3.1017 \(\int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=179 \[ -\frac {a^2 (A-B)}{32 d (a \sin (c+d x)+a)^4}+\frac {5 A+3 B}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 A+B}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {5 (3 A+B) \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac {a (3 A-B)}{48 d (a \sin (c+d x)+a)^3}+\frac {A+B}{64 d (a-a \sin (c+d x))^2}-\frac {3 A}{32 d (a \sin (c+d x)+a)^2} \]

[Out]

5/64*(3*A+B)*arctanh(sin(d*x+c))/a^2/d+1/64*(A+B)/d/(a-a*sin(d*x+c))^2-1/32*a^2*(A-B)/d/(a+a*sin(d*x+c))^4-1/4
8*a*(3*A-B)/d/(a+a*sin(d*x+c))^3-3/32*A/d/(a+a*sin(d*x+c))^2+1/64*(5*A+3*B)/d/(a^2-a^2*sin(d*x+c))+1/32*(-5*A-
B)/d/(a^2+a^2*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2836, 77, 206} \[ -\frac {a^2 (A-B)}{32 d (a \sin (c+d x)+a)^4}+\frac {5 A+3 B}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 A+B}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {5 (3 A+B) \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac {a (3 A-B)}{48 d (a \sin (c+d x)+a)^3}+\frac {A+B}{64 d (a-a \sin (c+d x))^2}-\frac {3 A}{32 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(3*A + B)*ArcTanh[Sin[c + d*x]])/(64*a^2*d) + (A + B)/(64*d*(a - a*Sin[c + d*x])^2) - (a^2*(A - B))/(32*d*(
a + a*Sin[c + d*x])^4) - (a*(3*A - B))/(48*d*(a + a*Sin[c + d*x])^3) - (3*A)/(32*d*(a + a*Sin[c + d*x])^2) + (
5*A + 3*B)/(64*d*(a^2 - a^2*Sin[c + d*x])) - (5*A + B)/(32*d*(a^2 + a^2*Sin[c + d*x]))

Rule 77

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 206

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

Rule 2836

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*x)/b
)^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*} \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^3 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \operatorname {Subst}\left (\int \left (\frac {A+B}{32 a^5 (a-x)^3}+\frac {5 A+3 B}{64 a^6 (a-x)^2}+\frac {A-B}{8 a^3 (a+x)^5}+\frac {3 A-B}{16 a^4 (a+x)^4}+\frac {3 A}{16 a^5 (a+x)^3}+\frac {5 A+B}{32 a^6 (a+x)^2}+\frac {5 (3 A+B)}{64 a^6 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {A+B}{64 d (a-a \sin (c+d x))^2}-\frac {a^2 (A-B)}{32 d (a+a \sin (c+d x))^4}-\frac {a (3 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {3 A}{32 d (a+a \sin (c+d x))^2}+\frac {5 A+3 B}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 A+B}{32 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {(5 (3 A+B)) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{64 a d}\\ &=\frac {5 (3 A+B) \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}+\frac {A+B}{64 d (a-a \sin (c+d x))^2}-\frac {a^2 (A-B)}{32 d (a+a \sin (c+d x))^4}-\frac {a (3 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {3 A}{32 d (a+a \sin (c+d x))^2}+\frac {5 A+3 B}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 A+B}{32 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.69, size = 123, normalized size = 0.69 \[ \frac {-\frac {3 (5 A+3 B)}{\sin (c+d x)-1}-\frac {6 (5 A+B)}{\sin (c+d x)+1}+\frac {3 (A+B)}{(\sin (c+d x)-1)^2}+\frac {4 (B-3 A)}{(\sin (c+d x)+1)^3}-\frac {6 (A-B)}{(\sin (c+d x)+1)^4}+15 (3 A+B) \tanh ^{-1}(\sin (c+d x))-\frac {18 A}{(\sin (c+d x)+1)^2}}{192 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(3*A + B)*ArcTanh[Sin[c + d*x]] + (3*(A + B))/(-1 + Sin[c + d*x])^2 - (3*(5*A + 3*B))/(-1 + Sin[c + d*x])
- (6*(A - B))/(1 + Sin[c + d*x])^4 + (4*(-3*A + B))/(1 + Sin[c + d*x])^3 - (18*A)/(1 + Sin[c + d*x])^2 - (6*(5
*A + B))/(1 + Sin[c + d*x]))/(192*a^2*d)

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 260, normalized size = 1.45 \[ \frac {60 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4} - 20 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4} - 20 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} - 36 \, A - 12 \, B\right )} \sin \left (d x + c\right ) - 24 \, A - 72 \, B}{384 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(60*(3*A + B)*cos(d*x + c)^4 - 20*(3*A + B)*cos(d*x + c)^2 + 15*((3*A + B)*cos(d*x + c)^6 - 2*(3*A + B)*
cos(d*x + c)^4*sin(d*x + c) - 2*(3*A + B)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - 15*((3*A + B)*cos(d*x + c)^6
- 2*(3*A + B)*cos(d*x + c)^4*sin(d*x + c) - 2*(3*A + B)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) + 2*(15*(3*A +
B)*cos(d*x + c)^4 - 20*(3*A + B)*cos(d*x + c)^2 - 36*A - 12*B)*sin(d*x + c) - 24*A - 72*B)/(a^2*d*cos(d*x + c
)^6 - 2*a^2*d*cos(d*x + c)^4*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^4)

________________________________________________________________________________________

giac [A]  time = 0.34, size = 214, normalized size = 1.20 \[ \frac {\frac {60 \, {\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {60 \, {\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (45 \, A \sin \left (d x + c\right )^{2} + 15 \, B \sin \left (d x + c\right )^{2} - 110 \, A \sin \left (d x + c\right ) - 42 \, B \sin \left (d x + c\right ) + 69 \, A + 31 \, B\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {375 \, A \sin \left (d x + c\right )^{4} + 125 \, B \sin \left (d x + c\right )^{4} + 1740 \, A \sin \left (d x + c\right )^{3} + 548 \, B \sin \left (d x + c\right )^{3} + 3114 \, A \sin \left (d x + c\right )^{2} + 894 \, B \sin \left (d x + c\right )^{2} + 2604 \, A \sin \left (d x + c\right ) + 612 \, B \sin \left (d x + c\right ) + 903 \, A + 93 \, B}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{1536 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1536*(60*(3*A + B)*log(abs(sin(d*x + c) + 1))/a^2 - 60*(3*A + B)*log(abs(sin(d*x + c) - 1))/a^2 + 6*(45*A*si
n(d*x + c)^2 + 15*B*sin(d*x + c)^2 - 110*A*sin(d*x + c) - 42*B*sin(d*x + c) + 69*A + 31*B)/(a^2*(sin(d*x + c)
- 1)^2) - (375*A*sin(d*x + c)^4 + 125*B*sin(d*x + c)^4 + 1740*A*sin(d*x + c)^3 + 548*B*sin(d*x + c)^3 + 3114*A
*sin(d*x + c)^2 + 894*B*sin(d*x + c)^2 + 2604*A*sin(d*x + c) + 612*B*sin(d*x + c) + 903*A + 93*B)/(a^2*(sin(d*
x + c) + 1)^4))/d

________________________________________________________________________________________

maple [A]  time = 0.76, size = 283, normalized size = 1.58 \[ -\frac {15 \ln \left (\sin \left (d x +c \right )-1\right ) A}{128 d \,a^{2}}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right ) B}{128 d \,a^{2}}+\frac {A}{64 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {B}{64 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5 A}{64 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 B}{64 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 A}{32 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {A}{32 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {B}{32 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {A}{16 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {B}{48 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5 A}{32 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {B}{32 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right ) A}{128 d \,a^{2}}+\frac {5 B \ln \left (1+\sin \left (d x +c \right )\right )}{128 a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/128/d/a^2*ln(sin(d*x+c)-1)*A-5/128/d/a^2*ln(sin(d*x+c)-1)*B+1/64/d/a^2/(sin(d*x+c)-1)^2*A+1/64/d/a^2/(sin(
d*x+c)-1)^2*B-5/64/d/a^2/(sin(d*x+c)-1)*A-3/64/d/a^2/(sin(d*x+c)-1)*B-3/32/d/a^2/(1+sin(d*x+c))^2*A-1/32/d/a^2
/(1+sin(d*x+c))^4*A+1/32/d/a^2/(1+sin(d*x+c))^4*B-1/16/d/a^2/(1+sin(d*x+c))^3*A+1/48/d/a^2/(1+sin(d*x+c))^3*B-
5/32/d/a^2/(1+sin(d*x+c))*A-1/32/d/a^2/(1+sin(d*x+c))*B+15/128/d/a^2*ln(1+sin(d*x+c))*A+5/128*B*ln(1+sin(d*x+c
))/a^2/d

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 207, normalized size = 1.16 \[ -\frac {\frac {2 \, {\left (15 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right )^{5} + 30 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right )^{4} - 10 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right )^{3} - 50 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right )^{2} - 17 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right ) + 48 \, A - 16 \, B\right )}}{a^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{2} \sin \left (d x + c\right )^{5} - a^{2} \sin \left (d x + c\right )^{4} - 4 \, a^{2} \sin \left (d x + c\right )^{3} - a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {15 \, {\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {15 \, {\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{384 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(2*(15*(3*A + B)*sin(d*x + c)^5 + 30*(3*A + B)*sin(d*x + c)^4 - 10*(3*A + B)*sin(d*x + c)^3 - 50*(3*A +
B)*sin(d*x + c)^2 - 17*(3*A + B)*sin(d*x + c) + 48*A - 16*B)/(a^2*sin(d*x + c)^6 + 2*a^2*sin(d*x + c)^5 - a^2
*sin(d*x + c)^4 - 4*a^2*sin(d*x + c)^3 - a^2*sin(d*x + c)^2 + 2*a^2*sin(d*x + c) + a^2) - 15*(3*A + B)*log(sin
(d*x + c) + 1)/a^2 + 15*(3*A + B)*log(sin(d*x + c) - 1)/a^2)/d

________________________________________________________________________________________

mupad [B]  time = 9.50, size = 193, normalized size = 1.08 \[ \frac {\left (-\frac {15\,A}{64}-\frac {5\,B}{64}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-\frac {15\,A}{32}-\frac {5\,B}{32}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {5\,A}{32}+\frac {5\,B}{96}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {25\,A}{32}+\frac {25\,B}{96}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {17\,A}{64}+\frac {17\,B}{192}\right )\,\sin \left (c+d\,x\right )-\frac {A}{4}+\frac {B}{12}}{d\,\left (a^2\,{\sin \left (c+d\,x\right )}^6+2\,a^2\,{\sin \left (c+d\,x\right )}^5-a^2\,{\sin \left (c+d\,x\right )}^4-4\,a^2\,{\sin \left (c+d\,x\right )}^3-a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )}+\frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (3\,A+B\right )}{64\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B/12 - A/4 + sin(c + d*x)*((17*A)/64 + (17*B)/192) - sin(c + d*x)^4*((15*A)/32 + (5*B)/32) + sin(c + d*x)^3*(
(5*A)/32 + (5*B)/96) - sin(c + d*x)^5*((15*A)/64 + (5*B)/64) + sin(c + d*x)^2*((25*A)/32 + (25*B)/96))/(d*(2*a
^2*sin(c + d*x) + a^2 - a^2*sin(c + d*x)^2 - 4*a^2*sin(c + d*x)^3 - a^2*sin(c + d*x)^4 + 2*a^2*sin(c + d*x)^5
+ a^2*sin(c + d*x)^6)) + (5*atanh(sin(c + d*x))*(3*A + B))/(64*a^2*d)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________