\(\int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [975]

Optimal result

Integrand size = 31, antiderivative size = 77 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (A-B) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4 (A+B)}{4 d (a-a \sin (c+d x))^2}+\frac {a^3 (A-B)}{4 d (a-a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 77, 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} \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^4 (A+B)}{4 d (a-a \sin (c+d x))^2}+\frac {a^3 (A-B)}{4 d (a-a \sin (c+d x))}+\frac {a^2 (A-B) \text {arctanh}(\sin (c+d x))}{4 d} \]

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(A - B)*ArcTanh[Sin[c + d*x]])/(4*d) + (a^4*(A + B))/(4*d*(a - a*Sin[c + d*x])^2) + (a^3*(A - B))/(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^5 \text {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^5 \text {Subst}\left (\int \left (\frac {A+B}{2 a (a-x)^3}+\frac {A-B}{4 a^2 (a-x)^2}+\frac {A-B}{4 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^4 (A+B)}{4 d (a-a \sin (c+d x))^2}+\frac {a^3 (A-B)}{4 d (a-a \sin (c+d x))}+\frac {\left (a^3 (A-B)\right ) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{4 d} \\ & = \frac {a^2 (A-B) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4 (A+B)}{4 d (a-a \sin (c+d x))^2}+\frac {a^3 (A-B)}{4 d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^5 \left (\frac {(A-B) \text {arctanh}(\sin (c+d x))}{4 a^3}+\frac {A+B}{4 a (a-a \sin (c+d x))^2}+\frac {A-B}{4 a^2 (a-a \sin (c+d x))}\right )}{d} \]

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^5*(((A - B)*ArcTanh[Sin[c + d*x]])/(4*a^3) + (A + B)/(4*a*(a - a*Sin[c + d*x])^2) + (A - B)/(4*a^2*(a - a*S
in[c + d*x]))))/d

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.75

method	result	size
parallelrisch	\(-\frac {\left (\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 A \cos \left (2 d x +2 c \right )+\left (2 B +6 A \right ) \sin \left (d x +c \right )-2 A \right ) a^{2}}{4 d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\)	\(135\)
risch	\(\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )} \left (4 i A \,{\mathrm e}^{i \left (d x +c \right )}-A \,{\mathrm e}^{2 i \left (d x +c \right )}+B \,{\mathrm e}^{2 i \left (d x +c \right )}+A -B \right )}{2 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{4 d}\)	\(167\)
derivativedivides	\(\frac {A \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+\frac {A \,a^{2}}{2 \cos \left (d x +c \right )^{4}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,a^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(238\)
default	\(\frac {A \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+\frac {A \,a^{2}}{2 \cos \left (d x +c \right )^{4}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,a^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(238\)
norman	\(\frac {\frac {\left (4 A \,a^{2}+2 B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 A \,a^{2}+2 B \,a^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (12 A \,a^{2}+10 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (12 A \,a^{2}+10 B \,a^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (16 A \,a^{2}+20 B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (16 A \,a^{2}+20 B \,a^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (7 A +5 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (7 A +5 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (3 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {a^{2} \left (3 A +B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 a^{2} \left (9 A +11 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (29 A +31 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a^{2} \left (29 A +31 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {a^{2} \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {a^{2} \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(426\)

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/4*((cos(2*d*x+2*c)-3+4*sin(d*x+c))*(A-B)*ln(tan(1/2*d*x+1/2*c)-1)-(cos(2*d*x+2*c)-3+4*sin(d*x+c))*(A-B)*ln(
tan(1/2*d*x+1/2*c)+1)+2*A*cos(2*d*x+2*c)+(2*B+6*A)*sin(d*x+c)-2*A)*a^2/d/(cos(2*d*x+2*c)-3+4*sin(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (73) = 146\).

Time = 0.29 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.09 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {2 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 4 \, A a^{2} + {\left ({\left (A - B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 2 \, {\left (A - B\right )} a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A - B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 2 \, {\left (A - B\right )} a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 2 \, A a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.69 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {2 \, {\left (A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \, {\left (A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {3 \, A a^{2} \sin \left (d x + c\right )^{2} - 3 \, B a^{2} \sin \left (d x + c\right )^{2} - 10 \, A a^{2} \sin \left (d x + c\right ) + 10 \, B a^{2} \sin \left (d x + c\right ) + 11 \, A a^{2} - 3 \, B a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*(2*(A*a^2 - B*a^2)*log(abs(sin(d*x + c) + 1)) - 2*(A*a^2 - B*a^2)*log(abs(sin(d*x + c) - 1)) + (3*A*a^2*s
in(d*x + c)^2 - 3*B*a^2*sin(d*x + c)^2 - 10*A*a^2*sin(d*x + c) + 10*B*a^2*sin(d*x + c) + 11*A*a^2 - 3*B*a^2)/(
sin(d*x + c) - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 9.91 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\frac {A\,a^2}{2}-\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{4}-\frac {B\,a^2}{4}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )}+\frac {a^2\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A-B\right )}{4\,d} \]

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^5,x)

[Out]

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