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

Optimal. Leaf size=82 \[ -\frac {1}{8 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {\tanh ^{-1}(\sin (c+d x))}{8 a^3 d}-\frac {1}{8 a d (a \sin (c+d x)+a)^2}+\frac {1}{6 d (a \sin (c+d x)+a)^3} \]

[Out]

1/8*arctanh(sin(d*x+c))/a^3/d+1/6/d/(a+a*sin(d*x+c))^3-1/8/a/d/(a+a*sin(d*x+c))^2-1/8/d/(a^3+a^3*sin(d*x+c))

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Rubi [A]  time = 0.06, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2707, 77, 206} \[ -\frac {1}{8 d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {\tanh ^{-1}(\sin (c+d x))}{8 a^3 d}-\frac {1}{8 a d (a \sin (c+d x)+a)^2}+\frac {1}{6 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[c + d*x]]/(8*a^3*d) + 1/(6*d*(a + a*Sin[c + d*x])^3) - 1/(8*a*d*(a + a*Sin[c + d*x])^2) - 1/(8*d*(
a^3 + a^3*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 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(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]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x}{(a-x) (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{2 (a+x)^4}+\frac {1}{4 a (a+x)^3}+\frac {1}{8 a^2 (a+x)^2}+\frac {1}{8 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {1}{6 d (a+a \sin (c+d x))^3}-\frac {1}{8 a d (a+a \sin (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 a^2 d}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{8 a^3 d}+\frac {1}{6 d (a+a \sin (c+d x))^3}-\frac {1}{8 a d (a+a \sin (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 52, normalized size = 0.63 \[ \frac {3 \tanh ^{-1}(\sin (c+d x))-\frac {3 \sin ^2(c+d x)+9 \sin (c+d x)+2}{(\sin (c+d x)+1)^3}}{24 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[Sin[c + d*x]] - (2 + 9*Sin[c + d*x] + 3*Sin[c + d*x]^2)/(1 + Sin[c + d*x])^3)/(24*a^3*d)

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fricas [B]  time = 0.44, size = 154, normalized size = 1.88 \[ -\frac {6 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 18 \, \sin \left (d x + c\right ) - 10}{48 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(6*cos(d*x + c)^2 - 3*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(sin(d*x + c) + 1) +
3*(3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 4)*sin(d*x + c) - 4)*log(-sin(d*x + c) + 1) - 18*sin(d*x + c) - 10)/(
3*a^3*d*cos(d*x + c)^2 - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 4*a^3*d)*sin(d*x + c))

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giac [A]  time = 0.79, size = 81, normalized size = 0.99 \[ \frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} - \frac {11 \, \sin \left (d x + c\right )^{3} + 45 \, \sin \left (d x + c\right )^{2} + 69 \, \sin \left (d x + c\right ) + 19}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(6*log(abs(sin(d*x + c) + 1))/a^3 - 6*log(abs(sin(d*x + c) - 1))/a^3 - (11*sin(d*x + c)^3 + 45*sin(d*x +
c)^2 + 69*sin(d*x + c) + 19)/(a^3*(sin(d*x + c) + 1)^3))/d

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maple [A]  time = 0.24, size = 90, normalized size = 1.10 \[ -\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{16 a^{3} d}+\frac {1}{6 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{8 a^{3} d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{8 a^{3} d \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{16 a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a^3/d*ln(sin(d*x+c)-1)+1/6/a^3/d/(1+sin(d*x+c))^3-1/8/a^3/d/(1+sin(d*x+c))^2-1/8/a^3/d/(1+sin(d*x+c))+1/
16*ln(1+sin(d*x+c))/a^3/d

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maxima [A]  time = 0.30, size = 98, normalized size = 1.20 \[ -\frac {\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) + 2\right )}}{a^{3} \sin \left (d x + c\right )^{3} + 3 \, a^{3} \sin \left (d x + c\right )^{2} + 3 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac {3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(2*(3*sin(d*x + c)^2 + 9*sin(d*x + c) + 2)/(a^3*sin(d*x + c)^3 + 3*a^3*sin(d*x + c)^2 + 3*a^3*sin(d*x +
c) + a^3) - 3*log(sin(d*x + c) + 1)/a^3 + 3*log(sin(d*x + c) - 1)/a^3)/d

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mupad [B]  time = 8.82, size = 186, normalized size = 2.27 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^3\,d}+\frac {-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+20\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(tan(c/2 + (d*x)/2))/(4*a^3*d) + (tan(c/2 + (d*x)/2)^2/2 - tan(c/2 + (d*x)/2)/4 + tan(c/2 + (d*x)/2)^3/6
+ tan(c/2 + (d*x)/2)^4/2 - tan(c/2 + (d*x)/2)^5/4)/(d*(15*a^3*tan(c/2 + (d*x)/2)^2 + 20*a^3*tan(c/2 + (d*x)/2)
^3 + 15*a^3*tan(c/2 + (d*x)/2)^4 + 6*a^3*tan(c/2 + (d*x)/2)^5 + a^3*tan(c/2 + (d*x)/2)^6 + a^3 + 6*a^3*tan(c/2
+ (d*x)/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+a*sin(d*x+c))**3,x)

[Out]

Integral(tan(c + d*x)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

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