3.26 \(\int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx\)

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

[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 = 0.07, 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 = {2707, 77} \[ \frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {2 a^4}{d (a-a \sin (c+d x))}+\frac {5 a^3 \sin (c+d x)}{d}+\frac {7 a^3 \log (1-\sin (c+d x))}{d} \]

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 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 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 (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^3 (a+x)}{(a-x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {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 = 0.16, size = 66, normalized size = 0.73 \[ \frac {a^3 \left (2 \sin ^3(c+d x)+9 \sin ^2(c+d x)+30 \sin (c+d x)+\frac {12}{1-\sin (c+d x)}+42 \log (1-\sin (c+d x))\right )}{6 d} \]

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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fricas [A]  time = 0.43, size = 104, normalized size = 1.14 \[ \frac {4 \, a^{3} \cos \left (d x + c\right )^{4} - 50 \, a^{3} \cos \left (d x + c\right )^{2} + 31 \, a^{3} + 84 \, {\left (a^{3} \sin \left (d x + c\right ) - a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (14 \, a^{3} \cos \left (d x + c\right )^{2} + 55 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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maple [B]  time = 0.18, size = 205, normalized size = 2.25 \[ \frac {a^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{2 d}+\frac {7 a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}+\frac {7 a^{3} \sin \left (d x +c \right )}{d}-\frac {7 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 a^{3} \left (\sin ^{6}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {3 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{d}+\frac {7 a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d} \]

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)

[Out]

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

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maxima [A]  time = 0.30, size = 72, normalized size = 0.79 \[ \frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 42 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, a^{3} \sin \left (d x + c\right ) - \frac {12 \, a^{3}}{\sin \left (d x + c\right ) - 1}}{6 \, d} \]

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

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

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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