\(\int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [958]

Optimal result

Integrand size = 29, antiderivative size = 47 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a (A-B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (A+B)}{2 d (a-a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 78, 212} \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a^2 (A+B)}{2 d (a-a \sin (c+d x))}+\frac {a (A-B) \text {arctanh}(\sin (c+d x))}{2 d} \]

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 2915

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

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

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left ((A-B) \text {arctanh}(\sin (c+d x))-\frac {A+B}{-1+\sin (c+d x)}\right )}{2 d} \]

[In]

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

[Out]

(a*((A - B)*ArcTanh[Sin[c + d*x]] - (A + B)/(-1 + Sin[c + d*x])))/(2*d)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.74

method	result	size
parallelrisch	\(\frac {\left (-\left (\sin \left (d x +c \right )-1\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\sin \left (d x +c \right )-1\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\left (A +B \right ) \sin \left (d x +c \right )\right ) a}{2 d \left (\sin \left (d x +c \right )-1\right )}\)	\(82\)
derivativedivides	\(\frac {\frac {a A}{2 \cos \left (d x +c \right )^{2}}+B a \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B a}{2 \cos \left (d x +c \right )^{2}}}{d}\)	\(110\)
default	\(\frac {\frac {a A}{2 \cos \left (d x +c \right )^{2}}+B a \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B a}{2 \cos \left (d x +c \right )^{2}}}{d}\)	\(110\)
risch	\(-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )} \left (A +B \right )}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\)	\(115\)
norman	\(\frac {\frac {\left (A +B \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (A +B \right ) a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 a A +2 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A +B \right ) a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (A +B \right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (A +B \right ) a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A +B \right ) a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {a \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(221\)

[In]

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

[Out]

1/2*(-(sin(d*x+c)-1)*(A-B)*ln(tan(1/2*d*x+1/2*c)-1)+(sin(d*x+c)-1)*(A-B)*ln(tan(1/2*d*x+1/2*c)+1)-(A+B)*sin(d*
x+c))*a/d/(sin(d*x+c)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (44) = 88\).

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.91 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {2 \, {\left (A + B\right )} a - {\left ({\left (A - B\right )} a \sin \left (d x + c\right ) - {\left (A - B\right )} a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A - B\right )} a \sin \left (d x + c\right ) - {\left (A - B\right )} a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{4 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.17 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (A - B\right )} a \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A - B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (A + B\right )} a}{\sin \left (d x + c\right ) - 1}}{4 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.79 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {A a \sin \left (d x + c\right ) - B a \sin \left (d x + c\right ) - 3 \, A a - B a}{\sin \left (d x + c\right ) - 1}}{4 \, d} \]

[In]

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

[Out]

1/4*((A*a - B*a)*log(abs(sin(d*x + c) + 1)) - (A*a - B*a)*log(abs(sin(d*x + c) - 1)) + (A*a*sin(d*x + c) - B*a
*sin(d*x + c) - 3*A*a - B*a)/(sin(d*x + c) - 1))/d

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A-B\right )}{2\,d}-\frac {\frac {A\,a}{2}+\frac {B\,a}{2}}{d\,\left (\sin \left (c+d\,x\right )-1\right )} \]

[In]

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

[Out]

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