\(\int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx\) [175]

Optimal result

Integrand size = 21, antiderivative size = 80 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{d}+\frac {a b \cos ^2(c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 80, 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, 908} \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {a b \cos ^2(c+d x)}{d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

[In]

Int[(a + b*Sec[c + d*x])^2*Sin[c + d*x]^3,x]

[Out]

-(((a^2 - b^2)*Cos[c + d*x])/d) + (a*b*Cos[c + d*x]^2)/d + (a^2*Cos[c + d*x]^3)/(3*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 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 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 (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 )}{x^2} \, dx,x,-a \cos (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \frac {(-b+x)^2 \left (a^2-x^2\right )}{x^2} \, dx,x,-a \cos (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \left (a^2 \left (1-\frac {b^2}{a^2}\right )+\frac {a^2 b^2}{x^2}-\frac {2 a^2 b}{x}+2 b x-x^2\right ) \, dx,x,-a \cos (c+d x)\right )}{a d} \\ & = -\frac {\left (a^2-b^2\right ) \cos (c+d x)}{d}+\frac {a b \cos ^2(c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 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 = 0.68 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\frac {\left (-9 a^2+12 b^2\right ) \cos (c+d x)+6 a b \cos (2 (c+d x))+a^2 \cos (3 (c+d x))-24 a b \log (\cos (c+d x))+12 b^2 \sec (c+d x)}{12 d} \]

[In]

Integrate[(a + b*Sec[c + d*x])^2*Sin[c + d*x]^3,x]

[Out]

((-9*a^2 + 12*b^2)*Cos[c + d*x] + 6*a*b*Cos[2*(c + d*x)] + a^2*Cos[3*(c + d*x)] - 24*a*b*Log[Cos[c + d*x]] + 1
2*b^2*Sec[c + d*x])/(12*d)

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+2 a b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(90\)
default	\(\frac {-\frac {a^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+2 a b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\)	\(90\)
parts	\(-\frac {a^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(95\)
parallelrisch	\(\frac {48 a b \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \cos \left (d x +c \right )-48 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-48 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\left (-8 a^{2}+12 b^{2}\right ) \cos \left (2 d x +2 c \right )+6 a b \cos \left (3 d x +3 c \right )+\cos \left (4 d x +4 c \right ) a^{2}+\left (-16 a^{2}-6 a b +48 b^{2}\right ) \cos \left (d x +c \right )-9 a^{2}+36 b^{2}}{24 d \cos \left (d x +c \right )}\)	\(160\)
norman	\(\frac {\frac {4 a^{2}-12 b^{2}}{3 d}-\frac {4 \left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {2 \left (4 a^{2}+6 a b -12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {2 a b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(181\)
risch	\(2 i a b x +\frac {a b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 d}-\frac {3 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {{\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}+\frac {a b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}+\frac {4 i a b c}{d}+\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {\cos \left (3 d x +3 c \right ) a^{2}}{12 d}\)	\(183\)

[In]

int((a+b*sec(d*x+c))^2*sin(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.15 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\frac {2 \, a^{2} \cos \left (d x + c\right )^{4} + 6 \, a b \cos \left (d x + c\right )^{3} - 12 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - 3 \, a b \cos \left (d x + c\right ) - 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, b^{2}}{6 \, d \cos \left (d x + c\right )} \]

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sin ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate((a+b*sec(d*x+c))**2*sin(d*x+c)**3,x)

[Out]

Integral((a + b*sec(c + d*x))**2*sin(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\frac {a^{2} \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )^{2} - 6 \, a b \log \left (\cos \left (d x + c\right )\right ) - 3 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right ) + \frac {3 \, b^{2}}{\cos \left (d x + c\right )}}{3 \, d} \]

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=-\frac {2 \, a b \log \left (\frac {{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac {b^{2}}{d \cos \left (d x + c\right )} + \frac {a^{2} d^{5} \cos \left (d x + c\right )^{3} + 3 \, a b d^{5} \cos \left (d x + c\right )^{2} - 3 \, a^{2} d^{5} \cos \left (d x + c\right ) + 3 \, b^{2} d^{5} \cos \left (d x + c\right )}{3 \, d^{6}} \]

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.86 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\frac {\frac {a^2\,{\cos \left (c+d\,x\right )}^3}{3}-\cos \left (c+d\,x\right )\,\left (a^2-b^2\right )+\frac {b^2}{\cos \left (c+d\,x\right )}+a\,b\,{\cos \left (c+d\,x\right )}^2-2\,a\,b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

[In]

int(sin(c + d*x)^3*(a + b/cos(c + d*x))^2,x)

[Out]

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