3.60 \(\int \frac{\sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=62 \[ \frac{4 \tan ^3(c+d x)}{15 a d}+\frac{4 \tan (c+d x)}{5 a d}-\frac{\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)} \]

[Out]

-Sec[c + d*x]^3/(5*d*(a + a*Sin[c + d*x])) + (4*Tan[c + d*x])/(5*a*d) + (4*Tan[c + d*x]^3)/(15*a*d)

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Rubi [A]  time = 0.0593119, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2672, 3767} \[ \frac{4 \tan ^3(c+d x)}{15 a d}+\frac{4 \tan (c+d x)}{5 a d}-\frac{\sec ^3(c+d x)}{5 d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sec[c + d*x]^3/(5*d*(a + a*Sin[c + d*x])) + (4*Tan[c + d*x])/(5*a*d) + (4*Tan[c + d*x]^3)/(15*a*d)

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac{\sec ^3(c+d x)}{5 d (a+a \sin (c+d x))}+\frac{4 \int \sec ^4(c+d x) \, dx}{5 a}\\ &=-\frac{\sec ^3(c+d x)}{5 d (a+a \sin (c+d x))}-\frac{4 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 a d}\\ &=-\frac{\sec ^3(c+d x)}{5 d (a+a \sin (c+d x))}+\frac{4 \tan (c+d x)}{5 a d}+\frac{4 \tan ^3(c+d x)}{15 a d}\\ \end{align*}

Mathematica [A]  time = 0.0969105, size = 66, normalized size = 1.06 \[ -\frac{\sec ^3(c+d x) (-2 (3 \sin (c+d x)+\sin (3 (c+d x)))+2 \cos (2 (c+d x))+\cos (4 (c+d x)))}{15 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^3*(2*Cos[2*(c + d*x)] + Cos[4*(c + d*x)] - 2*(3*Sin[c + d*x] + Sin[3*(c + d*x)])))/(15*a*d*(1 +
Sin[c + d*x]))

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Maple [B]  time = 0.059, size = 130, normalized size = 2.1 \begin{align*} 2\,{\frac{1}{da} \left ( -1/12\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-3}-1/8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-2}-{\frac{5}{16\,\tan \left ( 1/2\,dx+c/2 \right ) -16}}-1/5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-5}+1/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}-5/6\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-3}+3/4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}-{\frac{11}{16\,\tan \left ( 1/2\,dx+c/2 \right ) +16}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a*(-1/12/(tan(1/2*d*x+1/2*c)-1)^3-1/8/(tan(1/2*d*x+1/2*c)-1)^2-5/16/(tan(1/2*d*x+1/2*c)-1)-1/5/(tan(1/2*d*
x+1/2*c)+1)^5+1/2/(tan(1/2*d*x+1/2*c)+1)^4-5/6/(tan(1/2*d*x+1/2*c)+1)^3+3/4/(tan(1/2*d*x+1/2*c)+1)^2-11/16/(ta
n(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 0.966609, size = 397, normalized size = 6.4 \begin{align*} \frac{2 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{21 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{13 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{25 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{5 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 3\right )}}{15 \,{\left (a + \frac{2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 21*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 13*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 25*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 5*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)
^6/(cos(d*x + c) + 1)^6 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 3)/((a + 2*a*sin(d*x + c)/(cos(d*x + c) + 1
) - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6*a*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a*sin(
d*x + c)^8/(cos(d*x + c) + 1)^8)*d)

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Fricas [A]  time = 1.64399, size = 194, normalized size = 3.13 \begin{align*} -\frac{8 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} - 4 \,{\left (2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 1}{15 \,{\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(8*cos(d*x + c)^4 - 4*cos(d*x + c)^2 - 4*(2*cos(d*x + c)^2 + 1)*sin(d*x + c) - 1)/(a*d*cos(d*x + c)^3*si
n(d*x + c) + a*d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{4}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**4/(sin(c + d*x) + 1), x)/a

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Giac [B]  time = 1.14159, size = 161, normalized size = 2.6 \begin{align*} -\frac{\frac{5 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 13\right )}}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{165 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 480 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 650 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 400 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 113}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(5*(15*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) + 13)/(a*(tan(1/2*d*x + 1/2*c) - 1)^3) + (165*t
an(1/2*d*x + 1/2*c)^4 + 480*tan(1/2*d*x + 1/2*c)^3 + 650*tan(1/2*d*x + 1/2*c)^2 + 400*tan(1/2*d*x + 1/2*c) + 1
13)/(a*(tan(1/2*d*x + 1/2*c) + 1)^5))/d