\(\int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx\) [197]

Optimal result

Integrand size = 21, antiderivative size = 152 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a^5 d}-\frac {b \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{2 a^4 d}+\frac {\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 a^3 d}+\frac {b \cos ^4(c+d x)}{4 a^2 d}-\frac {\cos ^5(c+d x)}{5 a d}+\frac {b \left (a^2-b^2\right )^2 \log (b+a \cos (c+d x))}{a^6 d} \]

[Out]

-(a^2-b^2)^2*cos(d*x+c)/a^5/d-1/2*b*(2*a^2-b^2)*cos(d*x+c)^2/a^4/d+1/3*(2*a^2-b^2)*cos(d*x+c)^3/a^3/d+1/4*b*co
s(d*x+c)^4/a^2/d-1/5*cos(d*x+c)^5/a/d+b*(a^2-b^2)^2*ln(b+a*cos(d*x+c))/a^6/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3957, 2916, 12, 786} \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {b \cos ^4(c+d x)}{4 a^2 d}+\frac {b \left (a^2-b^2\right )^2 \log (a \cos (c+d x)+b)}{a^6 d}-\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a^5 d}-\frac {b \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{2 a^4 d}+\frac {\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 a^3 d}-\frac {\cos ^5(c+d x)}{5 a d} \]

[In]

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

[Out]

-(((a^2 - b^2)^2*Cos[c + d*x])/(a^5*d)) - (b*(2*a^2 - b^2)*Cos[c + d*x]^2)/(2*a^4*d) + ((2*a^2 - b^2)*Cos[c +
d*x]^3)/(3*a^3*d) + (b*Cos[c + d*x]^4)/(4*a^2*d) - Cos[c + d*x]^5/(5*a*d) + (b*(a^2 - b^2)^2*Log[b + a*Cos[c +
d*x]])/(a^6*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 786

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2916

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cos (c+d x) \sin ^5(c+d x)}{-b-a \cos (c+d x)} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {x \left (a^2-x^2\right )^2}{a (-b+x)} \, dx,x,-a \cos (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {x \left (a^2-x^2\right )^2}{-b+x} \, dx,x,-a \cos (c+d x)\right )}{a^6 d} \\ & = \frac {\text {Subst}\left (\int \left (\left (a^2-b^2\right )^2-\frac {b \left (-a^2+b^2\right )^2}{b-x}+b \left (-2 a^2+b^2\right ) x-\left (2 a^2-b^2\right ) x^2+b x^3+x^4\right ) \, dx,x,-a \cos (c+d x)\right )}{a^6 d} \\ & = -\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{a^5 d}-\frac {b \left (2 a^2-b^2\right ) \cos ^2(c+d x)}{2 a^4 d}+\frac {\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 a^3 d}+\frac {b \cos ^4(c+d x)}{4 a^2 d}-\frac {\cos ^5(c+d x)}{5 a d}+\frac {b \left (a^2-b^2\right )^2 \log (b+a \cos (c+d x))}{a^6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.13 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {-60 a \left (5 a^4-14 a^2 b^2+8 b^4\right ) \cos (c+d x)-60 \left (3 a^4 b-2 a^2 b^3\right ) \cos (2 (c+d x))+50 a^5 \cos (3 (c+d x))-40 a^3 b^2 \cos (3 (c+d x))+15 a^4 b \cos (4 (c+d x))-6 a^5 \cos (5 (c+d x))+480 a^4 b \log (b+a \cos (c+d x))-960 a^2 b^3 \log (b+a \cos (c+d x))+480 b^5 \log (b+a \cos (c+d x))}{480 a^6 d} \]

[In]

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

[Out]

(-60*a*(5*a^4 - 14*a^2*b^2 + 8*b^4)*Cos[c + d*x] - 60*(3*a^4*b - 2*a^2*b^3)*Cos[2*(c + d*x)] + 50*a^5*Cos[3*(c
+ d*x)] - 40*a^3*b^2*Cos[3*(c + d*x)] + 15*a^4*b*Cos[4*(c + d*x)] - 6*a^5*Cos[5*(c + d*x)] + 480*a^4*b*Log[b
+ a*Cos[c + d*x]] - 960*a^2*b^3*Log[b + a*Cos[c + d*x]] + 480*b^5*Log[b + a*Cos[c + d*x]])/(480*a^6*d)

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {-\frac {\frac {\cos \left (d x +c \right )^{5} a^{4}}{5}-\frac {b \cos \left (d x +c \right )^{4} a^{3}}{4}-\frac {2 \cos \left (d x +c \right )^{3} a^{4}}{3}+\frac {\cos \left (d x +c \right )^{3} a^{2} b^{2}}{3}+\cos \left (d x +c \right )^{2} a^{3} b -\frac {\cos \left (d x +c \right )^{2} a \,b^{3}}{2}+\cos \left (d x +c \right ) a^{4}-2 \cos \left (d x +c \right ) a^{2} b^{2}+\cos \left (d x +c \right ) b^{4}}{a^{5}}+\frac {b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{6}}}{d}\)	\(160\)
default	\(\frac {-\frac {\frac {\cos \left (d x +c \right )^{5} a^{4}}{5}-\frac {b \cos \left (d x +c \right )^{4} a^{3}}{4}-\frac {2 \cos \left (d x +c \right )^{3} a^{4}}{3}+\frac {\cos \left (d x +c \right )^{3} a^{2} b^{2}}{3}+\cos \left (d x +c \right )^{2} a^{3} b -\frac {\cos \left (d x +c \right )^{2} a \,b^{3}}{2}+\cos \left (d x +c \right ) a^{4}-2 \cos \left (d x +c \right ) a^{2} b^{2}+\cos \left (d x +c \right ) b^{4}}{a^{5}}+\frac {b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{6}}}{d}\)	\(160\)
parallelrisch	\(\frac {480 b \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right )-480 b \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-6 \left (\left (30 a^{3} b -20 a \,b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (-\frac {25}{3} a^{4}+\frac {20}{3} a^{2} b^{2}\right ) \cos \left (3 d x +3 c \right )-\frac {5 b \cos \left (4 d x +4 c \right ) a^{3}}{2}+\cos \left (5 d x +5 c \right ) a^{4}+\left (50 a^{4}-140 a^{2} b^{2}+80 b^{4}\right ) \cos \left (d x +c \right )+\frac {128 a^{4}}{3}-\frac {55 a^{3} b}{2}-\frac {400 a^{2} b^{2}}{3}+20 a \,b^{3}+80 b^{4}\right ) a}{480 d \,a^{6}}\)	\(209\)
norman	\(\frac {\frac {\left (2 a^{3} b +2 a^{2} b^{2}-2 a \,b^{3}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d \,a^{5}}+\frac {-16 a^{4}+50 a^{2} b^{2}-30 b^{4}}{15 d \,a^{5}}+\frac {2 \left (5 a^{3} b +6 a^{2} b^{2}-3 a \,b^{3}-4 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d \,a^{5}}+\frac {\left (-16 a^{4}+6 a^{3} b +44 a^{2} b^{2}-6 a \,b^{3}-24 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d \,a^{5}}+\frac {2 \left (-16 a^{4}+15 a^{3} b +32 a^{2} b^{2}-9 a \,b^{3}-18 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d \,a^{5}}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}+\frac {\left (a +b \right ) b \left (a^{3}-a^{2} b -a \,b^{2}+b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{d \,a^{6}}-\frac {\left (a +b \right ) b \left (a^{3}-a^{2} b -a \,b^{2}+b^{3}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{6}}\)	\(345\)
risch	\(-\frac {\cos \left (5 d x +5 c \right )}{80 d a}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )}}{16 a d}-\frac {2 i b^{5} c}{a^{6} d}+\frac {4 i b^{3} c}{a^{4} d}-\frac {i x \,b^{5}}{a^{6}}+\frac {5 \cos \left (3 d x +3 c \right )}{48 a d}-\frac {2 i b c}{a^{2} d}-\frac {i x b}{a^{2}}-\frac {5 \,{\mathrm e}^{i \left (d x +c \right )}}{16 d a}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{2} d}-\frac {2 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{4} d}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{6} d}-\frac {3 b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{2} d}+\frac {b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{4} d}+\frac {7 \,{\mathrm e}^{i \left (d x +c \right )} b^{2}}{8 a^{3} d}-\frac {{\mathrm e}^{i \left (d x +c \right )} b^{4}}{2 a^{5} d}+\frac {7 \,{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{8 a^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{4}}{2 a^{5} d}+\frac {b \cos \left (4 d x +4 c \right )}{32 d \,a^{2}}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{12 a^{3} d}+\frac {2 i x \,b^{3}}{a^{4}}-\frac {3 b \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 a^{2} d}+\frac {b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{4} d}\)	\(439\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {12 \, a^{5} \cos \left (d x + c\right )^{5} - 15 \, a^{4} b \cos \left (d x + c\right )^{4} - 20 \, {\left (2 \, a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \, {\left (2 \, a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) - 60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{60 \, a^{6} d} \]

[In]

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

[Out]

-1/60*(12*a^5*cos(d*x + c)^5 - 15*a^4*b*cos(d*x + c)^4 - 20*(2*a^5 - a^3*b^2)*cos(d*x + c)^3 + 30*(2*a^4*b - a
^2*b^3)*cos(d*x + c)^2 + 60*(a^5 - 2*a^3*b^2 + a*b^4)*cos(d*x + c) - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(a*cos(d*
x + c) + b))/(a^6*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {12 \, a^{4} \cos \left (d x + c\right )^{5} - 15 \, a^{3} b \cos \left (d x + c\right )^{4} - 20 \, {\left (2 \, a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )}{a^{5}} - \frac {60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6}}}{60 \, d} \]

[In]

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

[Out]

-1/60*((12*a^4*cos(d*x + c)^5 - 15*a^3*b*cos(d*x + c)^4 - 20*(2*a^4 - a^2*b^2)*cos(d*x + c)^3 + 30*(2*a^3*b -
a*b^3)*cos(d*x + c)^2 + 60*(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c))/a^5 - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(a*cos(
d*x + c) + b)/a^6)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 867 vs. \(2 (144) = 288\).

Time = 0.33 (sec) , antiderivative size = 867, normalized size of antiderivative = 5.70 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/60*(60*(a^5*b - a^4*b^2 - 2*a^3*b^3 + 2*a^2*b^4 + a*b^5 - b^6)*log(abs(a + b + a*(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)))/(a^7 - a^6*b) - 60*(a^4*b - 2*a^2*b^3 + b^5)*log(abs(-(
cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^6 + (64*a^5 - 137*a^4*b - 200*a^3*b^2 + 274*a^2*b^3 + 120*a*b^4 -
137*b^5 - 320*a^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 805*a^4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8
80*a^3*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1490*a^2*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 480*a*
b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 685*b^5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 640*a^5*(cos(d*x +
c) - 1)^2/(cos(d*x + c) + 1)^2 - 1970*a^4*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1280*a^3*b^2*(cos(d*x
+ c) - 1)^2/(cos(d*x + c) + 1)^2 + 3100*a^2*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 720*a*b^4*(cos(d*
x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1370*b^5*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1970*a^4*b*(cos(d*x
+ c) - 1)^3/(cos(d*x + c) + 1)^3 + 720*a^3*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 3100*a^2*b^3*(cos(d
*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 480*a*b^4*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 1370*b^5*(cos(d*x
+ c) - 1)^3/(cos(d*x + c) + 1)^3 - 805*a^4*b*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 120*a^3*b^2*(cos(d*x
+ c) - 1)^4/(cos(d*x + c) + 1)^4 + 1490*a^2*b^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 120*a*b^4*(cos(d*x
+ c) - 1)^4/(cos(d*x + c) + 1)^4 - 685*b^5*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 137*a^4*b*(cos(d*x + c
) - 1)^5/(cos(d*x + c) + 1)^5 - 274*a^2*b^3*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 137*b^5*(cos(d*x + c)
- 1)^5/(cos(d*x + c) + 1)^5)/(a^6*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5))/d

Mupad [B] (verification not implemented)

Time = 13.42 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.99 \[ \int \frac {\sin ^5(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\cos \left (c+d\,x\right )\,\left (\frac {1}{a}-\frac {b^2\,\left (\frac {2}{a}-\frac {b^2}{a^3}\right )}{a^2}\right )-{\cos \left (c+d\,x\right )}^3\,\left (\frac {2}{3\,a}-\frac {b^2}{3\,a^3}\right )+\frac {{\cos \left (c+d\,x\right )}^5}{5\,a}-\frac {b\,{\cos \left (c+d\,x\right )}^4}{4\,a^2}-\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (a^4\,b-2\,a^2\,b^3+b^5\right )}{a^6}+\frac {b\,{\cos \left (c+d\,x\right )}^2\,\left (\frac {2}{a}-\frac {b^2}{a^3}\right )}{2\,a}}{d} \]

[In]

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

[Out]

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