5,x)

[Out]

((3*b^5*log(1/cos(c/2 + (d*x)/2)^2))/8 - (3*b^5*log(-cos(c + d*x)/cos(c/2 + (d*x)/2)^2))/8 + b^5/32 + (3*a^5*a
tan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (5*a^2*b^3)/8 - (b^5*cos(2*c + 2*d*x))/8 + (3*b^5*cos(4*c + 4*
d*x))/32 + (15*a^4*b*log(1/cos(c/2 + (d*x)/2)^2))/8 + (15*a^2*b^3*log(-cos(c + d*x)/cos(c/2 + (d*x)/2)^2))/4 -
(b^5*log(-cos(c + d*x)/cos(c/2 + (d*x)/2)^2)*cos(2*c + 2*d*x))/2 - (b^5*log(-cos(c + d*x)/cos(c/2 + (d*x)/2)^
2)*cos(4*c + 4*d*x))/8 + (15*a*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 - (5*a*b^4*sin(2*c + 2*d*x))
/6 - (5*a*b^4*sin(4*c + 4*d*x))/6 - (15*a^2*b^3*log(1/cos(c/2 + (d*x)/2)^2))/4 + (b^5*log(1/cos(c/2 + (d*x)/2)
^2)*cos(2*c + 2*d*x))/2 + (b^5*log(1/cos(c/2 + (d*x)/2)^2)*cos(4*c + 4*d*x))/8 + a^5*atan(sin(c/2 + (d*x)/2)/c
os(c/2 + (d*x)/2))*cos(2*c + 2*d*x) + (a^5*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/4 - (
15*a^3*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - (5*a^2*b^3*cos(4*c + 4*d*x))/8 + (5*a^3*b^2*sin(2*
c + 2*d*x))/2 + (5*a^3*b^2*sin(4*c + 4*d*x))/4 - (15*a^4*b*log(-cos(c + d*x)/cos(c/2 + (d*x)/2)^2))/8 - 5*a^2*
b^3*log(1/cos(c/2 + (d*x)/2)^2)*cos(2*c + 2*d*x) - (5*a^2*b^3*log(1/cos(c/2 + (d*x)/2)^2)*cos(4*c + 4*d*x))/4
- 10*a^3*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) - (5*a^3*b^2*atan(sin(c/2 + (d*x)/2)
/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/2 - (5*a^4*b*log(-cos(c + d*x)/cos(c/2 + (d*x)/2)^2)*cos(2*c + 2*d*x))/
2 - (5*a^4*b*log(-cos(c + d*x)/cos(c/2 + (d*x)/2)^2)*cos(4*c + 4*d*x))/8 + (5*a^4*b*log(1/cos(c/2 + (d*x)/2)^2
)*cos(2*c + 2*d*x))/2 + (5*a^4*b*log(1/cos(c/2 + (d*x)/2)^2)*cos(4*c + 4*d*x))/8 + 5*a^2*b^3*log(-cos(c + d*x)
/cos(c/2 + (d*x)/2)^2)*cos(2*c + 2*d*x) + (5*a^2*b^3*log(-cos(c + d*x)/cos(c/2 + (d*x)/2)^2)*cos(4*c + 4*d*x))
/4 + 5*a*b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) + (5*a*b^4*atan(sin(c/2 + (d*x)/2)/c
os(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/4)/(d*(cos(2*c + 2*d*x)/2 + cos(4*c + 4*d*x)/8 + 3/8))

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

[Out]

Timed out

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