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

Optimal. Leaf size=73 \[ \frac{\sin ^3(c+d x)}{3 a d}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac{\sin (c+d x) \cos (c+d x)}{8 a d}-\frac{x}{8 a} \]

[Out]

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

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Rubi [A]  time = 0.15003, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3872, 2839, 2564, 30, 2568, 2635, 8} \[ \frac{\sin ^3(c+d x)}{3 a d}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a d}-\frac{\sin (c+d x) \cos (c+d x)}{8 a d}-\frac{x}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2839

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

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.578878, size = 83, normalized size = 1.14 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (24 \sin (c+d x)-8 \sin (3 (c+d x))+3 \left (\sin (4 (c+d x))+4 c-4 \tan \left (\frac{c}{2}\right )-4 d x\right )\right )}{48 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(24*Sin[c + d*x] - 8*Sin[3*(c + d*x)] + 3*(4*c - 4*d*x + Sin[4*(c + d*x)] - 4
*Tan[c/2])))/(48*a*d*(1 + Sec[c + d*x]))

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Maple [B]  time = 0.072, size = 154, normalized size = 2.1 \begin{align*} -{\frac{1}{4\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+{\frac{53}{12\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+{\frac{11}{12\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+{\frac{1}{4\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-{\frac{1}{4\,da}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7+53/12/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*
c)^5+11/12/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3+1/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+
1/2*c)-1/4/d/a*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.50441, size = 265, normalized size = 3.63 \begin{align*} \frac{\frac{\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{11 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{53 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a + \frac{4 \, 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 )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac{3 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((3*sin(d*x + c)/(cos(d*x + c) + 1) + 11*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 53*sin(d*x + c)^5/(cos(d*x
+ c) + 1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a*sin(d
*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x + c) + 1)
^8) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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Fricas [A]  time = 1.67449, size = 128, normalized size = 1.75 \begin{align*} -\frac{3 \, d x -{\left (6 \, \cos \left (d x + c\right )^{3} - 8 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{24 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*d*x - (6*cos(d*x + c)^3 - 8*cos(d*x + c)^2 - 3*cos(d*x + c) + 8)*sin(d*x + c))/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.3091, size = 117, normalized size = 1.6 \begin{align*} -\frac{\frac{3 \,{\left (d x + c\right )}}{a} + \frac{2 \,{\left (3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 53 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 11 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(d*x + c)/a + 2*(3*tan(1/2*d*x + 1/2*c)^7 - 53*tan(1/2*d*x + 1/2*c)^5 - 11*tan(1/2*d*x + 1/2*c)^3 - 3
*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a))/d