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

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

[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]))

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Rubi [A]  time = 0.0818173, antiderivative size = 119, 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} \[ -\frac{a^2 \sin ^2(c+d x)}{2 d}+\frac{a^4}{4 d (a-a \sin (c+d x))^2}-\frac{9 a^3}{4 d (a-a \sin (c+d x))}-\frac{2 a^2 \sin (c+d x)}{d}-\frac{31 a^2 \log (1-\sin (c+d x))}{8 d}-\frac{a^2 \log (\sin (c+d x)+1)}{8 d} \]

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 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))^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*}

Mathematica [A]  time = 0.244541, 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]

-(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]) + 1
6*Sin[c + d*x] + 4*Sin[c + d*x]^2))/(8*d)

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Maple [B]  time = 0.085, size = 261, normalized size = 2.2 \begin{align*}{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d}}-{\frac{3\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{3\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-4\,{\frac{{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d}}-{\frac{5\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{15\,{a}^{2}\sin \left ( dx+c \right ) }{4\,d}}+{\frac{15\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}} \end{align*}

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 = 1.12912, size = 130, normalized size = 1.09 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.55703, size = 412, normalized size = 3.46 \begin{align*} \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 )}} \end{align*}

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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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))**2*tan(d*x+c)**5,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))^2*tan(d*x+c)^5,x, algorithm="giac")

[Out]

Timed out