\(\int (a+b \sec (c+d x))^2 \sin ^5(c+d x) \, dx\) [174]

Optimal result

Integrand size = 21, antiderivative size = 124 \[ \int (a+b \sec (c+d x))^2 \sin ^5(c+d x) \, dx=-\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{d}+\frac {2 a b \cos ^2(c+d x)}{d}+\frac {\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {a b \cos ^4(c+d x)}{2 d}-\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 124, 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, 962} \[ \int (a+b \sec (c+d x))^2 \sin ^5(c+d x) \, dx=\frac {\left (2 a^2-b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{d}-\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {a b \cos ^4(c+d x)}{2 d}+\frac {2 a b \cos ^2(c+d x)}{d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

[In]

Int[(a + b*Sec[c + d*x])^2*Sin[c + d*x]^5,x]

[Out]

-(((a^2 - 2*b^2)*Cos[c + d*x])/d) + (2*a*b*Cos[c + d*x]^2)/d + ((2*a^2 - b^2)*Cos[c + d*x]^3)/(3*d) - (a*b*Cos
[c + d*x]^4)/(2*d) - (a^2*Cos[c + d*x]^5)/(5*d) - (2*a*b*Log[Cos[c + d*x]])/d + (b^2*Sec[c + d*x])/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 962

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]))

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

Mathematica [A] (verified)

Time = 2.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.90 \[ \int (a+b \sec (c+d x))^2 \sin ^5(c+d x) \, dx=-\frac {30 \left (5 a^2-14 b^2\right ) \cos (c+d x)-180 a b \cos (2 (c+d x))-25 a^2 \cos (3 (c+d x))+20 b^2 \cos (3 (c+d x))+15 a b \cos (4 (c+d x))+3 a^2 \cos (5 (c+d x))+480 a b \log (\cos (c+d x))-240 b^2 \sec (c+d x)}{240 d} \]

[In]

Integrate[(a + b*Sec[c + d*x])^2*Sin[c + d*x]^5,x]

[Out]

-1/240*(30*(5*a^2 - 14*b^2)*Cos[c + d*x] - 180*a*b*Cos[2*(c + d*x)] - 25*a^2*Cos[3*(c + d*x)] + 20*b^2*Cos[3*(
c + d*x)] + 15*a*b*Cos[4*(c + d*x)] + 3*a^2*Cos[5*(c + d*x)] + 480*a*b*Log[Cos[c + d*x]] - 240*b^2*Sec[c + d*x
])/d

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}+2 a b \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(120\)
default	\(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}+2 a b \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(120\)
parts	\(-\frac {a^{2} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5 d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(125\)
parallelrisch	\(\frac {-960 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+960 a b \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \cos \left (d x +c \right )-960 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-256 \cos \left (d x +c \right ) a^{2}-150 \cos \left (d x +c \right ) a b +1280 \cos \left (d x +c \right ) b^{2}-15 a b \cos \left (5 d x +5 c \right )+165 a b \cos \left (3 d x +3 c \right )+22 \cos \left (4 d x +4 c \right ) a^{2}-20 \cos \left (4 d x +4 c \right ) b^{2}-125 \cos \left (2 d x +2 c \right ) a^{2}+400 \cos \left (2 d x +2 c \right ) b^{2}-3 a^{2} \cos \left (6 d x +6 c \right )-150 a^{2}+900 b^{2}}{480 d \cos \left (d x +c \right )}\)	\(219\)
norman	\(\frac {\frac {16 a^{2}-80 b^{2}}{15 d}-\frac {32 \left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {16 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {4 \left (16 a^{2}+15 a b -80 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{15 d}+\frac {\left (16 a^{2}+48 a b -80 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {2 a b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(230\)
risch	\(2 i a b x +\frac {3 a b \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {5 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {7 \,{\mathrm e}^{i \left (d x +c \right )} b^{2}}{8 d}-\frac {5 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{16 d}+\frac {7 \,{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{8 d}+\frac {3 a b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {4 i a b c}{d}+\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {a^{2} \cos \left (5 d x +5 c \right )}{80 d}-\frac {a b \cos \left (4 d x +4 c \right )}{16 d}+\frac {5 \cos \left (3 d x +3 c \right ) a^{2}}{48 d}-\frac {\cos \left (3 d x +3 c \right ) b^{2}}{12 d}\)	\(233\)

[In]

int((a+b*sec(d*x+c))^2*sin(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/5*a^2*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)+2*a*b*(-1/4*sin(d*x+c)^4-1/2*sin(d*x+c)^2-ln(cos(
d*x+c)))+b^2*(sin(d*x+c)^6/cos(d*x+c)+(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.01 \[ \int (a+b \sec (c+d x))^2 \sin ^5(c+d x) \, dx=-\frac {48 \, a^{2} \cos \left (d x + c\right )^{6} + 120 \, a b \cos \left (d x + c\right )^{5} - 480 \, a b \cos \left (d x + c\right )^{3} - 80 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 480 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + 195 \, a b \cos \left (d x + c\right ) + 240 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 240 \, b^{2}}{240 \, d \cos \left (d x + c\right )} \]

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^5,x, algorithm="fricas")

[Out]

-1/240*(48*a^2*cos(d*x + c)^6 + 120*a*b*cos(d*x + c)^5 - 480*a*b*cos(d*x + c)^3 - 80*(2*a^2 - b^2)*cos(d*x + c
)^4 + 480*a*b*cos(d*x + c)*log(-cos(d*x + c)) + 195*a*b*cos(d*x + c) + 240*(a^2 - 2*b^2)*cos(d*x + c)^2 - 240*
b^2)/(d*cos(d*x + c))

Sympy [F]

\[ \int (a+b \sec (c+d x))^2 \sin ^5(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sin ^{5}{\left (c + d x \right )}\, dx \]

[In]

integrate((a+b*sec(d*x+c))**2*sin(d*x+c)**5,x)

[Out]

Integral((a + b*sec(c + d*x))**2*sin(c + d*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.85 \[ \int (a+b \sec (c+d x))^2 \sin ^5(c+d x) \, dx=-\frac {6 \, a^{2} \cos \left (d x + c\right )^{5} + 15 \, a b \cos \left (d x + c\right )^{4} - 60 \, a b \cos \left (d x + c\right )^{2} - 10 \, {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{3} + 60 \, a b \log \left (\cos \left (d x + c\right )\right ) + 30 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right ) - \frac {30 \, b^{2}}{\cos \left (d x + c\right )}}{30 \, d} \]

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^5,x, algorithm="maxima")

[Out]

-1/30*(6*a^2*cos(d*x + c)^5 + 15*a*b*cos(d*x + c)^4 - 60*a*b*cos(d*x + c)^2 - 10*(2*a^2 - b^2)*cos(d*x + c)^3
+ 60*a*b*log(cos(d*x + c)) + 30*(a^2 - 2*b^2)*cos(d*x + c) - 30*b^2/cos(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (118) = 236\).

Time = 0.39 (sec) , antiderivative size = 418, normalized size of antiderivative = 3.37 \[ \int (a+b \sec (c+d x))^2 \sin ^5(c+d x) \, dx=\frac {60 \, a b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {60 \, {\left (a b + b^{2} + \frac {a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1} + \frac {32 \, a^{2} + 137 \, a b - 100 \, b^{2} - \frac {160 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {805 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {440 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {320 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1970 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {640 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1970 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {360 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {805 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {60 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {137 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}}{30 \, d} \]

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^5,x, algorithm="giac")

[Out]

1/30*(60*a*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) - 1)) + 60*(a*b + b^2 + a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((cos(d*x + c) - 1)/(cos(d*x +
c) + 1) + 1) + (32*a^2 + 137*a*b - 100*b^2 - 160*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 805*a*b*(cos(d*x
+ c) - 1)/(cos(d*x + c) + 1) + 440*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 320*a^2*(cos(d*x + c) - 1)^2/(c
os(d*x + c) + 1)^2 + 1970*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 640*b^2*(cos(d*x + c) - 1)^2/(cos(d*
x + c) + 1)^2 - 1970*a*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 360*b^2*(cos(d*x + c) - 1)^3/(cos(d*x + c
) + 1)^3 + 805*a*b*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 60*b^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^
4 - 137*a*b*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 13.95 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.84 \[ \int (a+b \sec (c+d x))^2 \sin ^5(c+d x) \, dx=-\frac {\cos \left (c+d\,x\right )\,\left (a^2-2\,b^2\right )-{\cos \left (c+d\,x\right )}^3\,\left (\frac {2\,a^2}{3}-\frac {b^2}{3}\right )+\frac {a^2\,{\cos \left (c+d\,x\right )}^5}{5}-\frac {b^2}{\cos \left (c+d\,x\right )}-2\,a\,b\,{\cos \left (c+d\,x\right )}^2+\frac {a\,b\,{\cos \left (c+d\,x\right )}^4}{2}+2\,a\,b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

[In]

int(sin(c + d*x)^5*(a + b/cos(c + d*x))^2,x)

[Out]

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