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

Optimal. Leaf size=61 \[ \frac{a^3}{8 d (a-a \sin (c+d x))^2}+\frac{a^2}{8 d (a \sin (c+d x)+a)}-\frac{a \tanh ^{-1}(\sin (c+d x))}{8 d} \]

[Out]

-(a*ArcTanh[Sin[c + d*x]])/(8*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) + a^2/(8*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.0667725, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2836, 12, 77, 206} \[ \frac{a^3}{8 d (a-a \sin (c+d x))^2}+\frac{a^2}{8 d (a \sin (c+d x)+a)}-\frac{a \tanh ^{-1}(\sin (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*ArcTanh[Sin[c + d*x]])/(8*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) + 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 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]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx &=\frac{a^5 \operatorname{Subst}\left (\int \frac{x}{a (a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^4 \operatorname{Subst}\left (\int \frac{x}{(a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^4 \operatorname{Subst}\left (\int \left (\frac{1}{4 a (a-x)^3}-\frac{1}{8 a^2 (a+x)^2}-\frac{1}{8 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^3}{8 d (a-a \sin (c+d x))^2}+\frac{a^2}{8 d (a+a \sin (c+d x))}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=-\frac{a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3}{8 d (a-a \sin (c+d x))^2}+\frac{a^2}{8 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.0291179, size = 74, normalized size = 1.21 \[ \frac{a \sec ^4(c+d x)}{4 d}-\frac{a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac{a \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 92, normalized size = 1.5 \begin{align*}{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{a\sin \left ( dx+c \right ) }{8\,d}}-{\frac{a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{a}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05883, size = 113, normalized size = 1.85 \begin{align*} -\frac{a \log \left (\sin \left (d x + c\right ) + 1\right ) - a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) + 2 \, 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)*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.49932, size = 343, normalized size = 5.62 \begin{align*} \frac{2 \, a \cos \left (d x + c\right )^{2} -{\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 ) +{\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 ) + 2 \, a \sin \left (d x + c\right ) - 6 \, 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)*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.31221, size = 123, normalized size = 2.02 \begin{align*} -\frac{2 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (a \sin \left (d x + c\right ) + 3 \, a\right )}}{\sin \left (d x + c\right ) + 1} + \frac{3 \, a \sin \left (d x + c\right )^{2} - 6 \, a \sin \left (d x + c\right ) - 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)*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/32*(2*a*log(abs(sin(d*x + c) + 1)) - 2*a*log(abs(sin(d*x + c) - 1)) - 2*(a*sin(d*x + c) + 3*a)/(sin(d*x + c
) + 1) + (3*a*sin(d*x + c)^2 - 6*a*sin(d*x + c) - a)/(sin(d*x + c) - 1)^2)/d