3.14 \(\int (a+a \sin (c+d x))^2 \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=119 \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {9 a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \sin ^2(c+d x)}{2 d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {31 a^2 \log (1-\sin (c+d x))}{8 d}-\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

[Out]

-31/8*a^2*ln(1-sin(d*x+c))/d-1/8*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+1/4*a^4/d/(a
-a*sin(d*x+c))^2-9/4*a^3/d/(a-a*sin(d*x+c))

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Rubi [A]  time = 0.08, antiderivative size = 119, 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, 88} \[ -\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {9 a^3}{4 d (a-a \sin (c+d x))}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {31 a^2 \log (1-\sin (c+d x))}{8 d}-\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^5,x]

[Out]

(-31*a^2*Log[1 - Sin[c + d*x]])/(8*d) - (a^2*Log[1 + Sin[c + d*x]])/(8*d) - (2*a^2*Sin[c + d*x])/d - (a^2*Sin[
c + d*x]^2)/(2*d) + a^4/(4*d*(a - a*Sin[c + d*x])^2) - (9*a^3)/(4*d*(a - a*Sin[c + d*x]))

Rule 88

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 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 ^5(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{(a-x)^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-2 a+\frac {a^4}{2 (a-x)^3}-\frac {9 a^3}{4 (a-x)^2}+\frac {31 a^2}{8 (a-x)}-x-\frac {a^2}{8 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {31 a^2 \log (1-\sin (c+d x))}{8 d}-\frac {a^2 \log (1+\sin (c+d x))}{8 d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {9 a^3}{4 d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 75, normalized size = 0.63 \[ -\frac {a^2 \left (4 \sin ^2(c+d x)+16 \sin (c+d x)-\frac {18}{\sin (c+d x)-1}-\frac {2}{(\sin (c+d x)-1)^2}+31 \log (1-\sin (c+d x))+\log (\sin (c+d x)+1)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^5,x]

[Out]

-1/8*(a^2*(31*Log[1 - Sin[c + d*x]] + Log[1 + Sin[c + d*x]] - 2/(-1 + Sin[c + d*x])^2 - 18/(-1 + Sin[c + d*x])
+ 16*Sin[c + d*x] + 4*Sin[c + d*x]^2))/d

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fricas [A]  time = 0.47, size = 168, normalized size = 1.41 \[ \frac {4 \, a^{2} \cos \left (d x + c\right )^{4} + 22 \, a^{2} \cos \left (d x + c\right )^{2} - 12 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 31 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 5 \, a^{2}\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^5,x, algorithm="fricas")

[Out]

1/8*(4*a^2*cos(d*x + c)^4 + 22*a^2*cos(d*x + c)^2 - 12*a^2 - (a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)
*log(sin(d*x + c) + 1) - 31*(a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(-sin(d*x + c) + 1) - 2*(4*a^
2*cos(d*x + c)^2 - 5*a^2)*sin(d*x + c))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*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)^5,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.20, size = 261, normalized size = 2.19 \[ \frac {a^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {a^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}-\frac {a^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{2 d}-\frac {3 a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}-\frac {3 a^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {4 a^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{4}}-\frac {3 a^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{2}}-\frac {3 a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d}-\frac {5 a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d}-\frac {15 a^{2} \sin \left (d x +c \right )}{4 d}+\frac {15 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^2*tan(d*x+c)^5,x)

[Out]

1/4/d*a^2*sin(d*x+c)^8/cos(d*x+c)^4-1/2/d*a^2*sin(d*x+c)^8/cos(d*x+c)^2-1/2/d*a^2*sin(d*x+c)^6-3/4/d*a^2*sin(d
*x+c)^4-3/2*a^2*sin(d*x+c)^2/d-4/d*a^2*ln(cos(d*x+c))+1/2/d*a^2*sin(d*x+c)^7/cos(d*x+c)^4-3/4/d*a^2*sin(d*x+c)
^7/cos(d*x+c)^2-3/4/d*a^2*sin(d*x+c)^5-5/4/d*a^2*sin(d*x+c)^3-15/4*a^2*sin(d*x+c)/d+15/4/d*a^2*ln(sec(d*x+c)+t
an(d*x+c))+1/4/d*a^2*tan(d*x+c)^4-1/2/d*a^2*tan(d*x+c)^2

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maxima [A]  time = 0.31, size = 96, normalized size = 0.81 \[ -\frac {4 \, a^{2} \sin \left (d x + c\right )^{2} + a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 31 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a^{2} \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, a^{2} \sin \left (d x + c\right ) - 8 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^5,x, algorithm="maxima")

[Out]

-1/8*(4*a^2*sin(d*x + c)^2 + a^2*log(sin(d*x + c) + 1) + 31*a^2*log(sin(d*x + c) - 1) + 16*a^2*sin(d*x + c) -
2*(9*a^2*sin(d*x + c) - 8*a^2)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d

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mupad [B]  time = 6.57, size = 283, normalized size = 2.38 \[ \frac {4\,a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{4\,d}-\frac {\frac {15\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-22\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {61\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-36\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {61\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-22\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {15\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {31\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5*(a + a*sin(c + d*x))^2,x)

[Out]

(4*a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (a^2*log(tan(c/2 + (d*x)/2) + 1))/(4*d) - ((61*a^2*tan(c/2 + (d*x)/2
)^3)/2 - 22*a^2*tan(c/2 + (d*x)/2)^2 - 36*a^2*tan(c/2 + (d*x)/2)^4 + (61*a^2*tan(c/2 + (d*x)/2)^5)/2 - 22*a^2*
tan(c/2 + (d*x)/2)^6 + (15*a^2*tan(c/2 + (d*x)/2)^7)/2 + (15*a^2*tan(c/2 + (d*x)/2))/2)/(d*(8*tan(c/2 + (d*x)/
2)^2 - 4*tan(c/2 + (d*x)/2) - 12*tan(c/2 + (d*x)/2)^3 + 14*tan(c/2 + (d*x)/2)^4 - 12*tan(c/2 + (d*x)/2)^5 + 8*
tan(c/2 + (d*x)/2)^6 - 4*tan(c/2 + (d*x)/2)^7 + tan(c/2 + (d*x)/2)^8 + 1)) - (31*a^2*log(tan(c/2 + (d*x)/2) -
1))/(4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int 2 \sin {\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \tan ^{5}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))**2*tan(d*x+c)**5,x)

[Out]

a**2*(Integral(2*sin(c + d*x)*tan(c + d*x)**5, x) + Integral(sin(c + d*x)**2*tan(c + d*x)**5, x) + Integral(ta
n(c + d*x)**5, x))

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