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

Optimal. Leaf size=77 \[ -\frac{a}{8 d (a \sin (c+d x)+a)^2}+\frac{1}{8 d (a-a \sin (c+d x))}-\frac{1}{4 d (a \sin (c+d x)+a)}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{8 a d} \]

[Out]

(3*ArcTanh[Sin[c + d*x]])/(8*a*d) + 1/(8*d*(a - a*Sin[c + d*x])) - a/(8*d*(a + a*Sin[c + d*x])^2) - 1/(4*d*(a
+ a*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.0798467, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2667, 44, 206} \[ -\frac{a}{8 d (a \sin (c+d x)+a)^2}+\frac{1}{8 d (a-a \sin (c+d x))}-\frac{1}{4 d (a \sin (c+d x)+a)}+\frac{3 \tanh ^{-1}(\sin (c+d x))}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[Sin[c + d*x]])/(8*a*d) + 1/(8*d*(a - a*Sin[c + d*x])) - a/(8*d*(a + a*Sin[c + d*x])^2) - 1/(4*d*(a
+ a*Sin[c + d*x]))

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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

Mathematica [A]  time = 0.102742, size = 75, normalized size = 0.97 \[ -\frac{\sec ^2(c+d x) \left (-3 \sin ^2(c+d x)-3 \sin (c+d x)+3 (\sin (c+d x)-1) (\sin (c+d x)+1)^2 \tanh ^{-1}(\sin (c+d x))+2\right )}{8 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.058, size = 90, normalized size = 1.2 \begin{align*} -{\frac{1}{8\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{16\,da}}-{\frac{1}{8\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{4\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{16\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.966444, size = 123, normalized size = 1.6 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 2\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} - \frac{3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac{3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.7482, size = 336, normalized size = 4.36 \begin{align*} -\frac{6 \, \cos \left (d x + c\right )^{2} - 3 \,{\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, \sin \left (d x + c\right ) - 2}{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+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\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+a*sin(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.23488, size = 130, normalized size = 1.69 \begin{align*} \frac{\frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac{2 \,{\left (3 \, \sin \left (d x + c\right ) - 5\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}} - \frac{9 \, \sin \left (d x + c\right )^{2} + 26 \, \sin \left (d x + c\right ) + 21}{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+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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