\(\int (a+b \sec (c+d x))^2 \sin (c+d x) \, dx\) [176]

Optimal result

Integrand size = 19, antiderivative size = 42 \[ \int (a+b \sec (c+d x))^2 \sin (c+d x) \, dx=-\frac {a^2 \cos (c+d x)}{d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3957, 2912, 12, 45} \[ \int (a+b \sec (c+d x))^2 \sin (c+d x) \, dx=-\frac {a^2 \cos (c+d x)}{d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]

[In]

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

[Out]

-((a^2*Cos[c + d*x])/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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

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

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 \sec (c+d x) \tan (c+d x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {a^2 (-b+x)^2}{x^2} \, dx,x,-a \cos (c+d x)\right )}{a d} \\ & = \frac {a \text {Subst}\left (\int \frac {(-b+x)^2}{x^2} \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (1+\frac {b^2}{x^2}-\frac {2 b}{x}\right ) \, dx,x,-a \cos (c+d x)\right )}{d} \\ & = -\frac {a^2 \cos (c+d x)}{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 = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int (a+b \sec (c+d x))^2 \sin (c+d x) \, dx=\frac {-a^2 \cos (c+d x)+b (-2 a \log (\cos (c+d x))+b \sec (c+d x))}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\sec \left (d x +c \right ) b^{2}-\frac {a^{2}}{\sec \left (d x +c \right )}+2 a b \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(40\)
default	\(\frac {\sec \left (d x +c \right ) b^{2}-\frac {a^{2}}{\sec \left (d x +c \right )}+2 a b \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(40\)
parts	\(-\frac {a^{2} \cos \left (d x +c \right )}{d}+\frac {b^{2} \sec \left (d x +c \right )}{d}+\frac {2 a b \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(43\)
risch	\(2 i a b x -\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d}+\frac {4 i a b c}{d}+\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(100\)
parallelrisch	\(\frac {4 a b \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \cos \left (d x +c \right )-4 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-4 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-\cos \left (2 d x +2 c \right ) a^{2}+\left (-2 a^{2}+2 b^{2}\right ) \cos \left (d x +c \right )-a^{2}+2 b^{2}}{2 d \cos \left (d x +c \right )}\)	\(123\)
norman	\(\frac {\frac {2 a^{2}-2 b^{2}}{d}-\frac {2 \left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {2 a b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\)	\(131\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.19 \[ \int (a+b \sec (c+d x))^2 \sin (c+d x) \, dx=-\frac {a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - b^{2}}{d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int (a+b \sec (c+d x))^2 \sin (c+d x) \, dx=-\frac {a^{2} \cos \left (d x + c\right ) + 2 \, a b \log \left (\cos \left (d x + c\right )\right ) - \frac {b^{2}}{\cos \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.19 \[ \int (a+b \sec (c+d x))^2 \sin (c+d x) \, dx=-\frac {a^{2} \cos \left (d x + c\right )}{d} - \frac {2 \, a b \log \left (\frac {{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac {b^{2}}{d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int (a+b \sec (c+d x))^2 \sin (c+d x) \, dx=-\frac {a^2\,\cos \left (c+d\,x\right )-\frac {b^2}{\cos \left (c+d\,x\right )}+2\,a\,b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

[In]

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

[Out]

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