\(\int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\) [1008]

Optimal result

Integrand size = 31, antiderivative size = 91 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {(3 A+B) \text {arctanh}(\sin (c+d x))}{8 a d}+\frac {A+B}{8 d (a-a \sin (c+d x))}-\frac {a (A-B)}{8 d (a+a \sin (c+d x))^2}-\frac {A}{4 d (a+a \sin (c+d x))} \]

[Out]

1/8*(3*A+B)*arctanh(sin(d*x+c))/a/d+1/8*(A+B)/d/(a-a*sin(d*x+c))-1/8*a*(A-B)/d/(a+a*sin(d*x+c))^2-1/4*A/d/(a+a
*sin(d*x+c))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2915, 78, 212} \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {(3 A+B) \text {arctanh}(\sin (c+d x))}{8 a d}+\frac {A+B}{8 d (a-a \sin (c+d x))}-\frac {a (A-B)}{8 d (a \sin (c+d x)+a)^2}-\frac {A}{4 d (a \sin (c+d x)+a)} \]

[In]

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

[Out]

((3*A + B)*ArcTanh[Sin[c + d*x]])/(8*a*d) + (A + B)/(8*d*(a - a*Sin[c + d*x])) - (a*(A - B))/(8*d*(a + a*Sin[c
+ d*x])^2) - A/(4*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)^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^3 \text {Subst}\left (\int \left (\frac {A+B}{8 a^3 (a-x)^2}+\frac {A-B}{4 a^2 (a+x)^3}+\frac {A}{4 a^3 (a+x)^2}+\frac {3 A+B}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {A+B}{8 d (a-a \sin (c+d x))}-\frac {a (A-B)}{8 d (a+a \sin (c+d x))^2}-\frac {A}{4 d (a+a \sin (c+d x))}+\frac {(3 A+B) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d} \\ & = \frac {(3 A+B) \text {arctanh}(\sin (c+d x))}{8 a d}+\frac {A+B}{8 d (a-a \sin (c+d x))}-\frac {a (A-B)}{8 d (a+a \sin (c+d x))^2}-\frac {A}{4 d (a+a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {(3 A+B) \text {arctanh}(\sin (c+d x))}{a}+\frac {-A+B}{a (1+\sin (c+d x))^2}+\frac {A+B}{a-a \sin (c+d x)}-\frac {2 A}{a+a \sin (c+d x)}}{8 d} \]

[In]

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

[Out]

(((3*A + B)*ArcTanh[Sin[c + d*x]])/a + (-A + B)/(a*(1 + Sin[c + d*x])^2) + (A + B)/(a - a*Sin[c + d*x]) - (2*A
)/(a + a*Sin[c + d*x]))/(8*d)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {-\frac {A}{4 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{4}-\frac {B}{4}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {3 A}{16}+\frac {B}{16}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {3 A}{16}-\frac {B}{16}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {\frac {A}{8}+\frac {B}{8}}{\sin \left (d x +c \right )-1}}{d a}\)	\(94\)
default	\(\frac {-\frac {A}{4 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{4}-\frac {B}{4}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {3 A}{16}+\frac {B}{16}\right ) \ln \left (1+\sin \left (d x +c \right )\right )+\left (-\frac {3 A}{16}-\frac {B}{16}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {\frac {A}{8}+\frac {B}{8}}{\sin \left (d x +c \right )-1}}{d a}\)	\(94\)
parallelrisch	\(\frac {-3 \left (A +\frac {B}{3}\right ) \left (\frac {\sin \left (3 d x +3 c \right )}{2}+\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (A +\frac {B}{3}\right ) \left (\frac {\sin \left (3 d x +3 c \right )}{2}+\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-A -3 B \right ) \cos \left (2 d x +2 c \right )+\left (A -B \right ) \sin \left (3 d x +3 c \right )+\left (7 A +B \right ) \sin \left (d x +c \right )+A +3 B}{4 a d \left (\sin \left (3 d x +3 c \right )+\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )+2\right )}\)	\(186\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (6 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+3 A \,{\mathrm e}^{4 i \left (d x +c \right )}+2 i B \,{\mathrm e}^{3 i \left (d x +c \right )}+B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 i A \,{\mathrm e}^{i \left (d x +c \right )}+2 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i B \,{\mathrm e}^{i \left (d x +c \right )}-10 B \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A +B \right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d a}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 a d}\)	\(252\)
norman	\(\frac {\frac {\left (A +3 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (A +3 B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}+\frac {\left (A +3 B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}+\frac {\left (7 A +5 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {\left (7 A +5 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {\left (5 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (5 A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {\left (3 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 a d}+\frac {\left (3 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 a d}\)	\(268\)

[In]

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

[Out]

1/d/a*(-1/4*A/(1+sin(d*x+c))-1/2*(1/4*A-1/4*B)/(1+sin(d*x+c))^2+(3/16*A+1/16*B)*ln(1+sin(d*x+c))+(-3/16*A-1/16
*B)*ln(sin(d*x+c)-1)-(1/8*A+1/8*B)/(sin(d*x+c)-1))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.77 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right ) - 2 \, A - 6 \, B}{16 \, {\left (a d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

-1/16*(2*(3*A + B)*cos(d*x + c)^2 - ((3*A + B)*cos(d*x + c)^2*sin(d*x + c) + (3*A + B)*cos(d*x + c)^2)*log(sin
(d*x + c) + 1) + ((3*A + B)*cos(d*x + c)^2*sin(d*x + c) + (3*A + B)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2
*(3*A + B)*sin(d*x + c) - 2*A - 6*B)/(a*d*cos(d*x + c)^2*sin(d*x + c) + a*d*cos(d*x + c)^2)

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

(Integral(A*sec(c + d*x)**3/(sin(c + d*x) + 1), x) + Integral(B*sin(c + d*x)*sec(c + d*x)**3/(sin(c + d*x) + 1
), x))/a

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.24 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {{\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left ({\left (3 \, A + B\right )} \sin \left (d x + c\right )^{2} + {\left (3 \, A + B\right )} \sin \left (d x + c\right ) - 2 \, A + 2 \, B\right )}}{a \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a}}{16 \, d} \]

[In]

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

[Out]

1/16*((3*A + B)*log(sin(d*x + c) + 1)/a - (3*A + B)*log(sin(d*x + c) - 1)/a - 2*((3*A + B)*sin(d*x + c)^2 + (3
*A + B)*sin(d*x + c) - 2*A + 2*B)/(a*sin(d*x + c)^3 + a*sin(d*x + c)^2 - a*sin(d*x + c) - a))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {2 \, {\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (3 \, A \sin \left (d x + c\right ) + B \sin \left (d x + c\right ) - 5 \, A - 3 \, B\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {9 \, A \sin \left (d x + c\right )^{2} + 3 \, B \sin \left (d x + c\right )^{2} + 26 \, A \sin \left (d x + c\right ) + 6 \, B \sin \left (d x + c\right ) + 21 \, A - B}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{32 \, d} \]

[In]

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

[Out]

1/32*(2*(3*A + B)*log(abs(sin(d*x + c) + 1))/a - 2*(3*A + B)*log(abs(sin(d*x + c) - 1))/a + 2*(3*A*sin(d*x + c
) + B*sin(d*x + c) - 5*A - 3*B)/(a*(sin(d*x + c) - 1)) - (9*A*sin(d*x + c)^2 + 3*B*sin(d*x + c)^2 + 26*A*sin(d
*x + c) + 6*B*sin(d*x + c) + 21*A - B)/(a*(sin(d*x + c) + 1)^2))/d

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\left (\frac {3\,A}{8}+\frac {B}{8}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {3\,A}{8}+\frac {B}{8}\right )\,\sin \left (c+d\,x\right )-\frac {A}{4}+\frac {B}{4}}{d\,\left (-a\,{\sin \left (c+d\,x\right )}^3-a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (3\,A+B\right )}{8\,a\,d} \]

[In]

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

[Out]

(B/4 - A/4 + sin(c + d*x)*((3*A)/8 + B/8) + sin(c + d*x)^2*((3*A)/8 + B/8))/(d*(a + a*sin(c + d*x) - a*sin(c +
d*x)^2 - a*sin(c + d*x)^3)) + (atanh(sin(c + d*x))*(3*A + B))/(8*a*d)