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

Optimal. Leaf size=105 \[ \frac{a^3}{8 d (a-a \sin (c+d x))^2}-\frac{3 a^2}{4 d (a-a \sin (c+d x))}-\frac{a^2}{8 d (a \sin (c+d x)+a)}-\frac{11 a \log (1-\sin (c+d x))}{16 d}-\frac{5 a \log (\sin (c+d x)+1)}{16 d} \]

[Out]

(-11*a*Log[1 - Sin[c + d*x]])/(16*d) - (5*a*Log[1 + Sin[c + d*x]])/(16*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) -
(3*a^2)/(4*d*(a - a*Sin[c + d*x])) - a^2/(8*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.094418, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2836, 12, 88} \[ \frac{a^3}{8 d (a-a \sin (c+d x))^2}-\frac{3 a^2}{4 d (a-a \sin (c+d x))}-\frac{a^2}{8 d (a \sin (c+d x)+a)}-\frac{11 a \log (1-\sin (c+d x))}{16 d}-\frac{5 a \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*a*Log[1 - Sin[c + d*x]])/(16*d) - (5*a*Log[1 + Sin[c + d*x]])/(16*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) -
(3*a^2)/(4*d*(a - a*Sin[c + d*x])) - a^2/(8*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 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 \sec (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx &=\frac{a^5 \operatorname{Subst}\left (\int \frac{x^4}{a^4 (a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a \operatorname{Subst}\left (\int \frac{x^4}{(a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a \operatorname{Subst}\left (\int \left (\frac{a^2}{4 (a-x)^3}-\frac{3 a}{4 (a-x)^2}+\frac{11}{16 (a-x)}+\frac{a}{8 (a+x)^2}-\frac{5}{16 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{11 a \log (1-\sin (c+d x))}{16 d}-\frac{5 a \log (1+\sin (c+d x))}{16 d}+\frac{a^3}{8 d (a-a \sin (c+d x))^2}-\frac{3 a^2}{4 d (a-a \sin (c+d x))}-\frac{a^2}{8 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.230451, size = 106, normalized size = 1.01 \[ \frac{a \tan ^3(c+d x) \sec (c+d x)}{d}-\frac{a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d}-\frac{a \left (6 \tan (c+d x) \sec ^3(c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[c + d*x]*Tan[c + d*x]^3)/d - (a*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x]^2 - Tan[c + d*x]^4))/(4*d) - (a*(
6*Sec[c + d*x]^3*Tan[c + d*x] - 3*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x])))/(8*d)

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Maple [A]  time = 0.069, size = 133, normalized size = 1.3 \begin{align*}{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{3\,a\sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \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)),x)

[Out]

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

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Maxima [A]  time = 1.01102, size = 116, normalized size = 1.1 \begin{align*} -\frac{5 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 11 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (5 \, a \sin \left (d x + c\right )^{2} + 3 \, a \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, 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)),x, algorithm="maxima")

[Out]

-1/16*(5*a*log(sin(d*x + c) + 1) + 11*a*log(sin(d*x + c) - 1) - 2*(5*a*sin(d*x + c)^2 + 3*a*sin(d*x + c) - 6*a
)/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d

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Fricas [A]  time = 1.53321, size = 351, normalized size = 3.34 \begin{align*} \frac{10 \, a \cos \left (d x + c\right )^{2} - 5 \,{\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 11 \,{\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, a \sin \left (d x + c\right ) + 2 \, a}{16 \,{\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\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)),x, algorithm="fricas")

[Out]

1/16*(10*a*cos(d*x + c)^2 - 5*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 11*(a
*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 6*a*sin(d*x + c) + 2*a)/(d*cos(d*x +
c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

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

[Out]

Timed out

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Giac [A]  time = 1.49534, size = 126, normalized size = 1.2 \begin{align*} -\frac{10 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 22 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (5 \, a \sin \left (d x + c\right ) + 3 \, a\right )}}{\sin \left (d x + c\right ) + 1} - \frac{33 \, a \sin \left (d x + c\right )^{2} - 42 \, a \sin \left (d x + c\right ) + 13 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, 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)),x, algorithm="giac")

[Out]

-1/32*(10*a*log(abs(sin(d*x + c) + 1)) + 22*a*log(abs(sin(d*x + c) - 1)) - 2*(5*a*sin(d*x + c) + 3*a)/(sin(d*x
+ c) + 1) - (33*a*sin(d*x + c)^2 - 42*a*sin(d*x + c) + 13*a)/(sin(d*x + c) - 1)^2)/d