\(\int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx\) [104]

Optimal result

Integrand size = 28, antiderivative size = 30 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {(b+a \cot (c+d x))^6 \tan ^6(c+d x)}{6 b d} \]

[Out]

1/6*(b+a*cot(d*x+c))^6*tan(d*x+c)^6/b/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3167, 37} \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {\tan ^6(c+d x) (a \cot (c+d x)+b)^6}{6 b d} \]

[In]

Int[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

((b + a*Cot[c + d*x])^6*Tan[c + d*x]^6)/(6*b*d)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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)^5}{x^7} \, dx,x,\cot (c+d x)\right )}{d} \\ & = \frac {(b+a \cot (c+d x))^6 \tan ^6(c+d x)}{6 b d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(30)=60\).

Time = 0.54 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.97 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {\tan (c+d x) \left (6 a^5+15 a^4 b \tan (c+d x)+20 a^3 b^2 \tan ^2(c+d x)+15 a^2 b^3 \tan ^3(c+d x)+6 a b^4 \tan ^4(c+d x)+b^5 \tan ^5(c+d x)\right )}{6 d} \]

[In]

Integrate[Sec[c + d*x]^7*(a*Cos[c + d*x] + b*Sin[c + d*x])^5,x]

[Out]

(Tan[c + d*x]*(6*a^5 + 15*a^4*b*Tan[c + d*x] + 20*a^3*b^2*Tan[c + d*x]^2 + 15*a^2*b^3*Tan[c + d*x]^3 + 6*a*b^4
*Tan[c + d*x]^4 + b^5*Tan[c + d*x]^5))/(6*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(119\) vs. \(2(28)=56\).

Time = 1.60 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.00

method	result	size
derivativedivides	\(\frac {a^{5} \tan \left (d x +c \right )+\frac {5 a^{4} b}{2 \cos \left (d x +c \right )^{2}}+\frac {10 a^{3} b^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}+\frac {5 a^{2} b^{3} \sin \left (d x +c \right )^{4}}{2 \cos \left (d x +c \right )^{4}}+\frac {a \,b^{4} \sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )^{5}}+\frac {b^{5} \sin \left (d x +c \right )^{6}}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(120\)
default	\(\frac {a^{5} \tan \left (d x +c \right )+\frac {5 a^{4} b}{2 \cos \left (d x +c \right )^{2}}+\frac {10 a^{3} b^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}+\frac {5 a^{2} b^{3} \sin \left (d x +c \right )^{4}}{2 \cos \left (d x +c \right )^{4}}+\frac {a \,b^{4} \sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )^{5}}+\frac {b^{5} \sin \left (d x +c \right )^{6}}{6 \cos \left (d x +c \right )^{6}}}{d}\)	\(120\)
parts	\(\frac {a^{5} \tan \left (d x +c \right )}{d}+\frac {b^{5} \left (\frac {\sec \left (d x +c \right )^{6}}{6}-\frac {\sec \left (d x +c \right )^{4}}{2}+\frac {\sec \left (d x +c \right )^{2}}{2}\right )}{d}+\frac {10 a^{3} b^{2} \sin \left (d x +c \right )^{3}}{3 d \cos \left (d x +c \right )^{3}}+\frac {a \,b^{4} \sin \left (d x +c \right )^{5}}{d \cos \left (d x +c \right )^{5}}+\frac {10 a^{2} b^{3} \left (\frac {\sec \left (d x +c \right )^{4}}{4}-\frac {\sec \left (d x +c \right )^{2}}{2}\right )}{d}+\frac {5 a^{4} b \sec \left (d x +c \right )^{2}}{2 d}\)	\(153\)
parallelrisch	\(\frac {2 \left (\left (5 a^{5}-\frac {10}{3} a^{3} b^{2}-3 a \,b^{4}\right ) \cos \left (3 d x +3 c \right )+\left (a^{5}-\frac {10}{3} a^{3} b^{2}+a \,b^{4}\right ) \cos \left (5 d x +5 c \right )+\frac {5 \left (3 a^{4} b +a^{2} b^{3}-\frac {1}{3} b^{5}\right ) \sin \left (3 d x +3 c \right )}{2}+\frac {5 b \left (a^{4}-a^{2} b^{2}+\frac {1}{15} b^{4}\right ) \sin \left (5 d x +5 c \right )}{2}+5 \left (a^{4} b +a^{2} b^{3}+\frac {1}{3} b^{5}\right ) \sin \left (d x +c \right )+10 \left (a^{4}+\frac {2}{3} a^{2} b^{2}+\frac {1}{5} b^{4}\right ) a \cos \left (d x +c \right )\right ) \sin \left (d x +c \right )}{d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\)	\(216\)
risch	\(\frac {-\frac {200 i a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}}{3}+2 i a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-20 i a^{3} b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+10 i a \,b^{4} {\mathrm e}^{10 i \left (d x +c \right )}-60 i a^{3} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+10 i a \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-40 i a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+20 i a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-20 i a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+20 i a \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+2 b^{5} {\mathrm e}^{10 i \left (d x +c \right )}+2 b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+\frac {20 b^{5} {\mathrm e}^{6 i \left (d x +c \right )}}{3}+2 i a^{5}+2 i a^{5} {\mathrm e}^{10 i \left (d x +c \right )}+10 i a^{5} {\mathrm e}^{8 i \left (d x +c \right )}+20 i a^{5} {\mathrm e}^{6 i \left (d x +c \right )}+20 i a^{5} {\mathrm e}^{4 i \left (d x +c \right )}+10 i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}-\frac {20 i a^{3} b^{2}}{3}+2 i a \,b^{4}-20 a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-40 a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+10 a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}-40 a^{2} b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+40 a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}-40 a^{2} b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+60 a^{4} b \,{\mathrm e}^{6 i \left (d x +c \right )}-20 a^{2} b^{3} {\mathrm e}^{10 i \left (d x +c \right )}+40 a^{4} b \,{\mathrm e}^{8 i \left (d x +c \right )}+10 a^{4} b \,{\mathrm e}^{10 i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(489\)

[In]

int(sec(d*x+c)^7*(cos(d*x+c)*a+b*sin(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (28) = 56\).

Time = 0.26 (sec) , antiderivative size = 144, normalized size of antiderivative = 4.80 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {b^{5} + 3 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, a b^{4} \cos \left (d x + c\right ) + {\left (3 \, a^{5} - 10 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{6}} \]

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="fricas")

[Out]

1/6*(b^5 + 3*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^4 + 3*(5*a^2*b^3 - b^5)*cos(d*x + c)^2 + 2*(3*a*b^4*cos
(d*x + c) + (3*a^5 - 10*a^3*b^2 + 3*a*b^4)*cos(d*x + c)^5 + 2*(5*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^3)*sin(d*x +
c))/(d*cos(d*x + c)^6)

Sympy [F(-1)]

Timed out. \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**7*(a*cos(d*x+c)+b*sin(d*x+c))**5,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (28) = 56\).

Time = 0.22 (sec) , antiderivative size = 166, normalized size of antiderivative = 5.53 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {6 \, a b^{4} \tan \left (d x + c\right )^{5} + 20 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} + 6 \, a^{5} \tan \left (d x + c\right ) + \frac {15 \, {\left (2 \, \sin \left (d x + c\right )^{2} - 1\right )} a^{2} b^{3}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \frac {{\left (3 \, \sin \left (d x + c\right )^{4} - 3 \, \sin \left (d x + c\right )^{2} + 1\right )} b^{5}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - \frac {15 \, a^{4} b}{\sin \left (d x + c\right )^{2} - 1}}{6 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

1/6*(6*a*b^4*tan(d*x + c)^5 + 20*a^3*b^2*tan(d*x + c)^3 + 6*a^5*tan(d*x + c) + 15*(2*sin(d*x + c)^2 - 1)*a^2*b
^3/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - (3*sin(d*x + c)^4 - 3*sin(d*x + c)^2 + 1)*b^5/(sin(d*x + c)^6 - 3
*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*a^4*b/(sin(d*x + c)^2 - 1))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (28) = 56\).

Time = 0.58 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.97 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {b^{5} \tan \left (d x + c\right )^{6} + 6 \, a b^{4} \tan \left (d x + c\right )^{5} + 15 \, a^{2} b^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} b^{2} \tan \left (d x + c\right )^{3} + 15 \, a^{4} b \tan \left (d x + c\right )^{2} + 6 \, a^{5} \tan \left (d x + c\right )}{6 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="giac")

[Out]

1/6*(b^5*tan(d*x + c)^6 + 6*a*b^4*tan(d*x + c)^5 + 15*a^2*b^3*tan(d*x + c)^4 + 20*a^3*b^2*tan(d*x + c)^3 + 15*
a^4*b*tan(d*x + c)^2 + 6*a^5*tan(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 23.90 (sec) , antiderivative size = 169, normalized size of antiderivative = 5.63 \[ \int \sec ^7(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx=\frac {{\cos \left (c+d\,x\right )}^4\,\left (\frac {5\,a^4\,b}{2}-5\,a^2\,b^3+\frac {b^5}{2}\right )+{\cos \left (c+d\,x\right )}^5\,\left (\sin \left (c+d\,x\right )\,a^5-\frac {10\,\sin \left (c+d\,x\right )\,a^3\,b^2}{3}+\sin \left (c+d\,x\right )\,a\,b^4\right )-{\cos \left (c+d\,x\right )}^2\,\left (\frac {b^5}{2}-\frac {5\,a^2\,b^3}{2}\right )+\frac {b^5}{6}+{\cos \left (c+d\,x\right )}^3\,\left (\frac {10\,a^3\,b^2\,\sin \left (c+d\,x\right )}{3}-2\,a\,b^4\,\sin \left (c+d\,x\right )\right )+a\,b^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^6} \]

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^5/cos(c + d*x)^7,x)

[Out]

(cos(c + d*x)^4*((5*a^4*b)/2 + b^5/2 - 5*a^2*b^3) + cos(c + d*x)^5*(a^5*sin(c + d*x) - (10*a^3*b^2*sin(c + d*x
))/3 + a*b^4*sin(c + d*x)) - cos(c + d*x)^2*(b^5/2 - (5*a^2*b^3)/2) + b^5/6 + cos(c + d*x)^3*((10*a^3*b^2*sin(
c + d*x))/3 - 2*a*b^4*sin(c + d*x)) + a*b^4*cos(c + d*x)*sin(c + d*x))/(d*cos(c + d*x)^6)