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

Optimal. Leaf size=31 \[ \frac{a^5 \sin ^2(c+d x)}{2 d (a-a \sin (c+d x))^2} \]

[Out]

(a^5*Sin[c + d*x]^2)/(2*d*(a - a*Sin[c + d*x])^2)

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Rubi [A]  time = 0.059388, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2836, 12, 37} \[ \frac{a^5 \sin ^2(c+d x)}{2 d (a-a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^5*Sin[c + d*x]^2)/(2*d*(a - a*Sin[c + d*x])^2)

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0304367, size = 30, normalized size = 0.97 \[ \frac{a^3 \sin ^2(c+d x)}{2 d (1-\sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.073, size = 154, normalized size = 5. \begin{align*}{\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}}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{3}}{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))^3,x)

[Out]

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

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Maxima [A]  time = 1.10199, size = 57, normalized size = 1.84 \begin{align*} \frac{2 \, a^{3} \sin \left (d x + c\right ) - a^{3}}{2 \,{\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )} 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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.32256, size = 104, normalized size = 3.35 \begin{align*} -\frac{2 \, a^{3} \sin \left (d x + c\right ) - a^{3}}{2 \,{\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)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(2*a^3*sin(d*x + c) - a^3)/(d*cos(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)*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.19195, size = 43, normalized size = 1.39 \begin{align*} \frac{2 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{4}} \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))^3,x, algorithm="giac")

[Out]

2*a^3*tan(1/2*d*x + 1/2*c)^2/(d*(tan(1/2*d*x + 1/2*c) - 1)^4)