3.91 \(\int \sec ^{12}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=254 \[ \frac{b^2 \left (2 a^2+b^2\right ) \tan ^9(c+d x)}{3 d}+\frac{a b \left (a^2+3 b^2\right ) \tan ^8(c+d x)}{2 d}+\frac{\left (18 a^2 b^2+a^4+3 b^4\right ) \tan ^7(c+d x)}{7 d}+\frac{2 a b \left (a^2+b^2\right ) \tan ^6(c+d x)}{d}+\frac{\left (18 a^2 b^2+3 a^4+b^4\right ) \tan ^5(c+d x)}{5 d}+\frac{a b \left (3 a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac{a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{d}+\frac{2 a^3 b \tan ^2(c+d x)}{d}+\frac{a^4 \tan (c+d x)}{d}+\frac{2 a b^3 \tan ^{10}(c+d x)}{5 d}+\frac{b^4 \tan ^{11}(c+d x)}{11 d} \]

[Out]

(a^4*Tan[c + d*x])/d + (2*a^3*b*Tan[c + d*x]^2)/d + (a^2*(a^2 + 2*b^2)*Tan[c + d*x]^3)/d + (a*b*(3*a^2 + b^2)*
Tan[c + d*x]^4)/d + ((3*a^4 + 18*a^2*b^2 + b^4)*Tan[c + d*x]^5)/(5*d) + (2*a*b*(a^2 + b^2)*Tan[c + d*x]^6)/d +
((a^4 + 18*a^2*b^2 + 3*b^4)*Tan[c + d*x]^7)/(7*d) + (a*b*(a^2 + 3*b^2)*Tan[c + d*x]^8)/(2*d) + (b^2*(2*a^2 +
b^2)*Tan[c + d*x]^9)/(3*d) + (2*a*b^3*Tan[c + d*x]^10)/(5*d) + (b^4*Tan[c + d*x]^11)/(11*d)

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Rubi [A]  time = 0.219142, antiderivative size = 254, normalized size of antiderivative = 1., 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} \[ \frac{b^2 \left (2 a^2+b^2\right ) \tan ^9(c+d x)}{3 d}+\frac{a b \left (a^2+3 b^2\right ) \tan ^8(c+d x)}{2 d}+\frac{\left (18 a^2 b^2+a^4+3 b^4\right ) \tan ^7(c+d x)}{7 d}+\frac{2 a b \left (a^2+b^2\right ) \tan ^6(c+d x)}{d}+\frac{\left (18 a^2 b^2+3 a^4+b^4\right ) \tan ^5(c+d x)}{5 d}+\frac{a b \left (3 a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac{a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{d}+\frac{2 a^3 b \tan ^2(c+d x)}{d}+\frac{a^4 \tan (c+d x)}{d}+\frac{2 a b^3 \tan ^{10}(c+d x)}{5 d}+\frac{b^4 \tan ^{11}(c+d x)}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^12*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(a^4*Tan[c + d*x])/d + (2*a^3*b*Tan[c + d*x]^2)/d + (a^2*(a^2 + 2*b^2)*Tan[c + d*x]^3)/d + (a*b*(3*a^2 + b^2)*
Tan[c + d*x]^4)/d + ((3*a^4 + 18*a^2*b^2 + b^4)*Tan[c + d*x]^5)/(5*d) + (2*a*b*(a^2 + b^2)*Tan[c + d*x]^6)/d +
((a^4 + 18*a^2*b^2 + 3*b^4)*Tan[c + d*x]^7)/(7*d) + (a*b*(a^2 + 3*b^2)*Tan[c + d*x]^8)/(2*d) + (b^2*(2*a^2 +
b^2)*Tan[c + d*x]^9)/(3*d) + (2*a*b^3*Tan[c + d*x]^10)/(5*d) + (b^4*Tan[c + d*x]^11)/(11*d)

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])

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]))

Rubi steps

\begin{align*} \int \sec ^{12}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(b+a x)^4 \left (1+x^2\right )^3}{x^{12}} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{b^4}{x^{12}}+\frac{4 a b^3}{x^{11}}+\frac{3 \left (2 a^2 b^2+b^4\right )}{x^{10}}+\frac{4 a b \left (a^2+3 b^2\right )}{x^9}+\frac{a^4+18 a^2 b^2+3 b^4}{x^8}+\frac{12 a b \left (a^2+b^2\right )}{x^7}+\frac{3 a^4+18 a^2 b^2+b^4}{x^6}+\frac{4 a b \left (3 a^2+b^2\right )}{x^5}+\frac{3 \left (a^4+2 a^2 b^2\right )}{x^4}+\frac{4 a^3 b}{x^3}+\frac{a^4}{x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{a^4 \tan (c+d x)}{d}+\frac{2 a^3 b \tan ^2(c+d x)}{d}+\frac{a^2 \left (a^2+2 b^2\right ) \tan ^3(c+d x)}{d}+\frac{a b \left (3 a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac{\left (3 a^4+18 a^2 b^2+b^4\right ) \tan ^5(c+d x)}{5 d}+\frac{2 a b \left (a^2+b^2\right ) \tan ^6(c+d x)}{d}+\frac{\left (a^4+18 a^2 b^2+3 b^4\right ) \tan ^7(c+d x)}{7 d}+\frac{a b \left (a^2+3 b^2\right ) \tan ^8(c+d x)}{2 d}+\frac{b^2 \left (2 a^2+b^2\right ) \tan ^9(c+d x)}{3 d}+\frac{2 a b^3 \tan ^{10}(c+d x)}{5 d}+\frac{b^4 \tan ^{11}(c+d x)}{11 d}\\ \end{align*}

Mathematica [A]  time = 1.74173, size = 175, normalized size = 0.69 \[ \frac{\frac{1}{3} \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^9-\frac{1}{2} a \left (5 a^2+3 b^2\right ) (a+b \tan (c+d x))^8+\frac{3}{7} \left (a^2+b^2\right ) \left (5 a^2+b^2\right ) (a+b \tan (c+d x))^7-a \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^6+\frac{1}{5} \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^5+\frac{1}{11} (a+b \tan (c+d x))^{11}-\frac{3}{5} a (a+b \tan (c+d x))^{10}}{b^7 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^12*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(((a^2 + b^2)^3*(a + b*Tan[c + d*x])^5)/5 - a*(a^2 + b^2)^2*(a + b*Tan[c + d*x])^6 + (3*(a^2 + b^2)*(5*a^2 + b
^2)*(a + b*Tan[c + d*x])^7)/7 - (a*(5*a^2 + 3*b^2)*(a + b*Tan[c + d*x])^8)/2 + ((5*a^2 + b^2)*(a + b*Tan[c + d
*x])^9)/3 - (3*a*(a + b*Tan[c + d*x])^10)/5 + (a + b*Tan[c + d*x])^11/11)/(b^7*d)

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Maple [A]  time = 0.142, size = 300, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -{a}^{4} \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{35}} \right ) \tan \left ( dx+c \right ) +{\frac{{a}^{3}b}{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}+6\,{a}^{2}{b}^{2} \left ( 1/9\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{9}}}+2/21\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{16\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{315\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +4\,a{b}^{3} \left ( 1/10\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{10}}}+{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{40\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}+1/20\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+1/40\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \right ) +{b}^{4} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{11\, \left ( \cos \left ( dx+c \right ) \right ) ^{11}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{33\, \left ( \cos \left ( dx+c \right ) \right ) ^{9}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{231\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{16\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{1155\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^12*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

1/d*(-a^4*(-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)+1/2*a^3*b/cos(d*x+c)^8+6*a^
2*b^2*(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)+4*a*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/cos(d*x+c)^6+1/40*sin(d*x+c)^4/cos(d*x+c)^4)+b^4*(1/11*sin(d*x+c)^5/cos(d*x+c)^11+2/33*sin(d*x+c)^5/cos(d*
x+c)^9+8/231*sin(d*x+c)^5/cos(d*x+c)^7+16/1155*sin(d*x+c)^5/cos(d*x+c)^5))

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Maxima [A]  time = 1.23292, size = 315, normalized size = 1.24 \begin{align*} \frac{66 \,{\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^{4} + 44 \,{\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^{2} b^{2} + 2 \,{\left (105 \, \tan \left (d x + c\right )^{11} + 385 \, \tan \left (d x + c\right )^{9} + 495 \, \tan \left (d x + c\right )^{7} + 231 \, \tan \left (d x + c\right )^{5}\right )} b^{4} - \frac{231 \,{\left (5 \, \sin \left (d x + c\right )^{2} - 1\right )} a 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{1155 \, a^{3} b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{4}}}{2310 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/2310*(66*(5*tan(d*x + c)^7 + 21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 35*tan(d*x + c))*a^4 + 44*(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^2*b^2 + 2*(105*tan(d*x + c)^11 + 385*t
an(d*x + c)^9 + 495*tan(d*x + c)^7 + 231*tan(d*x + c)^5)*b^4 - 231*(5*sin(d*x + c)^2 - 1)*a*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) + 1155*a^3*b/(sin(d*x +
c)^2 - 1)^4)/d

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Fricas [A]  time = 0.579139, size = 471, normalized size = 1.85 \begin{align*} \frac{924 \, a b^{3} \cos \left (d x + c\right ) + 1155 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (16 \,{\left (33 \, a^{4} - 22 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{10} + 8 \,{\left (33 \, a^{4} - 22 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{8} + 6 \,{\left (33 \, a^{4} - 22 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} + 5 \,{\left (33 \, a^{4} - 22 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 105 \, b^{4} + 70 \,{\left (11 \, a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2310 \, d \cos \left (d x + c\right )^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/2310*(924*a*b^3*cos(d*x + c) + 1155*(a^3*b - a*b^3)*cos(d*x + c)^3 + 2*(16*(33*a^4 - 22*a^2*b^2 + b^4)*cos(d
*x + c)^10 + 8*(33*a^4 - 22*a^2*b^2 + b^4)*cos(d*x + c)^8 + 6*(33*a^4 - 22*a^2*b^2 + b^4)*cos(d*x + c)^6 + 5*(
33*a^4 - 22*a^2*b^2 + b^4)*cos(d*x + c)^4 + 105*b^4 + 70*(11*a^2*b^2 - 2*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d
*cos(d*x + c)^11)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**12*(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.21455, size = 383, normalized size = 1.51 \begin{align*} \frac{210 \, b^{4} \tan \left (d x + c\right )^{11} + 924 \, a b^{3} \tan \left (d x + c\right )^{10} + 1540 \, a^{2} b^{2} \tan \left (d x + c\right )^{9} + 770 \, b^{4} \tan \left (d x + c\right )^{9} + 1155 \, a^{3} b \tan \left (d x + c\right )^{8} + 3465 \, a b^{3} \tan \left (d x + c\right )^{8} + 330 \, a^{4} \tan \left (d x + c\right )^{7} + 5940 \, a^{2} b^{2} \tan \left (d x + c\right )^{7} + 990 \, b^{4} \tan \left (d x + c\right )^{7} + 4620 \, a^{3} b \tan \left (d x + c\right )^{6} + 4620 \, a b^{3} \tan \left (d x + c\right )^{6} + 1386 \, a^{4} \tan \left (d x + c\right )^{5} + 8316 \, a^{2} b^{2} \tan \left (d x + c\right )^{5} + 462 \, b^{4} \tan \left (d x + c\right )^{5} + 6930 \, a^{3} b \tan \left (d x + c\right )^{4} + 2310 \, a b^{3} \tan \left (d x + c\right )^{4} + 2310 \, a^{4} \tan \left (d x + c\right )^{3} + 4620 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 4620 \, a^{3} b \tan \left (d x + c\right )^{2} + 2310 \, a^{4} \tan \left (d x + c\right )}{2310 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/2310*(210*b^4*tan(d*x + c)^11 + 924*a*b^3*tan(d*x + c)^10 + 1540*a^2*b^2*tan(d*x + c)^9 + 770*b^4*tan(d*x +
c)^9 + 1155*a^3*b*tan(d*x + c)^8 + 3465*a*b^3*tan(d*x + c)^8 + 330*a^4*tan(d*x + c)^7 + 5940*a^2*b^2*tan(d*x +
c)^7 + 990*b^4*tan(d*x + c)^7 + 4620*a^3*b*tan(d*x + c)^6 + 4620*a*b^3*tan(d*x + c)^6 + 1386*a^4*tan(d*x + c)
^5 + 8316*a^2*b^2*tan(d*x + c)^5 + 462*b^4*tan(d*x + c)^5 + 6930*a^3*b*tan(d*x + c)^4 + 2310*a*b^3*tan(d*x + c
)^4 + 2310*a^4*tan(d*x + c)^3 + 4620*a^2*b^2*tan(d*x + c)^3 + 4620*a^3*b*tan(d*x + c)^2 + 2310*a^4*tan(d*x + c
))/d