3.25 \(\int (a+a \sin (c+d x))^3 \tan ^7(c+d x) \, dx\)

Optimal. Leaf size=160 \[ \frac {a^6}{6 d (a-a \sin (c+d x))^3}-\frac {13 a^5}{8 d (a-a \sin (c+d x))^2}+\frac {71 a^4}{8 d (a-a \sin (c+d x))}+\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {7 a^3 \sin (c+d x)}{d}+\frac {209 a^3 \log (1-\sin (c+d x))}{16 d}-\frac {a^3 \log (\sin (c+d x)+1)}{16 d} \]

[Out]

209/16*a^3*ln(1-sin(d*x+c))/d-1/16*a^3*ln(1+sin(d*x+c))/d+7*a^3*sin(d*x+c)/d+3/2*a^3*sin(d*x+c)^2/d+1/3*a^3*si
n(d*x+c)^3/d+1/6*a^6/d/(a-a*sin(d*x+c))^3-13/8*a^5/d/(a-a*sin(d*x+c))^2+71/8*a^4/d/(a-a*sin(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 160, 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} \[ \frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^6}{6 d (a-a \sin (c+d x))^3}-\frac {13 a^5}{8 d (a-a \sin (c+d x))^2}+\frac {71 a^4}{8 d (a-a \sin (c+d x))}+\frac {7 a^3 \sin (c+d x)}{d}+\frac {209 a^3 \log (1-\sin (c+d x))}{16 d}-\frac {a^3 \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^7,x]

[Out]

(209*a^3*Log[1 - Sin[c + d*x]])/(16*d) - (a^3*Log[1 + Sin[c + d*x]])/(16*d) + (7*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) + a^6/(6*d*(a - a*Sin[c + d*x])^3) - (13*a^5)/(8*d*(a - a*S
in[c + d*x])^2) + (71*a^4)/(8*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))^3 \tan ^7(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^7}{(a-x)^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (7 a^2+\frac {a^6}{2 (a-x)^4}-\frac {13 a^5}{4 (a-x)^3}+\frac {71 a^4}{8 (a-x)^2}-\frac {209 a^3}{16 (a-x)}+3 a x+x^2-\frac {a^3}{16 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {209 a^3 \log (1-\sin (c+d x))}{16 d}-\frac {a^3 \log (1+\sin (c+d x))}{16 d}+\frac {7 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 {a^6}{6 d (a-a \sin (c+d x))^3}-\frac {13 a^5}{8 d (a-a \sin (c+d x))^2}+\frac {71 a^4}{8 d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 99, normalized size = 0.62 \[ \frac {a^3 \left (16 \sin ^3(c+d x)+72 \sin ^2(c+d x)+336 \sin (c+d x)-\frac {426}{\sin (c+d x)-1}-\frac {78}{(\sin (c+d x)-1)^2}-\frac {8}{(\sin (c+d x)-1)^3}+627 \log (1-\sin (c+d x))-3 \log (\sin (c+d x)+1)\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^7,x]

[Out]

(a^3*(627*Log[1 - Sin[c + d*x]] - 3*Log[1 + Sin[c + d*x]] - 8/(-1 + Sin[c + d*x])^3 - 78/(-1 + Sin[c + d*x])^2
- 426/(-1 + Sin[c + d*x]) + 336*Sin[c + d*x] + 72*Sin[c + d*x]^2 + 16*Sin[c + d*x]^3))/(48*d)

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fricas [A]  time = 0.45, size = 240, normalized size = 1.50 \[ -\frac {16 \, a^{3} \cos \left (d x + c\right )^{6} - 216 \, a^{3} \cos \left (d x + c\right )^{4} + 1002 \, a^{3} \cos \left (d x + c\right )^{2} - 482 \, a^{3} + 3 \, {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 627 \, {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (12 \, a^{3} \cos \left (d x + c\right )^{4} + 398 \, a^{3} \cos \left (d x + c\right )^{2} - 245 \, a^{3}\right )} \sin \left (d x + c\right )}{48 \, {\left (3 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^7,x, algorithm="fricas")

[Out]

-1/48*(16*a^3*cos(d*x + c)^6 - 216*a^3*cos(d*x + c)^4 + 1002*a^3*cos(d*x + c)^2 - 482*a^3 + 3*(3*a^3*cos(d*x +
c)^2 - 4*a^3 - (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c))*log(sin(d*x + c) + 1) - 627*(3*a^3*cos(d*x + c)^2 -
4*a^3 - (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 2*(12*a^3*cos(d*x + c)^4 + 398*a^
3*cos(d*x + c)^2 - 245*a^3)*sin(d*x + c))/(3*d*cos(d*x + c)^2 - (d*cos(d*x + c)^2 - 4*d)*sin(d*x + c) - 4*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)^7,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.23, size = 445, normalized size = 2.78 \[ \frac {35 a^{3} \left (\sin ^{9}\left (d x +c \right )\right )}{48 d}+\frac {3 a^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{2 d}+\frac {15 a^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{8 d}+\frac {a^{3} \left (\sin ^{10}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{6}}-\frac {a^{3} \left (\sin ^{10}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{3} \left (\sin ^{10}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} \left (\sin ^{9}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{6}}-\frac {3 a^{3} \left (\sin ^{9}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {15 a^{3} \left (\sin ^{9}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} \left (\sin ^{11}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}-\frac {5 a^{3} \left (\sin ^{11}\left (d x +c \right )\right )}{24 d \cos \left (d x +c \right )^{4}}+\frac {35 a^{3} \left (\sin ^{11}\left (d x +c \right )\right )}{48 d \cos \left (d x +c \right )^{2}}+\frac {21 a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}+\frac {35 a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}+\frac {105 a^{3} \sin \left (d x +c \right )}{8 d}-\frac {105 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a^{3} \left (\tan ^{6}\left (d x +c \right )\right )}{6 d}-\frac {a^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {2 a^{3} \left (\sin ^{6}\left (d x +c \right )\right )}{d}+\frac {3 a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{d}+\frac {6 a^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{d}+\frac {13 a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^3*tan(d*x+c)^7,x)

[Out]

35/48/d*a^3*sin(d*x+c)^9+3/2/d*a^3*sin(d*x+c)^8+15/8/d*a^3*sin(d*x+c)^7+1/2/d*a^3*sin(d*x+c)^10/cos(d*x+c)^6-1
/2/d*a^3*sin(d*x+c)^10/cos(d*x+c)^4+3/2/d*a^3*sin(d*x+c)^10/cos(d*x+c)^2+1/2/d*a^3*sin(d*x+c)^9/cos(d*x+c)^6-3
/8/d*a^3*sin(d*x+c)^9/cos(d*x+c)^4+15/16/d*a^3*sin(d*x+c)^9/cos(d*x+c)^2+1/6/d*a^3*sin(d*x+c)^11/cos(d*x+c)^6-
5/24/d*a^3*sin(d*x+c)^11/cos(d*x+c)^4+35/48/d*a^3*sin(d*x+c)^11/cos(d*x+c)^2+21/8/d*a^3*sin(d*x+c)^5+35/8*a^3*
sin(d*x+c)^3/d+105/8*a^3*sin(d*x+c)/d-105/8/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/6/d*a^3*tan(d*x+c)^6-1/4/d*a^3*t
an(d*x+c)^4+1/2/d*a^3*tan(d*x+c)^2+2/d*a^3*sin(d*x+c)^6+3/d*a^3*sin(d*x+c)^4+6*a^3*sin(d*x+c)^2/d+13/d*a^3*ln(
cos(d*x+c))

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maxima [A]  time = 0.32, size = 133, normalized size = 0.83 \[ \frac {16 \, a^{3} \sin \left (d x + c\right )^{3} + 72 \, a^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 627 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 336 \, a^{3} \sin \left (d x + c\right ) - \frac {2 \, {\left (213 \, a^{3} \sin \left (d x + c\right )^{2} - 387 \, a^{3} \sin \left (d x + c\right ) + 178 \, a^{3}\right )}}{\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^7,x, algorithm="maxima")

[Out]

1/48*(16*a^3*sin(d*x + c)^3 + 72*a^3*sin(d*x + c)^2 - 3*a^3*log(sin(d*x + c) + 1) + 627*a^3*log(sin(d*x + c) -
1) + 336*a^3*sin(d*x + c) - 2*(213*a^3*sin(d*x + c)^2 - 387*a^3*sin(d*x + c) + 178*a^3)/(sin(d*x + c)^3 - 3*s
in(d*x + c)^2 + 3*sin(d*x + c) - 1))/d

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mupad [B]  time = 6.47, size = 398, normalized size = 2.49 \[ \frac {\frac {105\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{4}-\frac {263\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{2}+\frac {1301\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}-582\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {1657\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-\frac {2767\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {1657\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-582\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {1301\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-\frac {263\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {105\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-38\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+92\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-38\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}+\frac {209\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}-\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}-\frac {13\,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)^7*(a + a*sin(c + d*x))^3,x)

[Out]

((1301*a^3*tan(c/2 + (d*x)/2)^3)/4 - (263*a^3*tan(c/2 + (d*x)/2)^2)/2 - 582*a^3*tan(c/2 + (d*x)/2)^4 + (1657*a
^3*tan(c/2 + (d*x)/2)^5)/2 - (2767*a^3*tan(c/2 + (d*x)/2)^6)/3 + (1657*a^3*tan(c/2 + (d*x)/2)^7)/2 - 582*a^3*t
an(c/2 + (d*x)/2)^8 + (1301*a^3*tan(c/2 + (d*x)/2)^9)/4 - (263*a^3*tan(c/2 + (d*x)/2)^10)/2 + (105*a^3*tan(c/2
+ (d*x)/2)^11)/4 + (105*a^3*tan(c/2 + (d*x)/2))/4)/(d*(18*tan(c/2 + (d*x)/2)^2 - 6*tan(c/2 + (d*x)/2) - 38*ta
n(c/2 + (d*x)/2)^3 + 63*tan(c/2 + (d*x)/2)^4 - 84*tan(c/2 + (d*x)/2)^5 + 92*tan(c/2 + (d*x)/2)^6 - 84*tan(c/2
+ (d*x)/2)^7 + 63*tan(c/2 + (d*x)/2)^8 - 38*tan(c/2 + (d*x)/2)^9 + 18*tan(c/2 + (d*x)/2)^10 - 6*tan(c/2 + (d*x
)/2)^11 + tan(c/2 + (d*x)/2)^12 + 1)) + (209*a^3*log(tan(c/2 + (d*x)/2) - 1))/(8*d) - (a^3*log(tan(c/2 + (d*x)
/2) + 1))/(8*d) - (13*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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sympy [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)**7,x)

[Out]

Timed out

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