3.1.9 \(\int (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx\) [9]

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

[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 = 0.06, 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 = {2789, 3554, 8, 2670, 276} \begin {gather*} -\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 \end {gather*}

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 276

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 2670

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 2789

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 3554

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 \text {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 \text {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 \begin {gather*} \frac {a \tan ^{-1}(\tan (c+d x))}{d}-\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 {gather*}

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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Maple [A]
time = 0.18, size = 98, normalized size = 1.36

method	result	size
derivativedivides	\(\frac {a \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+a \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\)	\(98\)
default	\(\frac {a \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+a \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\)	\(98\)
risch	\(a x -\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {4 \left (-2 i a +a \,{\mathrm e}^{i \left (d x +c \right )}-3 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 a \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))*tan(d*x+c)^4,x,method=_RETURNVERBOSE)

[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 = 0.49, size = 65, normalized size = 0.90 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))*tan(d*x+c)^4,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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Fricas [A]
time = 0.35, size = 88, normalized size = 1.22 \begin {gather*} -\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))*tan(d*x+c)^4,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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int \sin {\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(sin(c + d*x)*tan(c + d*x)**4, x) + Integral(tan(c + d*x)**4, x))

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [B]
time = 9.42, size = 231, normalized size = 3.21 \begin {gather*} a\,x+\frac {\left (\frac {2\,a\,\left (3\,c+3\,d\,x\right )}{3}-\frac {a\,\left (6\,c+6\,d\,x-6\right )}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {a\,\left (3\,c+3\,d\,x-12\right )}{3}-\frac {a\,\left (3\,c+3\,d\,x\right )}{3}\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 (\frac {a\,\left (3\,c+3\,d\,x\right )}{3}-\frac {a\,\left (3\,c+3\,d\,x-4\right )}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (\frac {a\,\left (6\,c+6\,d\,x-26\right )}{3}-\frac {2\,a\,\left (3\,c+3\,d\,x\right )}{3}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,\left (3\,c+3\,d\,x\right )}{3}-\frac {a\,\left (3\,c+3\,d\,x-16\right )}{3}}{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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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