3.1008 \(\int \frac{\sec ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=91 \[ \frac{A+B}{8 d (a-a \sin (c+d x))}-\frac{a (A-B)}{8 d (a \sin (c+d x)+a)^2}+\frac{(3 A+B) \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac{A}{4 d (a \sin (c+d x)+a)} \]

[Out]

((3*A + B)*ArcTanh[Sin[c + d*x]])/(8*a*d) + (A + B)/(8*d*(a - a*Sin[c + d*x])) - (a*(A - B))/(8*d*(a + a*Sin[c
+ d*x])^2) - A/(4*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.138204, antiderivative size = 91, 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} \[ \frac{A+B}{8 d (a-a \sin (c+d x))}-\frac{a (A-B)}{8 d (a \sin (c+d x)+a)^2}+\frac{(3 A+B) \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac{A}{4 d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.25448, size = 75, normalized size = 0.82 \[ \frac{\frac{A+B}{a-a \sin (c+d x)}+\frac{B-A}{a (\sin (c+d x)+1)^2}+\frac{(3 A+B) \tanh ^{-1}(\sin (c+d x))}{a}-\frac{2 A}{a \sin (c+d x)+a}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*A + B)*ArcTanh[Sin[c + d*x]])/a + (-A + B)/(a*(1 + Sin[c + d*x])^2) + (A + B)/(a - a*Sin[c + d*x]) - (2*A
)/(a + a*Sin[c + d*x]))/(8*d)

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Maple [B]  time = 0.134, size = 169, normalized size = 1.9 \begin{align*} -{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) A}{16\,da}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) B}{16\,da}}-{\frac{A}{8\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{B}{8\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{A}{4\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{A}{8\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{B}{8\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) A}{16\,da}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) B}{16\,da}} \end{align*}

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)),x)

[Out]

-3/16/d/a*ln(sin(d*x+c)-1)*A-1/16/d/a*ln(sin(d*x+c)-1)*B-1/8/d/a/(sin(d*x+c)-1)*A-1/8/d/a/(sin(d*x+c)-1)*B-1/4
/d/a/(1+sin(d*x+c))*A-1/8/d/a/(1+sin(d*x+c))^2*A+1/8/d/a/(1+sin(d*x+c))^2*B+3/16/d/a*ln(1+sin(d*x+c))*A+1/16/d
/a*ln(1+sin(d*x+c))*B

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Maxima [A]  time = 1.01793, size = 153, normalized size = 1.68 \begin{align*} \frac{\frac{{\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac{{\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac{2 \,{\left ({\left (3 \, A + B\right )} \sin \left (d x + c\right )^{2} +{\left (3 \, A + B\right )} \sin \left (d x + c\right ) - 2 \, A + 2 \, B\right )}}{a \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a}}{16 \, d} \end{align*}

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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.50304, size = 423, normalized size = 4.65 \begin{align*} -\frac{2 \,{\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} -{\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) +{\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) +{\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (3 \, A + B\right )} \sin \left (d x + c\right ) - 2 \, A - 6 \, B}{16 \,{\left (a d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

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)),x, algorithm="fricas")

[Out]

-1/16*(2*(3*A + B)*cos(d*x + c)^2 - ((3*A + B)*cos(d*x + c)^2*sin(d*x + c) + (3*A + B)*cos(d*x + c)^2)*log(sin
(d*x + c) + 1) + ((3*A + B)*cos(d*x + c)^2*sin(d*x + c) + (3*A + B)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2
*(3*A + B)*sin(d*x + c) - 2*A - 6*B)/(a*d*cos(d*x + c)^2*sin(d*x + c) + a*d*cos(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{3}{\left (c + d x \right )}}{\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{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

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)),x)

[Out]

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

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Giac [A]  time = 1.33967, size = 198, normalized size = 2.18 \begin{align*} \frac{\frac{2 \,{\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{2 \,{\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac{2 \,{\left (3 \, A \sin \left (d x + c\right ) + B \sin \left (d x + c\right ) - 5 \, A - 3 \, B\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}} - \frac{9 \, A \sin \left (d x + c\right )^{2} + 3 \, B \sin \left (d x + c\right )^{2} + 26 \, A \sin \left (d x + c\right ) + 6 \, B \sin \left (d x + c\right ) + 21 \, A - B}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{32 \, d} \end{align*}

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)),x, algorithm="giac")

[Out]

1/32*(2*(3*A + B)*log(abs(sin(d*x + c) + 1))/a - 2*(3*A + B)*log(abs(sin(d*x + c) - 1))/a + 2*(3*A*sin(d*x + c
) + B*sin(d*x + c) - 5*A - 3*B)/(a*(sin(d*x + c) - 1)) - (9*A*sin(d*x + c)^2 + 3*B*sin(d*x + c)^2 + 26*A*sin(d
*x + c) + 6*B*sin(d*x + c) + 21*A - B)/(a*(sin(d*x + c) + 1)^2))/d