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

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

[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]))

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Rubi [A]  time = 0.109078, antiderivative size = 160, normalized size of antiderivative = 1., 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 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]

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]))

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*}

Mathematica [A]  time = 0.563704, 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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Maple [B]  time = 0.105, size = 445, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

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+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))+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))+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+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*tan(d*x+c)^6-1/4/d*a^3*tan(d*x+c)^4+1/2/d*a^3*t
an(d*x+c)^2

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Maxima [A]  time = 1.07702, size = 180, normalized size = 1.12 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.57514, size = 594, normalized size = 3.71 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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