3.36 \(\int (a+a \sin (c+d x))^4 \tan (c+d x) \, dx\)

Optimal. Leaf size=88 \[ -\frac{a^4 \sin ^4(c+d x)}{4 d}-\frac{4 a^4 \sin ^3(c+d x)}{3 d}-\frac{7 a^4 \sin ^2(c+d x)}{2 d}-\frac{8 a^4 \sin (c+d x)}{d}-\frac{8 a^4 \log (1-\sin (c+d x))}{d} \]

[Out]

(-8*a^4*Log[1 - Sin[c + d*x]])/d - (8*a^4*Sin[c + d*x])/d - (7*a^4*Sin[c + d*x]^2)/(2*d) - (4*a^4*Sin[c + d*x]
^3)/(3*d) - (a^4*Sin[c + d*x]^4)/(4*d)

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Rubi [A]  time = 0.0526452, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2707, 77} \[ -\frac{a^4 \sin ^4(c+d x)}{4 d}-\frac{4 a^4 \sin ^3(c+d x)}{3 d}-\frac{7 a^4 \sin ^2(c+d x)}{2 d}-\frac{8 a^4 \sin (c+d x)}{d}-\frac{8 a^4 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^4*Tan[c + d*x],x]

[Out]

(-8*a^4*Log[1 - Sin[c + d*x]])/d - (8*a^4*Sin[c + d*x])/d - (7*a^4*Sin[c + d*x]^2)/(2*d) - (4*a^4*Sin[c + d*x]
^3)/(3*d) - (a^4*Sin[c + d*x]^4)/(4*d)

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int (a+a \sin (c+d x))^4 \tan (c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x (a+x)^3}{a-x} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-8 a^3+\frac{8 a^4}{a-x}-7 a^2 x-4 a x^2-x^3\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{8 a^4 \log (1-\sin (c+d x))}{d}-\frac{8 a^4 \sin (c+d x)}{d}-\frac{7 a^4 \sin ^2(c+d x)}{2 d}-\frac{4 a^4 \sin ^3(c+d x)}{3 d}-\frac{a^4 \sin ^4(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.0699065, size = 62, normalized size = 0.7 \[ -\frac{a^4 \left (3 \sin ^4(c+d x)+16 \sin ^3(c+d x)+42 \sin ^2(c+d x)+96 \sin (c+d x)+96 \log (1-\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^4*Tan[c + d*x],x]

[Out]

-(a^4*(96*Log[1 - Sin[c + d*x]] + 96*Sin[c + d*x] + 42*Sin[c + d*x]^2 + 16*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4))
/(12*d)

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Maple [A]  time = 0.052, size = 101, normalized size = 1.2 \begin{align*} -{\frac{{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{7\,{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-8\,{\frac{{a}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{4\,{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-8\,{\frac{{a}^{4}\sin \left ( dx+c \right ) }{d}}+8\,{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^4*tan(d*x+c),x)

[Out]

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

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Maxima [A]  time = 1.07582, size = 95, normalized size = 1.08 \begin{align*} -\frac{3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 42 \, a^{4} \sin \left (d x + c\right )^{2} + 96 \, a^{4} \log \left (\sin \left (d x + c\right ) - 1\right ) + 96 \, a^{4} \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^4*tan(d*x+c),x, algorithm="maxima")

[Out]

-1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 42*a^4*sin(d*x + c)^2 + 96*a^4*log(sin(d*x + c) - 1) + 9
6*a^4*sin(d*x + c))/d

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Fricas [A]  time = 1.53773, size = 182, normalized size = 2.07 \begin{align*} -\frac{3 \, a^{4} \cos \left (d x + c\right )^{4} - 48 \, a^{4} \cos \left (d x + c\right )^{2} + 96 \, a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 16 \,{\left (a^{4} \cos \left (d x + c\right )^{2} - 7 \, a^{4}\right )} \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^4*tan(d*x+c),x, algorithm="fricas")

[Out]

-1/12*(3*a^4*cos(d*x + c)^4 - 48*a^4*cos(d*x + c)^2 + 96*a^4*log(-sin(d*x + c) + 1) - 16*(a^4*cos(d*x + c)^2 -
7*a^4)*sin(d*x + c))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{4} \left (\int 4 \sin{\left (c + d x \right )} \tan{\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \tan{\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \tan{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \tan{\left (c + d x \right )}\, dx + \int \tan{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))**4*tan(d*x+c),x)

[Out]

a**4*(Integral(4*sin(c + d*x)*tan(c + d*x), x) + Integral(6*sin(c + d*x)**2*tan(c + d*x), x) + Integral(4*sin(
c + d*x)**3*tan(c + d*x), x) + Integral(sin(c + d*x)**4*tan(c + d*x), x) + Integral(tan(c + d*x), x))

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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))^4*tan(d*x+c),x, algorithm="giac")

[Out]

Timed out