3.1.73 \(\int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [73]

Optimal. Leaf size=126 \[ -\frac {\tanh ^{-1}(\sin (c+d x))}{32 a^3 d}+\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{6 d (a+a \sin (c+d x))^3}+\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac {1}{16 d \left (a^3+a^3 \sin (c+d x)\right )} \]

[Out]

-1/32*arctanh(sin(d*x+c))/a^3/d+1/16*a/d/(a+a*sin(d*x+c))^4-1/6/d/(a+a*sin(d*x+c))^3+3/32/a/d/(a+a*sin(d*x+c))
^2+1/32/d/(a^3-a^3*sin(d*x+c))+1/16/d/(a^3+a^3*sin(d*x+c))

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Rubi [A]
time = 0.07, antiderivative size = 126, normalized size of antiderivative = 1.00, 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 = {2786, 90, 212} \begin {gather*} \frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac {1}{16 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {\tanh ^{-1}(\sin (c+d x))}{32 a^3 d}+\frac {a}{16 d (a \sin (c+d x)+a)^4}-\frac {1}{6 d (a \sin (c+d x)+a)^3}+\frac {3}{32 a d (a \sin (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*ArcTanh[Sin[c + d*x]]/(a^3*d) + a/(16*d*(a + a*Sin[c + d*x])^4) - 1/(6*d*(a + a*Sin[c + d*x])^3) + 3/(32
*a*d*(a + a*Sin[c + d*x])^2) + 1/(32*d*(a^3 - a^3*Sin[c + d*x])) + 1/(16*d*(a^3 + a^3*Sin[c + d*x]))

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2786

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 ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\text {Subst}\left (\int \frac {x^3}{(a-x)^2 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{32 a^2 (a-x)^2}-\frac {a}{4 (a+x)^5}+\frac {1}{2 (a+x)^4}-\frac {3}{16 a (a+x)^3}-\frac {1}{16 a^2 (a+x)^2}-\frac {1}{32 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{6 d (a+a \sin (c+d x))^3}+\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac {1}{16 d \left (a^3+a^3 \sin (c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{32 a^2 d}\\ &=-\frac {\tanh ^{-1}(\sin (c+d x))}{32 a^3 d}+\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{6 d (a+a \sin (c+d x))^3}+\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac {1}{16 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 82, normalized size = 0.65 \begin {gather*} -\frac {3 \tanh ^{-1}(\sin (c+d x))-\frac {3}{1-\sin (c+d x)}-\frac {6}{(1+\sin (c+d x))^4}+\frac {16}{(1+\sin (c+d x))^3}-\frac {9}{(1+\sin (c+d x))^2}-\frac {6}{1+\sin (c+d x)}}{96 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(3*ArcTanh[Sin[c + d*x]] - 3/(1 - Sin[c + d*x]) - 6/(1 + Sin[c + d*x])^4 + 16/(1 + Sin[c + d*x])^3 - 9/(
1 + Sin[c + d*x])^2 - 6/(1 + Sin[c + d*x]))/(a^3*d)

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Maple [A]
time = 0.28, size = 91, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{64}+\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{16+16 \sin \left (d x +c \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{64}}{d \,a^{3}}\)	\(91\)
default	\(\frac {-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{64}+\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{16+16 \sin \left (d x +c \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{64}}{d \,a^{3}}\)	\(91\)
risch	\(\frac {i \left (-310 \,{\mathrm e}^{5 i \left (d x +c \right )}-162 i {\mathrm e}^{4 i \left (d x +c \right )}+88 \,{\mathrm e}^{3 i \left (d x +c \right )}-18 i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}+88 \,{\mathrm e}^{7 i \left (d x +c \right )}+162 i {\mathrm e}^{6 i \left (d x +c \right )}+18 i {\mathrm e}^{8 i \left (d x +c \right )}+3 \,{\mathrm e}^{9 i \left (d x +c \right )}\right )}{48 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{32 a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{32 a^{3} d}\)	\(185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a^3*(-1/32/(sin(d*x+c)-1)+1/64*ln(sin(d*x+c)-1)+1/16/(1+sin(d*x+c))^4-1/6/(1+sin(d*x+c))^3+3/32/(1+sin(d*x
+c))^2+1/16/(1+sin(d*x+c))-1/64*ln(1+sin(d*x+c)))

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Maxima [A]
time = 0.28, size = 146, normalized size = 1.16 \begin {gather*} \frac {\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{4} + 9 \, \sin \left (d x + c\right )^{3} - 25 \, \sin \left (d x + c\right )^{2} - 27 \, \sin \left (d x + c\right ) - 8\right )}}{a^{3} \sin \left (d x + c\right )^{5} + 3 \, a^{3} \sin \left (d x + c\right )^{4} + 2 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, 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}}}{192 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(2*(3*sin(d*x + c)^4 + 9*sin(d*x + c)^3 - 25*sin(d*x + c)^2 - 27*sin(d*x + c) - 8)/(a^3*sin(d*x + c)^5 +
3*a^3*sin(d*x + c)^4 + 2*a^3*sin(d*x + c)^3 - 2*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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Fricas [A]
time = 0.38, size = 226, normalized size = 1.79 \begin {gather*} \frac {6 \, \cos \left (d x + c\right )^{4} + 38 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 18 \, {\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 60}{192 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2} + {\left (a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(6*cos(d*x + c)^4 + 38*cos(d*x + c)^2 - 3*(3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - 4*cos
(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) + 3*(3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 -
4*cos(d*x + c)^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 18*(cos(d*x + c)^2 + 2)*sin(d*x + c) - 60)/(3*a^3*d*c
os(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2 + (a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2)*sin(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\tan ^{3}{\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 9.47, size = 114, normalized size = 0.90 \begin {gather*} -\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} + \frac {12 \, {\left (\sin \left (d x + c\right ) + 1\right )}}{a^{3} {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {25 \, \sin \left (d x + c\right )^{4} + 148 \, \sin \left (d x + c\right )^{3} + 366 \, \sin \left (d x + c\right )^{2} + 260 \, \sin \left (d x + c\right ) + 65}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{768 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*(12*log(abs(sin(d*x + c) + 1))/a^3 - 12*log(abs(sin(d*x + c) - 1))/a^3 + 12*(sin(d*x + c) + 1)/(a^3*(si
n(d*x + c) - 1)) - (25*sin(d*x + c)^4 + 148*sin(d*x + c)^3 + 366*sin(d*x + c)^2 + 260*sin(d*x + c) + 65)/(a^3*
(sin(d*x + c) + 1)^4))/d

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Mupad [B]
time = 9.93, size = 302, normalized size = 2.40 \begin {gather*} \frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{16}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{8}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {37\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{8}+\frac {101\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}+\frac {37\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{8}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+13\,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 )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{16\,a^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)/16 + (3*tan(c/2 + (d*x)/2)^2)/8 + (5*tan(c/2 + (d*x)/2)^3)/6 + (37*tan(c/2 + (d*x)/2)^4)/8
+ (101*tan(c/2 + (d*x)/2)^5)/24 + (37*tan(c/2 + (d*x)/2)^6)/8 + (5*tan(c/2 + (d*x)/2)^7)/6 + (3*tan(c/2 + (d*
x)/2)^8)/8 + tan(c/2 + (d*x)/2)^9/16)/(d*(13*a^3*tan(c/2 + (d*x)/2)^2 + 8*a^3*tan(c/2 + (d*x)/2)^3 - 14*a^3*ta
n(c/2 + (d*x)/2)^4 - 28*a^3*tan(c/2 + (d*x)/2)^5 - 14*a^3*tan(c/2 + (d*x)/2)^6 + 8*a^3*tan(c/2 + (d*x)/2)^7 +
13*a^3*tan(c/2 + (d*x)/2)^8 + 6*a^3*tan(c/2 + (d*x)/2)^9 + a^3*tan(c/2 + (d*x)/2)^10 + a^3 + 6*a^3*tan(c/2 + (
d*x)/2))) - atanh(tan(c/2 + (d*x)/2))/(16*a^3*d)

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