3.106 \(\int \sec ^9(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\)

Optimal. Leaf size=177 \[ \frac {a^5 \tan (c+d x)}{d}+\frac {5 a^4 b \tan ^2(c+d x)}{2 d}+\frac {a b^2 \left (2 a^2+b^2\right ) \tan ^5(c+d x)}{d}+\frac {5 a^2 b \left (a^2+2 b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {b^3 \left (10 a^2+b^2\right ) \tan ^6(c+d x)}{6 d}+\frac {a^3 \left (a^2+10 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {5 a b^4 \tan ^7(c+d x)}{7 d}+\frac {b^5 \tan ^8(c+d x)}{8 d} \]

[Out]

a^5*tan(d*x+c)/d+5/2*a^4*b*tan(d*x+c)^2/d+1/3*a^3*(a^2+10*b^2)*tan(d*x+c)^3/d+5/4*a^2*b*(a^2+2*b^2)*tan(d*x+c)
^4/d+a*b^2*(2*a^2+b^2)*tan(d*x+c)^5/d+1/6*b^3*(10*a^2+b^2)*tan(d*x+c)^6/d+5/7*a*b^4*tan(d*x+c)^7/d+1/8*b^5*tan
(d*x+c)^8/d

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Rubi [A]  time = 0.15, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3088, 894} \[ \frac {b^3 \left (10 a^2+b^2\right ) \tan ^6(c+d x)}{6 d}+\frac {a b^2 \left (2 a^2+b^2\right ) \tan ^5(c+d x)}{d}+\frac {5 a^2 b \left (a^2+2 b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {a^3 \left (a^2+10 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {5 a^4 b \tan ^2(c+d x)}{2 d}+\frac {a^5 \tan (c+d x)}{d}+\frac {5 a b^4 \tan ^7(c+d x)}{7 d}+\frac {b^5 \tan ^8(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(a^5*Tan[c + d*x])/d + (5*a^4*b*Tan[c + d*x]^2)/(2*d) + (a^3*(a^2 + 10*b^2)*Tan[c + d*x]^3)/(3*d) + (5*a^2*b*(
a^2 + 2*b^2)*Tan[c + d*x]^4)/(4*d) + (a*b^2*(2*a^2 + b^2)*Tan[c + d*x]^5)/d + (b^3*(10*a^2 + b^2)*Tan[c + d*x]
^6)/(6*d) + (5*a*b^4*Tan[c + d*x]^7)/(7*d) + (b^5*Tan[c + d*x]^8)/(8*d)

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 ^9(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 \left (1+x^2\right )}{x^9} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {b^5}{x^9}+\frac {5 a b^4}{x^8}+\frac {10 a^2 b^3+b^5}{x^7}+\frac {5 a b^2 \left (2 a^2+b^2\right )}{x^6}+\frac {5 a^2 b \left (a^2+2 b^2\right )}{x^5}+\frac {a^5+10 a^3 b^2}{x^4}+\frac {5 a^4 b}{x^3}+\frac {a^5}{x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {a^5 \tan (c+d x)}{d}+\frac {5 a^4 b \tan ^2(c+d x)}{2 d}+\frac {a^3 \left (a^2+10 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {5 a^2 b \left (a^2+2 b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {a b^2 \left (2 a^2+b^2\right ) \tan ^5(c+d x)}{d}+\frac {b^3 \left (10 a^2+b^2\right ) \tan ^6(c+d x)}{6 d}+\frac {5 a b^4 \tan ^7(c+d x)}{7 d}+\frac {b^5 \tan ^8(c+d x)}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 54, normalized size = 0.31 \[ \frac {(a+b \tan (c+d x))^6 \left (a^2-6 a b \tan (c+d x)+21 b^2 \tan ^2(c+d x)+28 b^2\right )}{168 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

((a + b*Tan[c + d*x])^6*(a^2 + 28*b^2 - 6*a*b*Tan[c + d*x] + 21*b^2*Tan[c + d*x]^2))/(168*b^3*d)

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fricas [A]  time = 0.45, size = 176, normalized size = 0.99 \[ \frac {21 \, b^{5} + 42 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 56 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, {\left (7 \, a^{5} - 14 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{7} + 15 \, a b^{4} \cos \left (d x + c\right ) + {\left (7 \, a^{5} - 14 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (7 \, a^{3} b^{2} - 4 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{168 \, d \cos \left (d x + c\right )^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

1/168*(21*b^5 + 42*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4 + 56*(5*a^2*b^3 - b^5)*cos(d*x + c)^2 + 8*(2*(7
*a^5 - 14*a^3*b^2 + 3*a*b^4)*cos(d*x + c)^7 + 15*a*b^4*cos(d*x + c) + (7*a^5 - 14*a^3*b^2 + 3*a*b^4)*cos(d*x +
c)^5 + 6*(7*a^3*b^2 - 4*a*b^4)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^8)

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giac [A]  time = 2.94, size = 176, normalized size = 0.99 \[ \frac {21 \, b^{5} \tan \left (d x + c\right )^{8} + 120 \, a b^{4} \tan \left (d x + c\right )^{7} + 280 \, a^{2} b^{3} \tan \left (d x + c\right )^{6} + 28 \, b^{5} \tan \left (d x + c\right )^{6} + 336 \, a^{3} b^{2} \tan \left (d x + c\right )^{5} + 168 \, a b^{4} \tan \left (d x + c\right )^{5} + 210 \, a^{4} b \tan \left (d x + c\right )^{4} + 420 \, a^{2} b^{3} \tan \left (d x + c\right )^{4} + 56 \, a^{5} \tan \left (d x + c\right )^{3} + 560 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} + 420 \, a^{4} b \tan \left (d x + c\right )^{2} + 168 \, a^{5} \tan \left (d x + c\right )}{168 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

1/168*(21*b^5*tan(d*x + c)^8 + 120*a*b^4*tan(d*x + c)^7 + 280*a^2*b^3*tan(d*x + c)^6 + 28*b^5*tan(d*x + c)^6 +
336*a^3*b^2*tan(d*x + c)^5 + 168*a*b^4*tan(d*x + c)^5 + 210*a^4*b*tan(d*x + c)^4 + 420*a^2*b^3*tan(d*x + c)^4
+ 56*a^5*tan(d*x + c)^3 + 560*a^3*b^2*tan(d*x + c)^3 + 420*a^4*b*tan(d*x + c)^2 + 168*a^5*tan(d*x + c))/d

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maple [A]  time = 0.28, size = 217, normalized size = 1.23 \[ \frac {-a^{5} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {5 a^{4} b}{4 \cos \left (d x +c \right )^{4}}+10 a^{3} b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+10 a^{2} b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+5 a \,b^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}\right )+b^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{6}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{6}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^5,x)

[Out]

1/d*(-a^5*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+5/4*a^4*b/cos(d*x+c)^4+10*a^3*b^2*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+
2/15*sin(d*x+c)^3/cos(d*x+c)^3)+10*a^2*b^3*(1/6*sin(d*x+c)^4/cos(d*x+c)^6+1/12*sin(d*x+c)^4/cos(d*x+c)^4)+5*a*
b^4*(1/7*sin(d*x+c)^5/cos(d*x+c)^7+2/35*sin(d*x+c)^5/cos(d*x+c)^5)+b^5*(1/8*sin(d*x+c)^6/cos(d*x+c)^8+1/24*sin
(d*x+c)^6/cos(d*x+c)^6))

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maxima [A]  time = 0.96, size = 223, normalized size = 1.26 \[ \frac {56 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{5} + 112 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a^{3} b^{2} + 24 \, {\left (5 \, \tan \left (d x + c\right )^{7} + 7 \, \tan \left (d x + c\right )^{5}\right )} a b^{4} - \frac {140 \, {\left (3 \, \sin \left (d x + c\right )^{2} - 1\right )} a^{2} b^{3}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + \frac {7 \, {\left (6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1\right )} b^{5}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} + \frac {210 \, a^{4} b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{168 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

1/168*(56*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^5 + 112*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*a^3*b^2 + 24*(5*ta
n(d*x + c)^7 + 7*tan(d*x + c)^5)*a*b^4 - 140*(3*sin(d*x + c)^2 - 1)*a^2*b^3/(sin(d*x + c)^6 - 3*sin(d*x + c)^4
+ 3*sin(d*x + c)^2 - 1) + 7*(6*sin(d*x + c)^4 - 4*sin(d*x + c)^2 + 1)*b^5/(sin(d*x + c)^8 - 4*sin(d*x + c)^6
+ 6*sin(d*x + c)^4 - 4*sin(d*x + c)^2 + 1) + 210*a^4*b/(sin(d*x + c)^2 - 1)^2)/d

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mupad [B]  time = 4.27, size = 419, normalized size = 2.37 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {86\,a^5}{3}-\frac {208\,a^3\,b^2}{3}+32\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (40\,a^4\,b-40\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (40\,a^4\,b-40\,a^2\,b^3\right )-2\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {86\,a^5}{3}-\frac {208\,a^3\,b^2}{3}+32\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {130\,a^5}{3}-\frac {224\,a^3\,b^2}{3}+\frac {32\,a\,b^4}{7}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {130\,a^5}{3}-\frac {224\,a^3\,b^2}{3}+\frac {32\,a\,b^4}{7}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (-80\,a^4\,b+\frac {80\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (70\,a^4\,b-\frac {160\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (70\,a^4\,b-\frac {160\,a^2\,b^3}{3}+\frac {32\,b^5}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {34\,a^5}{3}-\frac {80\,a^3\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {34\,a^5}{3}-\frac {80\,a^3\,b^2}{3}\right )+2\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^8} \]

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)^9,x)

[Out]

(tan(c/2 + (d*x)/2)^5*(32*a*b^4 + (86*a^5)/3 - (208*a^3*b^2)/3) - tan(c/2 + (d*x)/2)^4*(40*a^4*b - 40*a^2*b^3)
- tan(c/2 + (d*x)/2)^12*(40*a^4*b - 40*a^2*b^3) - 2*a^5*tan(c/2 + (d*x)/2)^15 - tan(c/2 + (d*x)/2)^11*(32*a*b
^4 + (86*a^5)/3 - (208*a^3*b^2)/3) - tan(c/2 + (d*x)/2)^7*((32*a*b^4)/7 + (130*a^5)/3 - (224*a^3*b^2)/3) + tan
(c/2 + (d*x)/2)^9*((32*a*b^4)/7 + (130*a^5)/3 - (224*a^3*b^2)/3) + tan(c/2 + (d*x)/2)^8*((32*b^5)/3 - 80*a^4*b
+ (80*a^2*b^3)/3) + tan(c/2 + (d*x)/2)^6*(70*a^4*b + (32*b^5)/3 - (160*a^2*b^3)/3) + tan(c/2 + (d*x)/2)^10*(7
0*a^4*b + (32*b^5)/3 - (160*a^2*b^3)/3) - tan(c/2 + (d*x)/2)^3*((34*a^5)/3 - (80*a^3*b^2)/3) + tan(c/2 + (d*x)
/2)^13*((34*a^5)/3 - (80*a^3*b^2)/3) + 2*a^5*tan(c/2 + (d*x)/2) + 10*a^4*b*tan(c/2 + (d*x)/2)^2 + 10*a^4*b*tan
(c/2 + (d*x)/2)^14)/(d*(tan(c/2 + (d*x)/2)^2 - 1)^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)**9*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Timed out

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