3.31 \(\int (a+a \sec (c+d x))^2 \sin ^4(c+d x) \, dx\)

Optimal. Leaf size=115 \[ -\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {9 a^2 x}{8} \]

[Out]

-9/8*a^2*x+2*a^2*arctanh(sin(d*x+c))/d-2*a^2*sin(d*x+c)/d-1/8*a^2*cos(d*x+c)*sin(d*x+c)/d+1/4*a^2*cos(d*x+c)^3
*sin(d*x+c)/d-2/3*a^2*sin(d*x+c)^3/d+a^2*tan(d*x+c)/d

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Rubi [A]  time = 0.27, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3872, 2872, 2637, 2635, 8, 2633, 3770, 3767} \[ -\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {9 a^2 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^4,x]

[Out]

(-9*a^2*x)/8 + (2*a^2*ArcTanh[Sin[c + d*x]])/d - (2*a^2*Sin[c + d*x])/d - (a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d
) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (2*a^2*Sin[c + d*x]^3)/(3*d) + (a^2*Tan[c + d*x])/d

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 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]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^2 \sin ^4(c+d x) \, dx &=\int (-a-a \cos (c+d x))^2 \sin ^2(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {\int \left (-a^6-4 a^6 \cos (c+d x)-a^6 \cos ^2(c+d x)+2 a^6 \cos ^3(c+d x)+a^6 \cos ^4(c+d x)+2 a^6 \sec (c+d x)+a^6 \sec ^2(c+d x)\right ) \, dx}{a^4}\\ &=-a^2 x-a^2 \int \cos ^2(c+d x) \, dx+a^2 \int \cos ^4(c+d x) \, dx+a^2 \int \sec ^2(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^3(c+d x) \, dx+\left (2 a^2\right ) \int \sec (c+d x) \, dx-\left (4 a^2\right ) \int \cos (c+d x) \, dx\\ &=-a^2 x+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {4 a^2 \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {1}{2} a^2 \int 1 \, dx+\frac {1}{4} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx-\frac {a^2 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=-\frac {3 a^2 x}{2}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {1}{8} \left (3 a^2\right ) \int 1 \, dx\\ &=-\frac {9 a^2 x}{8}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 94, normalized size = 0.82 \[ -\frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (64 \sin ^3(c+d x)+192 \sin (c+d x)-3 \sin (4 (c+d x))+60 \tan ^{-1}(\tan (c+d x))-96 \tan (c+d x)-192 \tanh ^{-1}(\sin (c+d x))+48 c+48 d x\right )}{384 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^4,x]

[Out]

-1/384*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(48*c + 48*d*x + 60*ArcTan[Tan[c + d*x]] - 192*ArcTanh[Sin
[c + d*x]] + 192*Sin[c + d*x] + 64*Sin[c + d*x]^3 - 3*Sin[4*(c + d*x)] - 96*Tan[c + d*x]))/d

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fricas [A]  time = 0.95, size = 133, normalized size = 1.16 \[ -\frac {27 \, a^{2} d x \cos \left (d x + c\right ) - 24 \, a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + 24 \, a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (6 \, a^{2} \cos \left (d x + c\right )^{4} + 16 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )^{2} - 64 \, a^{2} \cos \left (d x + c\right ) + 24 \, a^{2}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^4,x, algorithm="fricas")

[Out]

-1/24*(27*a^2*d*x*cos(d*x + c) - 24*a^2*cos(d*x + c)*log(sin(d*x + c) + 1) + 24*a^2*cos(d*x + c)*log(-sin(d*x
+ c) + 1) - (6*a^2*cos(d*x + c)^4 + 16*a^2*cos(d*x + c)^3 - 3*a^2*cos(d*x + c)^2 - 64*a^2*cos(d*x + c) + 24*a^
2)*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 1.21, size = 161, normalized size = 1.40 \[ -\frac {27 \, {\left (d x + c\right )} a^{2} - 48 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 48 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {48 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (51 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 187 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 229 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 45 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^4,x, algorithm="giac")

[Out]

-1/24*(27*(d*x + c)*a^2 - 48*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 48*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)
) + 48*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(51*a^2*tan(1/2*d*x + 1/2*c)^7 + 187*a^2*tan(
1/2*d*x + 1/2*c)^5 + 229*a^2*tan(1/2*d*x + 1/2*c)^3 + 45*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1
)^4)/d

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maple [A]  time = 0.70, size = 134, normalized size = 1.17 \[ \frac {3 a^{2} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}-\frac {9 a^{2} x}{8}-\frac {9 a^{2} c}{8 d}-\frac {2 a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 a^{2} \sin \left (d x +c \right )}{d}+\frac {2 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*sin(d*x+c)^4,x)

[Out]

3/4/d*a^2*cos(d*x+c)*sin(d*x+c)^3+9/8*a^2*cos(d*x+c)*sin(d*x+c)/d-9/8*a^2*x-9/8/d*a^2*c-2/3*a^2*sin(d*x+c)^3/d
-2*a^2*sin(d*x+c)/d+2/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^2*sin(d*x+c)^5/cos(d*x+c)

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maxima [A]  time = 0.47, size = 126, normalized size = 1.10 \[ -\frac {32 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{2} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} + 48 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{2}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^4,x, algorithm="maxima")

[Out]

-1/96*(32*(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))*a^2 - 3*(12*
d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*a^2 + 48*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1)
- 2*tan(d*x + c))*a^2)/d

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mupad [B]  time = 1.83, size = 177, normalized size = 1.54 \[ \frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {9\,a^2\,x}{8}+\frac {\frac {25\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {58\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {31\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-\frac {22\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^4*(a + a/cos(c + d*x))^2,x)

[Out]

(4*a^2*atanh(tan(c/2 + (d*x)/2)))/d - (9*a^2*x)/8 + ((31*a^2*tan(c/2 + (d*x)/2)^5)/2 - (22*a^2*tan(c/2 + (d*x)
/2)^3)/3 + (58*a^2*tan(c/2 + (d*x)/2)^7)/3 + (25*a^2*tan(c/2 + (d*x)/2)^9)/4 - (7*a^2*tan(c/2 + (d*x)/2))/4)/(
d*(3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 - tan(c/2
+ (d*x)/2)^10 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int 2 \sin ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**2*sin(d*x+c)**4,x)

[Out]

a**2*(Integral(2*sin(c + d*x)**4*sec(c + d*x), x) + Integral(sin(c + d*x)**4*sec(c + d*x)**2, x) + Integral(si
n(c + d*x)**4, x))

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