3.1009 \(\int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=146 \[ -\frac {a^2 (A-B)}{24 d (a \sin (c+d x)+a)^3}+\frac {a (A+B)}{32 d (a-a \sin (c+d x))^2}-\frac {a (3 A-B)}{32 d (a \sin (c+d x)+a)^2}+\frac {2 A+B}{16 d (a-a \sin (c+d x))}+\frac {(5 A+B) \tanh ^{-1}(\sin (c+d x))}{16 a d}-\frac {3 A}{16 d (a \sin (c+d x)+a)} \]

[Out]

1/16*(5*A+B)*arctanh(sin(d*x+c))/a/d+1/32*a*(A+B)/d/(a-a*sin(d*x+c))^2+1/16*(2*A+B)/d/(a-a*sin(d*x+c))-1/24*a^
2*(A-B)/d/(a+a*sin(d*x+c))^3-1/32*a*(3*A-B)/d/(a+a*sin(d*x+c))^2-3/16*A/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.19, antiderivative size = 146, 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} \[ -\frac {a^2 (A-B)}{24 d (a \sin (c+d x)+a)^3}+\frac {a (A+B)}{32 d (a-a \sin (c+d x))^2}-\frac {a (3 A-B)}{32 d (a \sin (c+d x)+a)^2}+\frac {2 A+B}{16 d (a-a \sin (c+d x))}+\frac {(5 A+B) \tanh ^{-1}(\sin (c+d x))}{16 a d}-\frac {3 A}{16 d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*A + B)*ArcTanh[Sin[c + d*x]])/(16*a*d) + (a*(A + B))/(32*d*(a - a*Sin[c + d*x])^2) + (2*A + B)/(16*d*(a -
a*Sin[c + d*x])) - (a^2*(A - B))/(24*d*(a + a*Sin[c + d*x])^3) - (a*(3*A - B))/(32*d*(a + a*Sin[c + d*x])^2) -
(3*A)/(16*d*(a + a*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)} \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^3 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \operatorname {Subst}\left (\int \left (\frac {A+B}{16 a^4 (a-x)^3}+\frac {2 A+B}{16 a^5 (a-x)^2}+\frac {A-B}{8 a^3 (a+x)^4}+\frac {3 A-B}{16 a^4 (a+x)^3}+\frac {3 A}{16 a^5 (a+x)^2}+\frac {5 A+B}{16 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a (A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {2 A+B}{16 d (a-a \sin (c+d x))}-\frac {a^2 (A-B)}{24 d (a+a \sin (c+d x))^3}-\frac {a (3 A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {3 A}{16 d (a+a \sin (c+d x))}+\frac {(5 A+B) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{16 d}\\ &=\frac {(5 A+B) \tanh ^{-1}(\sin (c+d x))}{16 a d}+\frac {a (A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {2 A+B}{16 d (a-a \sin (c+d x))}-\frac {a^2 (A-B)}{24 d (a+a \sin (c+d x))^3}-\frac {a (3 A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {3 A}{16 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 105, normalized size = 0.72 \[ \frac {-\frac {6 (2 A+B)}{\sin (c+d x)-1}+\frac {3 (A+B)}{(\sin (c+d x)-1)^2}+\frac {3 B-9 A}{(\sin (c+d x)+1)^2}-\frac {4 (A-B)}{(\sin (c+d x)+1)^3}+6 (5 A+B) \tanh ^{-1}(\sin (c+d x))-\frac {18 A}{\sin (c+d x)+1}}{96 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(5*A + B)*ArcTanh[Sin[c + d*x]] + (3*(A + B))/(-1 + Sin[c + d*x])^2 - (6*(2*A + B))/(-1 + Sin[c + d*x]) - (
4*(A - B))/(1 + Sin[c + d*x])^3 + (-9*A + 3*B)/(1 + Sin[c + d*x])^2 - (18*A)/(1 + Sin[c + d*x]))/(96*a*d)

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fricas [A]  time = 0.72, size = 194, normalized size = 1.33 \[ -\frac {6 \, {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{2} + 10 \, A + 2 \, B\right )} \sin \left (d x + c\right ) - 4 \, A - 20 \, B}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \]

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)),x, algorithm="fricas")

[Out]

-1/96*(6*(5*A + B)*cos(d*x + c)^4 - 2*(5*A + B)*cos(d*x + c)^2 - 3*((5*A + B)*cos(d*x + c)^4*sin(d*x + c) + (5
*A + B)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) + 3*((5*A + B)*cos(d*x + c)^4*sin(d*x + c) + (5*A + B)*cos(d*x +
c)^4)*log(-sin(d*x + c) + 1) - 2*(3*(5*A + B)*cos(d*x + c)^2 + 10*A + 2*B)*sin(d*x + c) - 4*A - 20*B)/(a*d*co
s(d*x + c)^4*sin(d*x + c) + a*d*cos(d*x + c)^4)

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giac [A]  time = 0.26, size = 192, normalized size = 1.32 \[ \frac {\frac {6 \, {\left (5 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {6 \, {\left (5 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {3 \, {\left (15 \, A \sin \left (d x + c\right )^{2} + 3 \, B \sin \left (d x + c\right )^{2} - 38 \, A \sin \left (d x + c\right ) - 10 \, B \sin \left (d x + c\right ) + 25 \, A + 9 \, B\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {55 \, A \sin \left (d x + c\right )^{3} + 11 \, B \sin \left (d x + c\right )^{3} + 201 \, A \sin \left (d x + c\right )^{2} + 33 \, B \sin \left (d x + c\right )^{2} + 255 \, A \sin \left (d x + c\right ) + 27 \, B \sin \left (d x + c\right ) + 117 \, A - 3 \, B}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{192 \, d} \]

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)),x, algorithm="giac")

[Out]

1/192*(6*(5*A + B)*log(abs(sin(d*x + c) + 1))/a - 6*(5*A + B)*log(abs(sin(d*x + c) - 1))/a + 3*(15*A*sin(d*x +
c)^2 + 3*B*sin(d*x + c)^2 - 38*A*sin(d*x + c) - 10*B*sin(d*x + c) + 25*A + 9*B)/(a*(sin(d*x + c) - 1)^2) - (5
5*A*sin(d*x + c)^3 + 11*B*sin(d*x + c)^3 + 201*A*sin(d*x + c)^2 + 33*B*sin(d*x + c)^2 + 255*A*sin(d*x + c) + 2
7*B*sin(d*x + c) + 117*A - 3*B)/(a*(sin(d*x + c) + 1)^3))/d

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maple [A]  time = 0.53, size = 245, normalized size = 1.68 \[ -\frac {5 \ln \left (\sin \left (d x +c \right )-1\right ) A}{32 a d}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) B}{32 a d}+\frac {A}{32 a d \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {B}{32 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {A}{8 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {B}{16 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {3 A}{16 a d \left (1+\sin \left (d x +c \right )\right )}-\frac {A}{24 a d \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {B}{24 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3 A}{32 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {B}{32 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right ) A}{32 d a}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) B}{32 d a} \]

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)),x)

[Out]

-5/32/a/d*ln(sin(d*x+c)-1)*A-1/32/a/d*ln(sin(d*x+c)-1)*B+1/32/a/d/(sin(d*x+c)-1)^2*A+1/32/a/d/(sin(d*x+c)-1)^2
*B-1/8/a/d/(sin(d*x+c)-1)*A-1/16/a/d/(sin(d*x+c)-1)*B-3/16/a/d/(1+sin(d*x+c))*A-1/24/a/d/(1+sin(d*x+c))^3*A+1/
24/a/d/(1+sin(d*x+c))^3*B-3/32/a/d/(1+sin(d*x+c))^2*A+1/32/a/d/(1+sin(d*x+c))^2*B+5/32/d/a*ln(1+sin(d*x+c))*A+
1/32/d/a*ln(1+sin(d*x+c))*B

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maxima [A]  time = 0.35, size = 165, normalized size = 1.13 \[ \frac {\frac {3 \, {\left (5 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {3 \, {\left (5 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left (3 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{4} + 3 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{3} - 5 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{2} - 5 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right ) + 8 \, A - 8 \, B\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a}}{96 \, d} \]

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)),x, algorithm="maxima")

[Out]

1/96*(3*(5*A + B)*log(sin(d*x + c) + 1)/a - 3*(5*A + B)*log(sin(d*x + c) - 1)/a - 2*(3*(5*A + B)*sin(d*x + c)^
4 + 3*(5*A + B)*sin(d*x + c)^3 - 5*(5*A + B)*sin(d*x + c)^2 - 5*(5*A + B)*sin(d*x + c) + 8*A - 8*B)/(a*sin(d*x
+ c)^5 + a*sin(d*x + c)^4 - 2*a*sin(d*x + c)^3 - 2*a*sin(d*x + c)^2 + a*sin(d*x + c) + a))/d

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mupad [B]  time = 9.18, size = 151, normalized size = 1.03 \[ \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (5\,A+B\right )}{16\,a\,d}-\frac {\left (\frac {5\,A}{16}+\frac {B}{16}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {5\,A}{16}+\frac {B}{16}\right )\,{\sin \left (c+d\,x\right )}^3+\left (-\frac {25\,A}{48}-\frac {5\,B}{48}\right )\,{\sin \left (c+d\,x\right )}^2+\left (-\frac {25\,A}{48}-\frac {5\,B}{48}\right )\,\sin \left (c+d\,x\right )+\frac {A}{6}-\frac {B}{6}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,a\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \]

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))),x)

[Out]

(atanh(sin(c + d*x))*(5*A + B))/(16*a*d) - (A/6 - B/6 - sin(c + d*x)*((25*A)/48 + (5*B)/48) + sin(c + d*x)^3*(
(5*A)/16 + B/16) + sin(c + d*x)^4*((5*A)/16 + B/16) - sin(c + d*x)^2*((25*A)/48 + (5*B)/48))/(d*(a + a*sin(c +
d*x) - 2*a*sin(c + d*x)^2 - 2*a*sin(c + d*x)^3 + a*sin(c + d*x)^4 + a*sin(c + d*x)^5))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

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)),x)

[Out]

(Integral(A*sec(c + d*x)**5/(sin(c + d*x) + 1), x) + Integral(B*sin(c + d*x)*sec(c + d*x)**5/(sin(c + d*x) + 1
), x))/a

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