3.73 \(\int \sec ^{11}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=213 \[ \frac {a^3 \tan (c+d x)}{d}+\frac {3 b \left (a^2+b^2\right ) \tan ^8(c+d x)}{8 d}+\frac {a \left (a^2+9 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {b \left (3 a^2+b^2\right ) \tan ^6(c+d x)}{2 d}+\frac {3 a \left (a^2+3 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b \left (9 a^2+b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {a \left (a^2+b^2\right ) \tan ^3(c+d x)}{d}+\frac {3 a^2 b \tan ^2(c+d x)}{2 d}+\frac {a b^2 \tan ^9(c+d x)}{3 d}+\frac {b^3 \tan ^{10}(c+d x)}{10 d} \]

[Out]

a^3*tan(d*x+c)/d+3/2*a^2*b*tan(d*x+c)^2/d+a*(a^2+b^2)*tan(d*x+c)^3/d+1/4*b*(9*a^2+b^2)*tan(d*x+c)^4/d+3/5*a*(a
^2+3*b^2)*tan(d*x+c)^5/d+1/2*b*(3*a^2+b^2)*tan(d*x+c)^6/d+1/7*a*(a^2+9*b^2)*tan(d*x+c)^7/d+3/8*b*(a^2+b^2)*tan
(d*x+c)^8/d+1/3*a*b^2*tan(d*x+c)^9/d+1/10*b^3*tan(d*x+c)^10/d

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Rubi [A]  time = 0.18, antiderivative size = 213, 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, 948} \[ \frac {3 b \left (a^2+b^2\right ) \tan ^8(c+d x)}{8 d}+\frac {a \left (a^2+9 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {b \left (3 a^2+b^2\right ) \tan ^6(c+d x)}{2 d}+\frac {3 a \left (a^2+3 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b \left (9 a^2+b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {a \left (a^2+b^2\right ) \tan ^3(c+d x)}{d}+\frac {3 a^2 b \tan ^2(c+d x)}{2 d}+\frac {a^3 \tan (c+d x)}{d}+\frac {a b^2 \tan ^9(c+d x)}{3 d}+\frac {b^3 \tan ^{10}(c+d x)}{10 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^11*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(a^3*Tan[c + d*x])/d + (3*a^2*b*Tan[c + d*x]^2)/(2*d) + (a*(a^2 + b^2)*Tan[c + d*x]^3)/d + (b*(9*a^2 + b^2)*Ta
n[c + d*x]^4)/(4*d) + (3*a*(a^2 + 3*b^2)*Tan[c + d*x]^5)/(5*d) + (b*(3*a^2 + b^2)*Tan[c + d*x]^6)/(2*d) + (a*(
a^2 + 9*b^2)*Tan[c + d*x]^7)/(7*d) + (3*b*(a^2 + b^2)*Tan[c + d*x]^8)/(8*d) + (a*b^2*Tan[c + d*x]^9)/(3*d) + (
b^3*Tan[c + d*x]^10)/(10*d)

Rule 948

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] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 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 ^{11}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(b+a x)^3 \left (1+x^2\right )^3}{x^{11}} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {b^3}{x^{11}}+\frac {3 a b^2}{x^{10}}+\frac {3 b \left (a^2+b^2\right )}{x^9}+\frac {a^3+9 a b^2}{x^8}+\frac {3 \left (3 a^2 b+b^3\right )}{x^7}+\frac {3 \left (a^3+3 a b^2\right )}{x^6}+\frac {9 a^2 b+b^3}{x^5}+\frac {3 a \left (a^2+b^2\right )}{x^4}+\frac {3 a^2 b}{x^3}+\frac {a^3}{x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {a^3 \tan (c+d x)}{d}+\frac {3 a^2 b \tan ^2(c+d x)}{2 d}+\frac {a \left (a^2+b^2\right ) \tan ^3(c+d x)}{d}+\frac {b \left (9 a^2+b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {3 a \left (a^2+3 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b \left (3 a^2+b^2\right ) \tan ^6(c+d x)}{2 d}+\frac {a \left (a^2+9 b^2\right ) \tan ^7(c+d x)}{7 d}+\frac {3 b \left (a^2+b^2\right ) \tan ^8(c+d x)}{8 d}+\frac {a b^2 \tan ^9(c+d x)}{3 d}+\frac {b^3 \tan ^{10}(c+d x)}{10 d}\\ \end {align*}

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Mathematica [A]  time = 2.07, size = 177, normalized size = 0.83 \[ \frac {\frac {3}{8} \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^8-\frac {4}{7} a \left (5 a^2+3 b^2\right ) (a+b \tan (c+d x))^7+\frac {1}{2} \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^6-\frac {6}{5} a \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^5+\frac {1}{4} \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^4+\frac {1}{10} (a+b \tan (c+d x))^{10}-\frac {2}{3} a (a+b \tan (c+d x))^9}{b^7 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^11*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(((a^2 + b^2)^3*(a + b*Tan[c + d*x])^4)/4 - (6*a*(a^2 + b^2)^2*(a + b*Tan[c + d*x])^5)/5 + ((a^2 + b^2)*(5*a^2
+ b^2)*(a + b*Tan[c + d*x])^6)/2 - (4*a*(5*a^2 + 3*b^2)*(a + b*Tan[c + d*x])^7)/7 + (3*(5*a^2 + b^2)*(a + b*T
an[c + d*x])^8)/8 - (2*a*(a + b*Tan[c + d*x])^9)/3 + (a + b*Tan[c + d*x])^10/10)/(b^7*d)

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fricas [A]  time = 0.58, size = 150, normalized size = 0.70 \[ \frac {84 \, b^{3} + 105 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (16 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{9} + 8 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{7} + 6 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{5} + 35 \, a b^{2} \cos \left (d x + c\right ) + 5 \, {\left (3 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/840*(84*b^3 + 105*(3*a^2*b - b^3)*cos(d*x + c)^2 + 8*(16*(3*a^3 - a*b^2)*cos(d*x + c)^9 + 8*(3*a^3 - a*b^2)*
cos(d*x + c)^7 + 6*(3*a^3 - a*b^2)*cos(d*x + c)^5 + 35*a*b^2*cos(d*x + c) + 5*(3*a^3 - a*b^2)*cos(d*x + c)^3)*
sin(d*x + c))/(d*cos(d*x + c)^10)

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giac [A]  time = 0.45, size = 220, normalized size = 1.03 \[ \frac {84 \, b^{3} \tan \left (d x + c\right )^{10} + 280 \, a b^{2} \tan \left (d x + c\right )^{9} + 315 \, a^{2} b \tan \left (d x + c\right )^{8} + 315 \, b^{3} \tan \left (d x + c\right )^{8} + 120 \, a^{3} \tan \left (d x + c\right )^{7} + 1080 \, a b^{2} \tan \left (d x + c\right )^{7} + 1260 \, a^{2} b \tan \left (d x + c\right )^{6} + 420 \, b^{3} \tan \left (d x + c\right )^{6} + 504 \, a^{3} \tan \left (d x + c\right )^{5} + 1512 \, a b^{2} \tan \left (d x + c\right )^{5} + 1890 \, a^{2} b \tan \left (d x + c\right )^{4} + 210 \, b^{3} \tan \left (d x + c\right )^{4} + 840 \, a^{3} \tan \left (d x + c\right )^{3} + 840 \, a b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 840 \, a^{3} \tan \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/840*(84*b^3*tan(d*x + c)^10 + 280*a*b^2*tan(d*x + c)^9 + 315*a^2*b*tan(d*x + c)^8 + 315*b^3*tan(d*x + c)^8 +
120*a^3*tan(d*x + c)^7 + 1080*a*b^2*tan(d*x + c)^7 + 1260*a^2*b*tan(d*x + c)^6 + 420*b^3*tan(d*x + c)^6 + 504
*a^3*tan(d*x + c)^5 + 1512*a*b^2*tan(d*x + c)^5 + 1890*a^2*b*tan(d*x + c)^4 + 210*b^3*tan(d*x + c)^4 + 840*a^3
*tan(d*x + c)^3 + 840*a*b^2*tan(d*x + c)^3 + 1260*a^2*b*tan(d*x + c)^2 + 840*a^3*tan(d*x + c))/d

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maple [A]  time = 20.74, size = 219, normalized size = 1.03 \[ \frac {-a^{3} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )+\frac {3 a^{2} b}{8 \cos \left (d x +c \right )^{8}}+3 b^{2} a \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{40 \cos \left (d x +c \right )^{4}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^11*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

1/d*(-a^3*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c)+3/8*a^2*b/cos(d*x+c)^8+3*b^
2*a*(1/9*sin(d*x+c)^3/cos(d*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x+c)^3/cos(d*x+c)^5+16/315*sin(d
*x+c)^3/cos(d*x+c)^3)+b^3*(1/10*sin(d*x+c)^4/cos(d*x+c)^10+3/40*sin(d*x+c)^4/cos(d*x+c)^8+1/20*sin(d*x+c)^4/co
s(d*x+c)^6+1/40*sin(d*x+c)^4/cos(d*x+c)^4))

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maxima [A]  time = 0.33, size = 184, normalized size = 0.86 \[ \frac {24 \, {\left (5 \, \tan \left (d x + c\right )^{7} + 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} a^{3} + 8 \, {\left (35 \, \tan \left (d x + c\right )^{9} + 135 \, \tan \left (d x + c\right )^{7} + 189 \, \tan \left (d x + c\right )^{5} + 105 \, \tan \left (d x + c\right )^{3}\right )} a b^{2} - \frac {21 \, {\left (5 \, \sin \left (d x + c\right )^{2} - 1\right )} b^{3}}{\sin \left (d x + c\right )^{10} - 5 \, \sin \left (d x + c\right )^{8} + 10 \, \sin \left (d x + c\right )^{6} - 10 \, \sin \left (d x + c\right )^{4} + 5 \, \sin \left (d x + c\right )^{2} - 1} + \frac {315 \, a^{2} b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{4}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^11*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/840*(24*(5*tan(d*x + c)^7 + 21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 35*tan(d*x + c))*a^3 + 8*(35*tan(d*x + c
)^9 + 135*tan(d*x + c)^7 + 189*tan(d*x + c)^5 + 105*tan(d*x + c)^3)*a*b^2 - 21*(5*sin(d*x + c)^2 - 1)*b^3/(sin
(d*x + c)^10 - 5*sin(d*x + c)^8 + 10*sin(d*x + c)^6 - 10*sin(d*x + c)^4 + 5*sin(d*x + c)^2 - 1) + 315*a^2*b/(s
in(d*x + c)^2 - 1)^4)/d

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mupad [B]  time = 1.58, size = 189, normalized size = 0.89 \[ \frac {{\cos \left (c+d\,x\right )}^3\,\left (\frac {a^3\,\sin \left (c+d\,x\right )}{7}-\frac {a\,b^2\,\sin \left (c+d\,x\right )}{21}\right )+{\cos \left (c+d\,x\right )}^5\,\left (\frac {6\,a^3\,\sin \left (c+d\,x\right )}{35}-\frac {2\,a\,b^2\,\sin \left (c+d\,x\right )}{35}\right )+{\cos \left (c+d\,x\right )}^7\,\left (\frac {8\,a^3\,\sin \left (c+d\,x\right )}{35}-\frac {8\,a\,b^2\,\sin \left (c+d\,x\right )}{105}\right )+{\cos \left (c+d\,x\right )}^9\,\left (\frac {16\,a^3\,\sin \left (c+d\,x\right )}{35}-\frac {16\,a\,b^2\,\sin \left (c+d\,x\right )}{105}\right )+{\cos \left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{8}-\frac {b^3}{8}\right )+\frac {b^3}{10}+\frac {a\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{3}}{d\,{\cos \left (c+d\,x\right )}^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^3/cos(c + d*x)^11,x)

[Out]

(cos(c + d*x)^3*((a^3*sin(c + d*x))/7 - (a*b^2*sin(c + d*x))/21) + cos(c + d*x)^5*((6*a^3*sin(c + d*x))/35 - (
2*a*b^2*sin(c + d*x))/35) + cos(c + d*x)^7*((8*a^3*sin(c + d*x))/35 - (8*a*b^2*sin(c + d*x))/105) + cos(c + d*
x)^9*((16*a^3*sin(c + d*x))/35 - (16*a*b^2*sin(c + d*x))/105) + cos(c + d*x)^2*((3*a^2*b)/8 - b^3/8) + b^3/10
+ (a*b^2*cos(c + d*x)*sin(c + d*x))/3)/(d*cos(c + d*x)^10)

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

[Out]

Timed out

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