\(\int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [1533]

Optimal result

Integrand size = 29, antiderivative size = 59 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(a A-b B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\sec ^2(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{2 d} \]

[Out]

1/2*(A*a-B*b)*arctanh(sin(d*x+c))/d+1/2*sec(d*x+c)^2*(A*b+B*a+(A*a+B*b)*sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 792, 212} \[ \int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(a A-b B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\sec ^2(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{2 d} \]

[In]

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

[Out]

((a*A - b*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (Sec[c + d*x]^2*(A*b + a*B + (a*A + b*B)*Sin[c + d*x]))/(2*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {b^3 \text {Subst}\left (\int \frac {(a+x) \left (A+\frac {B x}{b}\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^2(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}+\frac {(b (a A-b B)) \text {Subst}\left (\int \frac {1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 d} \\ & = \frac {(a A-b B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\sec ^2(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {(a A-b B) \text {arctanh}(\sin (c+d x))+\sec ^2(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{2 d} \]

[In]

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

[Out]

((a*A - b*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]^2*(A*b + a*B + (a*A + b*B)*Sin[c + d*x]))/(2*d)

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.86

method	result	size
derivativedivides	\(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B a}{2 \cos \left (d x +c \right )^{2}}+\frac {A b}{2 \cos \left (d x +c \right )^{2}}+B b \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(110\)
default	\(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B a}{2 \cos \left (d x +c \right )^{2}}+\frac {A b}{2 \cos \left (d x +c \right )^{2}}+B b \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(110\)
parallelrisch	\(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-A b -B a \right ) \cos \left (2 d x +2 c \right )+\left (2 a A +2 B b \right ) \sin \left (d x +c \right )+A b +B a}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(126\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (A a \,{\mathrm e}^{2 i \left (d x +c \right )}+B b \,{\mathrm e}^{2 i \left (d x +c \right )}-a A +2 i A b \,{\mathrm e}^{i \left (d x +c \right )}-B b +2 i B a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 d}\)	\(177\)
norman	\(\frac {\frac {\left (a A +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (a A +B b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (a A +B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (a A +B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 A b +2 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\left (a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (a A -B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(243\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.56 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (A a - B b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a - B b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a + 2 \, A b + 2 \, {\left (A a + B b\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.32 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (A a - B b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a - B b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (B a + A b + {\left (A a + B b\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

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

[Out]

1/4*((A*a - B*b)*log(sin(d*x + c) + 1) - (A*a - B*b)*log(sin(d*x + c) - 1) - 2*(B*a + A*b + (A*a + B*b)*sin(d*
x + c))/(sin(d*x + c)^2 - 1))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.42 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (A a \sin \left (d x + c\right ) + B b \sin \left (d x + c\right ) + B a + A b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

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

[Out]

1/4*((A*a - B*b)*log(abs(sin(d*x + c) + 1)) - (A*a - B*b)*log(abs(sin(d*x + c) - 1)) - 2*(A*a*sin(d*x + c) + B
*b*sin(d*x + c) + B*a + A*b)/(sin(d*x + c)^2 - 1))/d

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.07 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {A\,a}{2}-\frac {B\,b}{2}\right )}{d}-\frac {\frac {A\,b}{2}+\frac {B\,a}{2}+\sin \left (c+d\,x\right )\,\left (\frac {A\,a}{2}+\frac {B\,b}{2}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \]

[In]

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

[Out]

(atanh(sin(c + d*x))*((A*a)/2 - (B*b)/2))/d - ((A*b)/2 + (B*a)/2 + sin(c + d*x)*((A*a)/2 + (B*b)/2))/(d*(sin(c
+ d*x)^2 - 1))