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]

((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]))

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Rubi [A]  time = 0.18878, antiderivative size = 146, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.523395, 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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Maple [A]  time = 0.099, size = 245, normalized size = 1.7 \begin{align*} -{\frac{5\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) A}{32\,da}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) B}{32\,da}}+{\frac{A}{32\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{B}{32\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{A}{8\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{B}{16\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{3\,A}{16\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{A}{24\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{B}{24\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{3\,A}{32\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{B}{32\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{5\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) A}{32\,da}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) B}{32\,da}} \end{align*}

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/d/a*ln(sin(d*x+c)-1)*A-1/32/d/a*ln(sin(d*x+c)-1)*B+1/32/d/a/(sin(d*x+c)-1)^2*A+1/32/d/a/(sin(d*x+c)-1)^2
*B-1/8/d/a/(sin(d*x+c)-1)*A-1/16/d/a/(sin(d*x+c)-1)*B-3/16/d/a/(1+sin(d*x+c))*A-1/24/d/a/(1+sin(d*x+c))^3*A+1/
24/d/a/(1+sin(d*x+c))^3*B-3/32/d/a/(1+sin(d*x+c))^2*A+1/32/d/a/(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 = 1.09045, size = 223, normalized size = 1.53 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.64501, size = 512, normalized size = 3.51 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.39077, size = 259, normalized size = 1.77 \begin{align*} \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} \end{align*}

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