3.56 \(\int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=125 \[ \frac{\left (a^2+2 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac{\left (2 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}+\frac{a b \tan ^6(c+d x)}{3 d}+\frac{a b \tan ^4(c+d x)}{d}+\frac{a b \tan ^2(c+d x)}{d}+\frac{b^2 \tan ^7(c+d x)}{7 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.104404, antiderivative size = 125, 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} \[ \frac{\left (a^2+2 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac{\left (2 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}+\frac{a b \tan ^6(c+d x)}{3 d}+\frac{a b \tan ^4(c+d x)}{d}+\frac{a b \tan ^2(c+d x)}{d}+\frac{b^2 \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(a^2*Tan[c + d*x])/d + (a*b*Tan[c + d*x]^2)/d + ((2*a^2 + b^2)*Tan[c + d*x]^3)/(3*d) + (a*b*Tan[c + d*x]^4)/d
+ ((a^2 + 2*b^2)*Tan[c + d*x]^5)/(5*d) + (a*b*Tan[c + d*x]^6)/(3*d) + (b^2*Tan[c + d*x]^7)/(7*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 ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(b+a x)^2 \left (1+x^2\right )^2}{x^8} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{b^2}{x^8}+\frac{2 a b}{x^7}+\frac{a^2+2 b^2}{x^6}+\frac{4 a b}{x^5}+\frac{2 a^2+b^2}{x^4}+\frac{2 a b}{x^3}+\frac{a^2}{x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{a^2 \tan (c+d x)}{d}+\frac{a b \tan ^2(c+d x)}{d}+\frac{\left (2 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a b \tan ^4(c+d x)}{d}+\frac{\left (a^2+2 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac{a b \tan ^6(c+d x)}{3 d}+\frac{b^2 \tan ^7(c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.654685, size = 104, normalized size = 0.83 \[ \frac{\tan (c+d x) \left (21 \left (a^2+2 b^2\right ) \tan ^4(c+d x)+35 \left (2 a^2+b^2\right ) \tan ^2(c+d x)+105 a^2+35 a b \tan ^5(c+d x)+105 a b \tan ^3(c+d x)+105 a b \tan (c+d x)+15 b^2 \tan ^6(c+d x)\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(Tan[c + d*x]*(105*a^2 + 105*a*b*Tan[c + d*x] + 35*(2*a^2 + b^2)*Tan[c + d*x]^2 + 105*a*b*Tan[c + d*x]^3 + 21*
(a^2 + 2*b^2)*Tan[c + d*x]^4 + 35*a*b*Tan[c + d*x]^5 + 15*b^2*Tan[c + d*x]^6))/(105*d)

________________________________________________________________________________________

Maple [A]  time = 0.114, size = 110, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( -{a}^{2} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) +{\frac{ab}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^2,x)

[Out]

1/d*(-a^2*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+1/3*a*b/cos(d*x+c)^6+b^2*(1/7*sin(d*x+c)^3/cos
(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3))

________________________________________________________________________________________

Maxima [A]  time = 1.12142, size = 123, normalized size = 0.98 \begin{align*} \frac{7 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{2} +{\left (15 \, \tan \left (d x + c\right )^{7} + 42 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3}\right )} b^{2} - \frac{35 \, a b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/105*(7*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^2 + (15*tan(d*x + c)^7 + 42*tan(d*x + c)^5
+ 35*tan(d*x + c)^3)*b^2 - 35*a*b/(sin(d*x + c)^2 - 1)^3)/d

________________________________________________________________________________________

Fricas [A]  time = 0.487539, size = 231, normalized size = 1.85 \begin{align*} \frac{35 \, a b \cos \left (d x + c\right ) +{\left (8 \,{\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 4 \,{\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/105*(35*a*b*cos(d*x + c) + (8*(7*a^2 - b^2)*cos(d*x + c)^6 + 4*(7*a^2 - b^2)*cos(d*x + c)^4 + 3*(7*a^2 - b^2
)*cos(d*x + c)^2 + 15*b^2)*sin(d*x + c))/(d*cos(d*x + c)^7)

________________________________________________________________________________________

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)**8*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.17727, size = 159, normalized size = 1.27 \begin{align*} \frac{15 \, b^{2} \tan \left (d x + c\right )^{7} + 35 \, a b \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} + 42 \, b^{2} \tan \left (d x + c\right )^{5} + 105 \, a b \tan \left (d x + c\right )^{4} + 70 \, a^{2} \tan \left (d x + c\right )^{3} + 35 \, b^{2} \tan \left (d x + c\right )^{3} + 105 \, a b \tan \left (d x + c\right )^{2} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/105*(15*b^2*tan(d*x + c)^7 + 35*a*b*tan(d*x + c)^6 + 21*a^2*tan(d*x + c)^5 + 42*b^2*tan(d*x + c)^5 + 105*a*b
*tan(d*x + c)^4 + 70*a^2*tan(d*x + c)^3 + 35*b^2*tan(d*x + c)^3 + 105*a*b*tan(d*x + c)^2 + 105*a^2*tan(d*x + c
))/d