3.169 \(\int (a+b \sec (c+d x)) \sin ^4(c+d x) \, dx\)

Optimal. Leaf size=89 \[ -\frac{a \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a x}{8}-\frac{b \sin ^3(c+d x)}{3 d}-\frac{b \sin (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.110673, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {3872, 2838, 2592, 302, 206, 2635, 8} \[ -\frac{a \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a x}{8}-\frac{b \sin ^3(c+d x)}{3 d}-\frac{b \sin (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x)/8 + (b*ArcTanh[Sin[c + d*x]])/d - (b*Sin[c + d*x])/d - (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (b*Sin[
c + d*x]^3)/(3*d) - (a*Cos[c + d*x]*Sin[c + d*x]^3)/(4*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 2838

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

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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])

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

Mathematica [A]  time = 0.149137, size = 86, normalized size = 0.97 \[ \frac{3 a (c+d x)}{8 d}-\frac{a \sin (2 (c+d x))}{4 d}+\frac{a \sin (4 (c+d x))}{32 d}-\frac{b \sin ^3(c+d x)}{3 d}-\frac{b \sin (c+d x)}{d}+\frac{b \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(c + d*x))/(8*d) + (b*ArcTanh[Sin[c + d*x]])/d - (b*Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/(3*d) - (a*Sin[2
*(c + d*x)])/(4*d) + (a*Sin[4*(c + d*x)])/(32*d)

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Maple [A]  time = 0.035, size = 96, normalized size = 1.1 \begin{align*} -{\frac{a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,a\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,ax}{8}}+{\frac{3\,ac}{8\,d}}-{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{b\sin \left ( dx+c \right ) }{d}}+{\frac{b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.957505, size = 109, normalized size = 1.22 \begin{align*} \frac{3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a - 16 \,{\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} b}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*a - 16*(2*sin(d*x + c)^3 - 3*log(sin(d*x + c)
+ 1) + 3*log(sin(d*x + c) - 1) + 6*sin(d*x + c))*b)/d

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Fricas [A]  time = 1.81185, size = 217, normalized size = 2.44 \begin{align*} \frac{9 \, a d x + 12 \, b \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, b \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (6 \, a \cos \left (d x + c\right )^{3} + 8 \, b \cos \left (d x + c\right )^{2} - 15 \, a \cos \left (d x + c\right ) - 32 \, b\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*a*d*x + 12*b*log(sin(d*x + c) + 1) - 12*b*log(-sin(d*x + c) + 1) + (6*a*cos(d*x + c)^3 + 8*b*cos(d*x +
c)^2 - 15*a*cos(d*x + c) - 32*b)*sin(d*x + c))/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((a+b*sec(d*x+c))*sin(d*x+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.31595, size = 232, normalized size = 2.61 \begin{align*} \frac{9 \,{\left (d x + c\right )} a + 24 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (9 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 33 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 104 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 33 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 104 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, b \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}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*(d*x + c)*a + 24*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 24*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(9
*a*tan(1/2*d*x + 1/2*c)^7 - 24*b*tan(1/2*d*x + 1/2*c)^7 + 33*a*tan(1/2*d*x + 1/2*c)^5 - 104*b*tan(1/2*d*x + 1/
2*c)^5 - 33*a*tan(1/2*d*x + 1/2*c)^3 - 104*b*tan(1/2*d*x + 1/2*c)^3 - 9*a*tan(1/2*d*x + 1/2*c) - 24*b*tan(1/2*
d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d