3.870 \(\int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=114 \[ -\frac{a^3 \sin ^3(c+d x)}{3 d}-\frac{3 a^3 \sin ^2(c+d x)}{2 d}+\frac{a^5}{2 d (a-a \sin (c+d x))^2}-\frac{5 a^4}{d (a-a \sin (c+d x))}-\frac{6 a^3 \sin (c+d x)}{d}-\frac{10 a^3 \log (1-\sin (c+d x))}{d} \]

[Out]

(-10*a^3*Log[1 - Sin[c + d*x]])/d - (6*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^5/(2*d*(a - a*Sin[c + d*x])^2) - (5*a^4)/(d*(a - a*Sin[c + d*x]))

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Rubi [A]  time = 0.0840697, antiderivative size = 114, 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, 43} \[ -\frac{a^3 \sin ^3(c+d x)}{3 d}-\frac{3 a^3 \sin ^2(c+d x)}{2 d}+\frac{a^5}{2 d (a-a \sin (c+d x))^2}-\frac{5 a^4}{d (a-a \sin (c+d x))}-\frac{6 a^3 \sin (c+d x)}{d}-\frac{10 a^3 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^5,x]

[Out]

(-10*a^3*Log[1 - Sin[c + d*x]])/d - (6*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^5/(2*d*(a - a*Sin[c + d*x])^2) - (5*a^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 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])

Rubi steps

\begin{align*} \int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-6 a^2+\frac{a^5}{(a-x)^3}-\frac{5 a^4}{(a-x)^2}+\frac{10 a^3}{a-x}-3 a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{10 a^3 \log (1-\sin (c+d x))}{d}-\frac{6 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^5}{2 d (a-a \sin (c+d x))^2}-\frac{5 a^4}{d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.306308, size = 73, normalized size = 0.64 \[ -\frac{a^3 \left (2 \sin ^3(c+d x)+9 \sin ^2(c+d x)+36 \sin (c+d x)+\frac{27-30 \sin (c+d x)}{(\sin (c+d x)-1)^2}+60 \log (1-\sin (c+d x))\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^5,x]

[Out]

-(a^3*(60*Log[1 - Sin[c + d*x]] + (27 - 30*Sin[c + d*x])/(-1 + Sin[c + d*x])^2 + 36*Sin[c + d*x] + 9*Sin[c + d
*x]^2 + 2*Sin[c + d*x]^3))/(6*d)

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Maple [B]  time = 0.1, size = 325, normalized size = 2.9 \begin{align*}{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{5\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d}}-2\,{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{d}}-{\frac{10\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-10\,{\frac{{a}^{3}\sin \left ( dx+c \right ) }{d}}+10\,{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d}}-{\frac{9\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{9\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-10\,{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{9\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c))^3,x)

[Out]

1/4/d*a^3*sin(d*x+c)^9/cos(d*x+c)^4-5/8/d*a^3*sin(d*x+c)^9/cos(d*x+c)^2-5/8*a^3*sin(d*x+c)^7/d-2*a^3*sin(d*x+c
)^5/d-10/3*a^3*sin(d*x+c)^3/d-10*a^3*sin(d*x+c)/d+10/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+3/4/d*a^3*sin(d*x+c)^8/co
s(d*x+c)^4-3/2/d*a^3*sin(d*x+c)^8/cos(d*x+c)^2-3/2*a^3*sin(d*x+c)^6/d-9/4*a^3*sin(d*x+c)^4/d-9/2*a^3*sin(d*x+c
)^2/d-10/d*a^3*ln(cos(d*x+c))+3/4/d*a^3*sin(d*x+c)^7/cos(d*x+c)^4-9/8/d*a^3*sin(d*x+c)^7/cos(d*x+c)^2+1/4/d*a^
3*tan(d*x+c)^4-1/2/d*a^3*tan(d*x+c)^2

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Maxima [A]  time = 1.07241, size = 130, normalized size = 1.14 \begin{align*} -\frac{2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 36 \, a^{3} \sin \left (d x + c\right ) - \frac{3 \,{\left (10 \, a^{3} \sin \left (d x + c\right ) - 9 \, a^{3}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/6*(2*a^3*sin(d*x + c)^3 + 9*a^3*sin(d*x + c)^2 + 60*a^3*log(sin(d*x + c) - 1) + 36*a^3*sin(d*x + c) - 3*(10
*a^3*sin(d*x + c) - 9*a^3)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d

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Fricas [A]  time = 1.44393, size = 351, normalized size = 3.08 \begin{align*} \frac{10 \, a^{3} \cos \left (d x + c\right )^{4} + 115 \, a^{3} \cos \left (d x + c\right )^{2} - 80 \, a^{3} - 120 \,{\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, a^{3} \cos \left (d x + c\right )^{4} - 24 \, a^{3} \cos \left (d x + c\right )^{2} + 37 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \,{\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(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(10*a^3*cos(d*x + c)^4 + 115*a^3*cos(d*x + c)^2 - 80*a^3 - 120*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) -
2*a^3)*log(-sin(d*x + c) + 1) + 2*(2*a^3*cos(d*x + c)^4 - 24*a^3*cos(d*x + c)^2 + 37*a^3)*sin(d*x + c))/(d*co
s(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(sec(d*x+c)**5*sin(d*x+c)**5*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.25699, size = 327, normalized size = 2.87 \begin{align*} \frac{30 \, a^{3} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 60 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{55 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 36 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 183 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 80 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 183 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 36 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 55 \, a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}} + \frac{125 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 524 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 804 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 524 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 125 \, a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{4}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/3*(30*a^3*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 60*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (55*a^3*tan(1/2*d*x
+ 1/2*c)^6 + 36*a^3*tan(1/2*d*x + 1/2*c)^5 + 183*a^3*tan(1/2*d*x + 1/2*c)^4 + 80*a^3*tan(1/2*d*x + 1/2*c)^3 +
183*a^3*tan(1/2*d*x + 1/2*c)^2 + 36*a^3*tan(1/2*d*x + 1/2*c) + 55*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3 + (125*a
^3*tan(1/2*d*x + 1/2*c)^4 - 524*a^3*tan(1/2*d*x + 1/2*c)^3 + 804*a^3*tan(1/2*d*x + 1/2*c)^2 - 524*a^3*tan(1/2*
d*x + 1/2*c) + 125*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d