\(\int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\) [54]

Optimal result

Integrand size = 28, antiderivative size = 85 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {a^2 \tan (c+d x)}{d}+\frac {a b \tan ^2(c+d x)}{d}+\frac {\left (a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a b \tan ^4(c+d x)}{2 d}+\frac {b^2 \tan ^5(c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3167, 908} \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {\left (a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {a b \tan ^4(c+d x)}{2 d}+\frac {a b \tan ^2(c+d x)}{d}+\frac {b^2 \tan ^5(c+d x)}{5 d} \]

[In]

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

Rule 908

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 3167

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*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {(b+a x)^2 \left (1+x^2\right )}{x^6} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {b^2}{x^6}+\frac {2 a b}{x^5}+\frac {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 (a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {a b \tan ^4(c+d x)}{2 d}+\frac {b^2 \tan ^5(c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {(a+b \tan (c+d x))^3 \left (a^2+10 b^2-3 a b \tan (c+d x)+6 b^2 \tan ^2(c+d x)\right )}{30 b^3 d} \]

[In]

Integrate[Sec[c + d*x]^6*(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

((a + b*Tan[c + d*x])^3*(a^2 + 10*b^2 - 3*a*b*Tan[c + d*x] + 6*b^2*Tan[c + d*x]^2))/(30*b^3*d)

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {a b}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )}{d}\)	\(82\)
default	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+\frac {a b}{2 \cos \left (d x +c \right )^{4}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )}{d}\)	\(82\)
parts	\(-\frac {a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )}{d}+\frac {a b \sec \left (d x +c \right )^{4}}{2 d}\)	\(87\)
risch	\(\frac {4 i \left (-30 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+15 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-15 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-30 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}+35 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+5 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+25 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+5 a^{2}-b^{2}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\)	\(143\)
parallelrisch	\(-\frac {2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a^{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a b +\frac {4 \left (-2 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a b +\frac {2 \left (5 a^{2}+\frac {4 b^{2}}{5}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a b +\frac {4 \left (-2 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a b +a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(186\)
norman	\(\frac {\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {4 \left (a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {4 \left (a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {2 \left (5 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d}+\frac {2 \left (5 a^{2}-16 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{5 d}-\frac {8 \left (5 a^{2}+14 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\)	\(272\)

[In]

int(sec(d*x+c)^6*(cos(d*x+c)*a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.93 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {15 \, a b \cos \left (d x + c\right ) + 2 \, {\left (2 \, {\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/30*(15*a*b*cos(d*x + c) + 2*(2*(5*a^2 - b^2)*cos(d*x + c)^4 + (5*a^2 - b^2)*cos(d*x + c)^2 + 3*b^2)*sin(d*x
+ c))/(d*cos(d*x + c)^5)

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**6*(a*cos(d*x+c)+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {10 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + 2 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} b^{2} + \frac {15 \, a b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{30 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/30*(10*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 + 2*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*b^2 + 15*a*b/(sin(d*x
+ c)^2 - 1)^2)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {6 \, b^{2} \tan \left (d x + c\right )^{5} + 15 \, a b \tan \left (d x + c\right )^{4} + 10 \, a^{2} \tan \left (d x + c\right )^{3} + 10 \, b^{2} \tan \left (d x + c\right )^{3} + 30 \, a b \tan \left (d x + c\right )^{2} + 30 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a*cos(d*x+c)+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/30*(6*b^2*tan(d*x + c)^5 + 15*a*b*tan(d*x + c)^4 + 10*a^2*tan(d*x + c)^3 + 10*b^2*tan(d*x + c)^3 + 30*a*b*ta
n(d*x + c)^2 + 30*a^2*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 22.48 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.15 \[ \int \sec ^6(c+d x) (a \cos (c+d x)+b \sin (c+d x))^2 \, dx=\frac {\frac {b^2\,\sin \left (c+d\,x\right )}{5}+{\cos \left (c+d\,x\right )}^2\,\left (\frac {a^2\,\sin \left (c+d\,x\right )}{3}-\frac {b^2\,\sin \left (c+d\,x\right )}{15}\right )+{\cos \left (c+d\,x\right )}^4\,\left (\frac {2\,a^2\,\sin \left (c+d\,x\right )}{3}-\frac {2\,b^2\,\sin \left (c+d\,x\right )}{15}\right )+\frac {a\,b\,\cos \left (c+d\,x\right )}{2}}{d\,{\cos \left (c+d\,x\right )}^5} \]

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^2/cos(c + d*x)^6,x)

[Out]

((b^2*sin(c + d*x))/5 + cos(c + d*x)^2*((a^2*sin(c + d*x))/3 - (b^2*sin(c + d*x))/15) + cos(c + d*x)^4*((2*a^2
*sin(c + d*x))/3 - (2*b^2*sin(c + d*x))/15) + (a*b*cos(c + d*x))/2)/(d*cos(c + d*x)^5)