3.797 \(\int (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=72 \[ -\frac {a \cos (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {2 a \sec (c+d x)}{d}+a x \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2710, 3473, 8, 2590, 270} \[ -\frac {a \cos (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {2 a \sec (c+d x)}{d}+a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*x - (a*Cos[c + d*x])/d - (2*a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) - (a*Tan[c + d*x])/d + (a*Tan[c + d
*x]^3)/(3*d)

Rule 8

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 2710

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 81, normalized size = 1.12 \[ -\frac {a \cos (c+d x)}{d}+\frac {a \tan ^{-1}(\tan (c+d x))}{d}+\frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d}-\frac {2 a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTan[Tan[c + d*x]])/d - (a*Cos[c + d*x])/d - (2*a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) - (a*Tan[c +
d*x])/d + (a*Tan[c + d*x]^3)/(3*d)

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fricas [A]  time = 0.48, size = 88, normalized size = 1.22 \[ -\frac {3 \, a d x \cos \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} - {\left (3 \, a d x \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) - a}{3 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*a*d*x*cos(d*x + c) - 7*a*cos(d*x + c)^2 - (3*a*d*x*cos(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a)*sin(d*x +
c) - a)/(d*cos(d*x + c)*sin(d*x + c) - d*cos(d*x + c))

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giac [A]  time = 0.19, size = 124, normalized size = 1.72 \[ \frac {6 \, {\left (d x + c\right )} a - \frac {3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 17 \, a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(d*x + c)*a - 3*(a*tan(1/2*d*x + 1/2*c)^2 + 4*a*tan(1/2*d*x + 1/2*c) + 5*a)/(tan(1/2*d*x + 1/2*c)^3 + t
an(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 1) + (15*a*tan(1/2*d*x + 1/2*c)^2 - 36*a*tan(1/2*d*x + 1/2*c) +
17*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

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maple [A]  time = 0.36, size = 98, normalized size = 1.36 \[ \frac {a \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+a \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(1/3*sin(d*x+c)^6/cos(d*x+c)^3-sin(d*x+c)^6/cos(d*x+c)-(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c))+
a*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c))

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maxima [A]  time = 0.42, size = 65, normalized size = 0.90 \[ \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a - a {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.72, size = 185, normalized size = 2.57 \[ a\,x+\frac {\left (2\,a\,d\,x-\frac {a\,\left (6\,d\,x-6\right )}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {a\,\left (3\,d\,x-12\right )}{3}-a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\left (a\,d\,x-\frac {a\,\left (3\,d\,x-4\right )}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (\frac {a\,\left (6\,d\,x-26\right )}{3}-2\,a\,d\,x\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a\,\left (3\,d\,x-16\right )}{3}+a\,d\,x}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + (tan(c/2 + (d*x)/2)*((a*(6*d*x - 26))/3 - 2*a*d*x) - (a*(3*d*x - 16))/3 - tan(c/2 + (d*x)/2)^2*((a*(3*d*
x - 4))/3 - a*d*x) - tan(c/2 + (d*x)/2)^5*((a*(6*d*x - 6))/3 - 2*a*d*x) + tan(c/2 + (d*x)/2)^4*((a*(3*d*x - 12
))/3 - a*d*x) + (4*a*tan(c/2 + (d*x)/2)^3)/3 + a*d*x)/(d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)*(
tan(c/2 + (d*x)/2)^2 + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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