3.15 \(\int (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=72 \[ \frac {a^3}{d (a-a \sin (c+d x))}+\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {3 a^2 \log (1-\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 72, 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, 43} \[ \frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {a^3}{d (a-a \sin (c+d x))}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {3 a^2 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^3}{(a-x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (2 a+\frac {a^3}{(a-x)^2}-\frac {3 a^2}{a-x}+x\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {3 a^2 \log (1-\sin (c+d x))}{d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {a^3}{d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 54, normalized size = 0.75 \[ \frac {a^2 \left (\sin ^2(c+d x)+4 \sin (c+d x)+\frac {2}{1-\sin (c+d x)}+6 \log (1-\sin (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

(a^2*(6*Log[1 - Sin[c + d*x]] + 2/(1 - Sin[c + d*x]) + 4*Sin[c + d*x] + Sin[c + d*x]^2))/(2*d)

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fricas [A]  time = 0.46, size = 90, normalized size = 1.25 \[ -\frac {6 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 12 \, {\left (a^{2} \sin \left (d x + c\right ) - a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} + 7 \, a^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/4*(6*a^2*cos(d*x + c)^2 - 3*a^2 - 12*(a^2*sin(d*x + c) - a^2)*log(-sin(d*x + c) + 1) + (2*a^2*cos(d*x + c)^
2 + 7*a^2)*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))^2*tan(d*x+c)^3,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.17, size = 162, normalized size = 2.25 \[ \frac {a^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{2 d}+\frac {a^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{d}+\frac {3 a^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{d}+\frac {3 a^{2} \sin \left (d x +c \right )}{d}-\frac {3 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 58, normalized size = 0.81 \[ \frac {a^{2} \sin \left (d x + c\right )^{2} + 6 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + 4 \, a^{2} \sin \left (d x + c\right ) - \frac {2 \, a^{2}}{\sin \left (d x + c\right ) - 1}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

1/2*(a^2*sin(d*x + c)^2 + 6*a^2*log(sin(d*x + c) - 1) + 4*a^2*sin(d*x + c) - 2*a^2/(sin(d*x + c) - 1))/d

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mupad [B]  time = 7.11, size = 204, normalized size = 2.83 \[ \frac {6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}+\frac {6\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d}-\frac {3\,a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*a^2*tan(c/2 + (d*x)/2)^3 - 6*a^2*tan(c/2 + (d*x)/2)^2 - 6*a^2*tan(c/2 + (d*x)/2)^4 + 6*a^2*tan(c/2 + (d*x)/
2)^5 + 6*a^2*tan(c/2 + (d*x)/2))/(d*(3*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2) - 4*tan(c/2 + (d*x)/2)^3 +
3*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^5 + tan(c/2 + (d*x)/2)^6 + 1)) + (6*a^2*log(tan(c/2 + (d*x)/2) -
1))/d - (3*a^2*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 \[ a^{2} \left (\int 2 \sin {\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{3}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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