3.1016 \(\int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=123 \[ \frac {A+B}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 A+B}{16 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {(2 A+B) \tanh ^{-1}(\sin (c+d x))}{8 a^2 d}-\frac {a (A-B)}{12 d (a \sin (c+d x)+a)^3}-\frac {A}{8 d (a \sin (c+d x)+a)^2} \]

[Out]

1/8*(2*A+B)*arctanh(sin(d*x+c))/a^2/d-1/12*a*(A-B)/d/(a+a*sin(d*x+c))^3-1/8*A/d/(a+a*sin(d*x+c))^2+1/16*(A+B)/
d/(a^2-a^2*sin(d*x+c))+1/16*(-3*A-B)/d/(a^2+a^2*sin(d*x+c))

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Rubi [A]  time = 0.15, antiderivative size = 123, 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} \[ \frac {A+B}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 A+B}{16 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {(2 A+B) \tanh ^{-1}(\sin (c+d x))}{8 a^2 d}-\frac {a (A-B)}{12 d (a \sin (c+d x)+a)^3}-\frac {A}{8 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

((2*A + B)*ArcTanh[Sin[c + d*x]])/(8*a^2*d) - (a*(A - B))/(12*d*(a + a*Sin[c + d*x])^3) - A/(8*d*(a + a*Sin[c
+ d*x])^2) + (A + B)/(16*d*(a^2 - a^2*Sin[c + d*x])) - (3*A + B)/(16*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 ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx &=\frac {a^3 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^2 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3 \operatorname {Subst}\left (\int \left (\frac {A+B}{16 a^4 (a-x)^2}+\frac {A-B}{4 a^2 (a+x)^4}+\frac {A}{4 a^3 (a+x)^3}+\frac {3 A+B}{16 a^4 (a+x)^2}+\frac {2 A+B}{8 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {a (A-B)}{12 d (a+a \sin (c+d x))^3}-\frac {A}{8 d (a+a \sin (c+d x))^2}+\frac {A+B}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 A+B}{16 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {(2 A+B) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 a d}\\ &=\frac {(2 A+B) \tanh ^{-1}(\sin (c+d x))}{8 a^2 d}-\frac {a (A-B)}{12 d (a+a \sin (c+d x))^3}-\frac {A}{8 d (a+a \sin (c+d x))^2}+\frac {A+B}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 A+B}{16 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 87, normalized size = 0.71 \[ -\frac {\frac {3 (A+B)}{\sin (c+d x)-1}+\frac {3 (3 A+B)}{\sin (c+d x)+1}+\frac {4 (A-B)}{(\sin (c+d x)+1)^3}-6 (2 A+B) \tanh ^{-1}(\sin (c+d x))+\frac {6 A}{(\sin (c+d x)+1)^2}}{48 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/48*(-6*(2*A + B)*ArcTanh[Sin[c + d*x]] + (3*(A + B))/(-1 + Sin[c + d*x]) + (4*(A - B))/(1 + Sin[c + d*x])^3
+ (6*A)/(1 + Sin[c + d*x])^2 + (3*(3*A + B))/(1 + Sin[c + d*x]))/(a^2*d)

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fricas [B]  time = 0.86, size = 230, normalized size = 1.87 \[ \frac {12 \, {\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (2 \, A + B\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, {\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A + B\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, {\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, {\left (2 \, A + B\right )} \cos \left (d x + c\right )^{2} - 8 \, A - 4 \, B\right )} \sin \left (d x + c\right ) - 8 \, A - 16 \, B}{48 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(12*(2*A + B)*cos(d*x + c)^2 + 3*((2*A + B)*cos(d*x + c)^4 - 2*(2*A + B)*cos(d*x + c)^2*sin(d*x + c) - 2*
(2*A + B)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 3*((2*A + B)*cos(d*x + c)^4 - 2*(2*A + B)*cos(d*x + c)^2*sin
(d*x + c) - 2*(2*A + B)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(3*(2*A + B)*cos(d*x + c)^2 - 8*A - 4*B)*si
n(d*x + c) - 8*A - 16*B)/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^2)

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giac [A]  time = 0.32, size = 169, normalized size = 1.37 \[ \frac {\frac {6 \, {\left (2 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, {\left (2 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (2 \, A \sin \left (d x + c\right ) + B \sin \left (d x + c\right ) - 3 \, A - 2 \, B\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {22 \, A \sin \left (d x + c\right )^{3} + 11 \, B \sin \left (d x + c\right )^{3} + 84 \, A \sin \left (d x + c\right )^{2} + 39 \, B \sin \left (d x + c\right )^{2} + 114 \, A \sin \left (d x + c\right ) + 45 \, B \sin \left (d x + c\right ) + 60 \, A + 9 \, B}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/96*(6*(2*A + B)*log(abs(sin(d*x + c) + 1))/a^2 - 6*(2*A + B)*log(abs(sin(d*x + c) - 1))/a^2 + 6*(2*A*sin(d*x
+ c) + B*sin(d*x + c) - 3*A - 2*B)/(a^2*(sin(d*x + c) - 1)) - (22*A*sin(d*x + c)^3 + 11*B*sin(d*x + c)^3 + 84
*A*sin(d*x + c)^2 + 39*B*sin(d*x + c)^2 + 114*A*sin(d*x + c) + 45*B*sin(d*x + c) + 60*A + 9*B)/(a^2*(sin(d*x +
c) + 1)^3))/d

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maple [A]  time = 0.76, size = 207, normalized size = 1.68 \[ -\frac {\ln \left (\sin \left (d x +c \right )-1\right ) A}{8 d \,a^{2}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) B}{16 d \,a^{2}}-\frac {A}{16 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {B}{16 d \,a^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {A}{8 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {A}{12 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {B}{12 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) A}{8 d \,a^{2}}+\frac {B \ln \left (1+\sin \left (d x +c \right )\right )}{16 a^{2} d}-\frac {3 A}{16 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {B}{16 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x)

[Out]

-1/8/d/a^2*ln(sin(d*x+c)-1)*A-1/16/d/a^2*ln(sin(d*x+c)-1)*B-1/16/d/a^2/(sin(d*x+c)-1)*A-1/16/d/a^2/(sin(d*x+c)
-1)*B-1/8/d/a^2/(1+sin(d*x+c))^2*A-1/12/d/a^2/(1+sin(d*x+c))^3*A+1/12/d/a^2/(1+sin(d*x+c))^3*B+1/8/d/a^2*ln(1+
sin(d*x+c))*A+1/16*B*ln(1+sin(d*x+c))/a^2/d-3/16/d/a^2/(1+sin(d*x+c))*A-1/16/d/a^2/(1+sin(d*x+c))*B

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maxima [A]  time = 0.42, size = 139, normalized size = 1.13 \[ -\frac {\frac {2 \, {\left (3 \, {\left (2 \, A + B\right )} \sin \left (d x + c\right )^{3} + 6 \, {\left (2 \, A + B\right )} \sin \left (d x + c\right )^{2} + {\left (2 \, A + B\right )} \sin \left (d x + c\right ) - 8 \, A + 2 \, B\right )}}{a^{2} \sin \left (d x + c\right )^{4} + 2 \, a^{2} \sin \left (d x + c\right )^{3} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac {3 \, {\left (2 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {3 \, {\left (2 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/48*(2*(3*(2*A + B)*sin(d*x + c)^3 + 6*(2*A + B)*sin(d*x + c)^2 + (2*A + B)*sin(d*x + c) - 8*A + 2*B)/(a^2*s
in(d*x + c)^4 + 2*a^2*sin(d*x + c)^3 - 2*a^2*sin(d*x + c) - a^2) - 3*(2*A + B)*log(sin(d*x + c) + 1)/a^2 + 3*(
2*A + B)*log(sin(d*x + c) - 1)/a^2)/d

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mupad [B]  time = 0.15, size = 121, normalized size = 0.98 \[ \frac {\left (\frac {A}{4}+\frac {B}{8}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {A}{2}+\frac {B}{4}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {A}{12}+\frac {B}{24}\right )\,\sin \left (c+d\,x\right )-\frac {A}{3}+\frac {B}{12}}{d\,\left (-a^2\,{\sin \left (c+d\,x\right )}^4-2\,a^2\,{\sin \left (c+d\,x\right )}^3+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (2\,A+B\right )}{8\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(c + d*x))/(cos(c + d*x)^3*(a + a*sin(c + d*x))^2),x)

[Out]

(B/12 - A/3 + sin(c + d*x)*(A/12 + B/24) + sin(c + d*x)^2*(A/2 + B/4) + sin(c + d*x)^3*(A/4 + B/8))/(d*(2*a^2*
sin(c + d*x) + a^2 - 2*a^2*sin(c + d*x)^3 - a^2*sin(c + d*x)^4)) + (atanh(sin(c + d*x))*(2*A + B))/(8*a^2*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)

[Out]

(Integral(A*sec(c + d*x)**3/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x) + Integral(B*sin(c + d*x)*sec(c + d*x)*
*3/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x))/a**2

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