3.871 \(\int \sec (c+d x) (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=96 \[ -\frac{a^3 \sin ^2(c+d x)}{2 d}+\frac{a^5}{2 d (a-a \sin (c+d x))^2}-\frac{4 a^4}{d (a-a \sin (c+d x))}-\frac{3 a^3 \sin (c+d x)}{d}-\frac{6 a^3 \log (1-\sin (c+d x))}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.110984, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2836, 12, 43} \[ -\frac{a^3 \sin ^2(c+d x)}{2 d}+\frac{a^5}{2 d (a-a \sin (c+d x))^2}-\frac{4 a^4}{d (a-a \sin (c+d x))}-\frac{3 a^3 \sin (c+d x)}{d}-\frac{6 a^3 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + a*Sin[c + d*x])^3*Tan[c + d*x]^4,x]

[Out]

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

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.236321, size = 61, normalized size = 0.64 \[ -\frac{a^3 \left (\sin ^2(c+d x)+6 \sin (c+d x)+\frac{7-8 \sin (c+d x)}{(\sin (c+d x)-1)^2}+12 \log (1-\sin (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^3*Tan[c + d*x]^4,x]

[Out]

-(a^3*(12*Log[1 - Sin[c + d*x]] + (7 - 8*Sin[c + d*x])/(-1 + Sin[c + d*x])^2 + 6*Sin[c + d*x] + Sin[c + d*x]^2
))/(2*d)

________________________________________________________________________________________

Maple [B]  time = 0.095, size = 309, normalized size = 3.2 \begin{align*}{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d}}-{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-6\,{\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{9\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d}}-2\,{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}-6\,{\frac{{a}^{3}\sin \left ( dx+c \right ) }{d}}+6\,{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{3\,{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.00627, size = 111, normalized size = 1.16 \begin{align*} -\frac{a^{3} \sin \left (d x + c\right )^{2} + 12 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, a^{3} \sin \left (d x + c\right ) - \frac{8 \, a^{3} \sin \left (d x + c\right ) - 7 \, a^{3}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(a^3*sin(d*x + c)^2 + 12*a^3*log(sin(d*x + c) - 1) + 6*a^3*sin(d*x + c) - (8*a^3*sin(d*x + c) - 7*a^3)/(s
in(d*x + c)^2 - 2*sin(d*x + c) + 1))/d

________________________________________________________________________________________

Fricas [A]  time = 1.49283, size = 311, normalized size = 3.24 \begin{align*} \frac{2 \, a^{3} \cos \left (d x + c\right )^{4} + 19 \, a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3} - 24 \,{\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 (4 \, a^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3}\right )} \sin \left (d x + c\right )}{4 \,{\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)^4*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(2*a^3*cos(d*x + c)^4 + 19*a^3*cos(d*x + c)^2 - 8*a^3 - 24*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) - 2*a^
3)*log(-sin(d*x + c) + 1) - 2*(4*a^3*cos(d*x + c)^2 - 3*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c
) - 2*d)

________________________________________________________________________________________

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)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.20934, size = 282, normalized size = 2.94 \begin{align*} \frac{6 \, a^{3} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 12 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac{25 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 106 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 164 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 106 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 25 \, a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{4}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

(6*a^3*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 12*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (9*a^3*tan(1/2*d*x + 1/2*
c)^4 + 6*a^3*tan(1/2*d*x + 1/2*c)^3 + 20*a^3*tan(1/2*d*x + 1/2*c)^2 + 6*a^3*tan(1/2*d*x + 1/2*c) + 9*a^3)/(tan
(1/2*d*x + 1/2*c)^2 + 1)^2 + (25*a^3*tan(1/2*d*x + 1/2*c)^4 - 106*a^3*tan(1/2*d*x + 1/2*c)^3 + 164*a^3*tan(1/2
*d*x + 1/2*c)^2 - 106*a^3*tan(1/2*d*x + 1/2*c) + 25*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d