\(\int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\) [67]

Optimal result

Integrand size = 28, antiderivative size = 30 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {(b+a \cot (c+d x))^4 \tan ^4(c+d x)}{4 b d} \]

[Out]

1/4*(b+a*cot(d*x+c))^4*tan(d*x+c)^4/b/d

Rubi [A] (verified)

Time = 0.06 (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} \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {\tan ^4(c+d x) (a \cot (c+d x)+b)^4}{4 b d} \]

[In]

Int[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

((b + a*Cot[c + d*x])^4*Tan[c + d*x]^4)/(4*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)^3}{x^5} \, dx,x,\cot (c+d x)\right )}{d} \\ & = \frac {(b+a \cot (c+d x))^4 \tan ^4(c+d x)}{4 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {\tan (c+d x) \left (4 a^3+6 a^2 b \tan (c+d x)+4 a b^2 \tan ^2(c+d x)+b^3 \tan ^3(c+d x)\right )}{4 d} \]

[In]

Integrate[Sec[c + d*x]^5*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(71\) vs. \(2(28)=56\).

Time = 1.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.40

method	result	size
derivativedivides	\(\frac {a^{3} \tan \left (d x +c \right )+\frac {3 a^{2} b}{2 \cos \left (d x +c \right )^{2}}+\frac {a \,b^{2} \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3}}+\frac {b^{3} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(72\)
default	\(\frac {a^{3} \tan \left (d x +c \right )+\frac {3 a^{2} b}{2 \cos \left (d x +c \right )^{2}}+\frac {a \,b^{2} \sin \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3}}+\frac {b^{3} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(72\)
parts	\(\frac {a^{3} \tan \left (d x +c \right )}{d}+\frac {b^{3} \left (\frac {\sec \left (d x +c \right )^{4}}{4}-\frac {\sec \left (d x +c \right )^{2}}{2}\right )}{d}+\frac {a \,b^{2} \sin \left (d x +c \right )^{3}}{d \cos \left (d x +c \right )^{3}}+\frac {3 a^{2} b \sec \left (d x +c \right )^{2}}{2 d}\)	\(84\)
parallelrisch	\(-\frac {2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a b +\left (-2 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a b +a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}\)	\(140\)
risch	\(-\frac {2 \left (-i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-i a^{3}+i a \,b^{2}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}\)	\(197\)
norman	\(\frac {\frac {6 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {6 a^{2} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {\left (6 a^{2} b +4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {\left (6 a^{2} b +4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {3 \left (4 a^{2} b -4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {3 \left (4 a^{2} b -4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {8 a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {8 a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {2 a \left (3 a^{2}-8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {2 a \left (3 a^{2}-8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\)	\(315\)

[In]

int(sec(d*x+c)^5*(cos(d*x+c)*a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (28) = 56\).

Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.60 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {b^{3} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (a b^{2} \cos \left (d x + c\right ) + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (28) = 56\).

Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.90 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {4 \, a b^{2} \tan \left (d x + c\right )^{3} + 4 \, a^{3} \tan \left (d x + c\right ) + \frac {{\left (2 \, \sin \left (d x + c\right )^{2} - 1\right )} b^{3}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \frac {6 \, a^{2} b}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*(4*a*b^2*tan(d*x + c)^3 + 4*a^3*tan(d*x + c) + (2*sin(d*x + c)^2 - 1)*b^3/(sin(d*x + c)^4 - 2*sin(d*x + c)
^2 + 1) - 6*a^2*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. 57 vs. \(2 (28) = 56\).

Time = 0.37 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {b^{3} \tan \left (d x + c\right )^{4} + 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \, a^{2} b \tan \left (d x + c\right )^{2} + 4 \, a^{3} \tan \left (d x + c\right )}{4 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 22.63 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.93 \[ \int \sec ^5(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx=\frac {{\cos \left (c+d\,x\right )}^3\,\left (a^3\,\sin \left (c+d\,x\right )-a\,b^2\,\sin \left (c+d\,x\right )\right )+{\cos \left (c+d\,x\right )}^2\,\left (\frac {3\,a^2\,b}{2}-\frac {b^3}{2}\right )+\frac {b^3}{4}+a\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^4} \]

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^3/cos(c + d*x)^5,x)

[Out]

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