3.1.26 \(\int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx\) [26]

Optimal. Leaf size=91 \[ \frac {7 a^3 \log (1-\sin (c+d x))}{d}+\frac {5 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {2 a^4}{d (a-a \sin (c+d x))} \]

[Out]

7*a^3*ln(1-sin(d*x+c))/d+5*a^3*sin(d*x+c)/d+3/2*a^3*sin(d*x+c)^2/d+1/3*a^3*sin(d*x+c)^3/d+2*a^4/d/(a-a*sin(d*x
+c))

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Rubi [A]
time = 0.05, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2786, 78} \begin {gather*} \frac {2 a^4}{d (a-a \sin (c+d x))}+\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {5 a^3 \sin (c+d x)}{d}+\frac {7 a^3 \log (1-\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^3*Log[1 - Sin[c + d*x]])/d + (5*a^3*Sin[c + d*x])/d + (3*a^3*Sin[c + d*x]^2)/(2*d) + (a^3*Sin[c + d*x]^3)
/(3*d) + (2*a^4)/(d*(a - a*Sin[c + d*x]))

Rule 78

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 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 (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx &=\frac {\text {Subst}\left (\int \frac {x^3 (a+x)}{(a-x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (5 a^2+\frac {2 a^4}{(a-x)^2}-\frac {7 a^3}{a-x}+3 a x+x^2\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {7 a^3 \log (1-\sin (c+d x))}{d}+\frac {5 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {2 a^4}{d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 66, normalized size = 0.73 \begin {gather*} \frac {a^3 \left (42 \log (1-\sin (c+d x))+\frac {12}{1-\sin (c+d x)}+30 \sin (c+d x)+9 \sin ^2(c+d x)+2 \sin ^3(c+d x)\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(42*Log[1 - Sin[c + d*x]] + 12/(1 - Sin[c + d*x]) + 30*Sin[c + d*x] + 9*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3
))/(6*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(203\) vs. \(2(87)=174\).
time = 0.17, size = 204, normalized size = 2.24

method	result	size
risch	\(-7 i a^{3} x -\frac {21 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {21 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {14 i a^{3} c}{d}-\frac {4 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d}+\frac {14 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}-\frac {3 a^{3} \cos \left (2 d x +2 c \right )}{4 d}\)	\(142\)
derivativedivides	\(\frac {a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(204\)
default	\(\frac {a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{2}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(204\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(1/2*sin(d*x+c)^7/cos(d*x+c)^2+1/2*sin(d*x+c)^5+5/6*sin(d*x+c)^3+5/2*sin(d*x+c)-5/2*ln(sec(d*x+c)+tan
(d*x+c)))+3*a^3*(1/2*sin(d*x+c)^6/cos(d*x+c)^2+1/2*sin(d*x+c)^4+sin(d*x+c)^2+2*ln(cos(d*x+c)))+3*a^3*(1/2*sin(
d*x+c)^5/cos(d*x+c)^2+1/2*sin(d*x+c)^3+3/2*sin(d*x+c)-3/2*ln(sec(d*x+c)+tan(d*x+c)))+a^3*(1/2*tan(d*x+c)^2+ln(
cos(d*x+c))))

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Maxima [A]
time = 0.28, size = 72, normalized size = 0.79 \begin {gather*} \frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 42 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, a^{3} \sin \left (d x + c\right ) - \frac {12 \, a^{3}}{\sin \left (d x + c\right ) - 1}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a^3*sin(d*x + c)^3 + 9*a^3*sin(d*x + c)^2 + 42*a^3*log(sin(d*x + c) - 1) + 30*a^3*sin(d*x + c) - 12*a^3
/(sin(d*x + c) - 1))/d

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Fricas [A]
time = 0.37, size = 104, normalized size = 1.14 \begin {gather*} \frac {4 \, a^{3} \cos \left (d x + c\right )^{4} - 50 \, a^{3} \cos \left (d x + c\right )^{2} + 31 \, a^{3} + 84 \, {\left (a^{3} \sin \left (d x + c\right ) - a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (14 \, a^{3} \cos \left (d x + c\right )^{2} + 55 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*a^3*cos(d*x + c)^4 - 50*a^3*cos(d*x + c)^2 + 31*a^3 + 84*(a^3*sin(d*x + c) - a^3)*log(-sin(d*x + c) +
1) - (14*a^3*cos(d*x + c)^2 + 55*a^3)*sin(d*x + c))/(d*sin(d*x + c) - d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int 3 \sin {\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{3}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [B]
time = 7.46, size = 262, normalized size = 2.88 \begin {gather*} \frac {14\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d}+\frac {14\,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+\frac {98\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {100\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {98\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+14\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {7\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{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]

(14*a^3*log(tan(c/2 + (d*x)/2) - 1))/d + ((98*a^3*tan(c/2 + (d*x)/2)^3)/3 - 14*a^3*tan(c/2 + (d*x)/2)^2 - (100
*a^3*tan(c/2 + (d*x)/2)^4)/3 + (98*a^3*tan(c/2 + (d*x)/2)^5)/3 - 14*a^3*tan(c/2 + (d*x)/2)^6 + 14*a^3*tan(c/2
+ (d*x)/2)^7 + 14*a^3*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2) - 6*tan(c/2 + (d*x
)/2)^3 + 6*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^5 + 4*tan(c/2 + (d*x)/2)^6 - 2*tan(c/2 + (d*x)/2)^7 + t
an(c/2 + (d*x)/2)^8 + 1)) - (7*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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