3.1.8 \(\int (a+a \sin (c+d x)) \tan ^6(c+d x) \, dx\) [8]

Optimal. Leaf size=101 \[ -a x+\frac {a \cos (c+d x)}{d}+\frac {3 a \sec (c+d x)}{d}-\frac {a \sec ^3(c+d x)}{d}+\frac {a \sec ^5(c+d x)}{5 d}+\frac {a \tan (c+d x)}{d}-\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \]

[Out]

-a*x+a*cos(d*x+c)/d+3*a*sec(d*x+c)/d-a*sec(d*x+c)^3/d+1/5*a*sec(d*x+c)^5/d+a*tan(d*x+c)/d-1/3*a*tan(d*x+c)^3/d
+1/5*a*tan(d*x+c)^5/d

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Rubi [A]
time = 0.07, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 9, 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 ^5(c+d x)}{5 d}-\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \sec ^5(c+d x)}{5 d}-\frac {a \sec ^3(c+d x)}{d}+\frac {3 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]^6,x]

[Out]

-(a*x) + (a*Cos[c + d*x])/d + (3*a*Sec[c + d*x])/d - (a*Sec[c + d*x]^3)/d + (a*Sec[c + d*x]^5)/(5*d) + (a*Tan[
c + d*x])/d - (a*Tan[c + d*x]^3)/(3*d) + (a*Tan[c + d*x]^5)/(5*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 ^6(c+d x) \, dx &=\int \left (a \tan ^6(c+d x)+a \sin (c+d x) \tan ^6(c+d x)\right ) \, dx\\ &=a \int \tan ^6(c+d x) \, dx+a \int \sin (c+d x) \tan ^6(c+d x) \, dx\\ &=\frac {a \tan ^5(c+d x)}{5 d}-a \int \tan ^4(c+d x) \, dx-\frac {a \text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^6} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d}+a \int \tan ^2(c+d x) \, dx-\frac {a \text {Subst}\left (\int \left (-1+\frac {1}{x^6}-\frac {3}{x^4}+\frac {3}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a \cos (c+d x)}{d}+\frac {3 a \sec (c+d x)}{d}-\frac {a \sec ^3(c+d x)}{d}+\frac {a \sec ^5(c+d x)}{5 d}+\frac {a \tan (c+d x)}{d}-\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d}-a \int 1 \, dx\\ &=-a x+\frac {a \cos (c+d x)}{d}+\frac {3 a \sec (c+d x)}{d}-\frac {a \sec ^3(c+d x)}{d}+\frac {a \sec ^5(c+d x)}{5 d}+\frac {a \tan (c+d x)}{d}-\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 110, normalized size = 1.09 \begin {gather*} -\frac {a \tan ^{-1}(\tan (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {3 a \sec (c+d x)}{d}-\frac {a \sec ^3(c+d x)}{d}+\frac {a \sec ^5(c+d x)}{5 d}+\frac {a \tan (c+d x)}{d}-\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])*Tan[c + d*x]^6,x]

[Out]

-((a*ArcTan[Tan[c + d*x]])/d) + (a*Cos[c + d*x])/d + (3*a*Sec[c + d*x])/d - (a*Sec[c + d*x]^3)/d + (a*Sec[c +
d*x]^5)/(5*d) + (a*Tan[c + d*x])/d - (a*Tan[c + d*x]^3)/(3*d) + (a*Tan[c + d*x]^5)/(5*d)

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Maple [A]
time = 0.19, size = 135, normalized size = 1.34

method	result	size
derivativedivides	\(\frac {a \left (\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+a \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(135\)
default	\(\frac {a \left (\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+a \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(135\)
risch	\(-a x +\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {-\frac {182 i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{15}+\frac {2 a \,{\mathrm e}^{i \left (d x +c \right )}}{15}+\frac {42 a \,{\mathrm e}^{3 i \left (d x +c \right )}}{5}-\frac {46 i a}{15}-14 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+10 a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+6 a \,{\mathrm e}^{7 i \left (d x +c \right )}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{5} d}\)	\(160\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))*tan(d*x+c)^6,x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(1/5*sin(d*x+c)^8/cos(d*x+c)^5-1/5*sin(d*x+c)^8/cos(d*x+c)^3+sin(d*x+c)^8/cos(d*x+c)+(16/5+sin(d*x+c)^6
+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c))+a*(1/5*tan(d*x+c)^5-1/3*tan(d*x+c)^3+tan(d*x+c)-d*x-c))

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Maxima [A]
time = 0.50, size = 87, normalized size = 0.86 \begin {gather*} \frac {{\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a + 3 \, a {\left (\frac {15 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 1}{\cos \left (d x + c\right )^{5}} + 5 \, \cos \left (d x + c\right )\right )}}{15 \, 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)^6,x, algorithm="maxima")

[Out]

1/15*((3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*a + 3*a*((15*cos(d*x + c)^4 - 5*
cos(d*x + c)^2 + 1)/cos(d*x + c)^5 + 5*cos(d*x + c)))/d

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Fricas [A]
time = 0.35, size = 116, normalized size = 1.15 \begin {gather*} \frac {15 \, a d x \cos \left (d x + c\right )^{3} - 38 \, a \cos \left (d x + c\right )^{4} - 11 \, a \cos \left (d x + c\right )^{2} - {\left (15 \, a d x \cos \left (d x + c\right )^{3} - 15 \, a \cos \left (d x + c\right )^{4} - 22 \, a \cos \left (d x + c\right )^{2} + 4 \, a\right )} \sin \left (d x + c\right ) + a}{15 \, {\left (d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{3}\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)^6,x, algorithm="fricas")

[Out]

1/15*(15*a*d*x*cos(d*x + c)^3 - 38*a*cos(d*x + c)^4 - 11*a*cos(d*x + c)^2 - (15*a*d*x*cos(d*x + c)^3 - 15*a*co
s(d*x + c)^4 - 22*a*cos(d*x + c)^2 + 4*a)*sin(d*x + c) + a)/(d*cos(d*x + c)^3*sin(d*x + c) - d*cos(d*x + c)^3)

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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 ^{6}{\left (c + d x \right )}\, dx + \int \tan ^{6}{\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)**6,x)

[Out]

a*(Integral(sin(c + d*x)*tan(c + d*x)**6, x) + Integral(tan(c + d*x)**6, 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)^6,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 11.22, size = 351, normalized size = 3.48 \begin {gather*} \frac {\left (\frac {a\,\left (30\,c+30\,d\,x-30\right )}{15}-2\,a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {a\,\left (15\,c+15\,d\,x+60\right )}{15}-a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (4\,a\,\left (c+d\,x\right )-\frac {a\,\left (60\,c+60\,d\,x-40\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {a\,\left (30\,c+30\,d\,x-140\right )}{15}-2\,a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {44\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\left (2\,a\,\left (c+d\,x\right )-\frac {a\,\left (30\,c+30\,d\,x-52\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {a\,\left (60\,c+60\,d\,x-344\right )}{15}-4\,a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a\,\left (c+d\,x\right )-\frac {a\,\left (15\,c+15\,d\,x-156\right )}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (2\,a\,\left (c+d\,x\right )-\frac {a\,\left (30\,c+30\,d\,x-162\right )}{15}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,\left (15\,c+15\,d\,x-96\right )}{15}-a\,\left (c+d\,x\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^5\,{\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 )}^2+1\right )}-a\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6*(a + a*sin(c + d*x)),x)

[Out]

((a*(15*c + 15*d*x - 96))/15 - tan(c/2 + (d*x)/2)*((a*(30*c + 30*d*x - 162))/15 - 2*a*(c + d*x)) - a*(c + d*x)
+ tan(c/2 + (d*x)/2)^8*((a*(15*c + 15*d*x + 60))/15 - a*(c + d*x)) + tan(c/2 + (d*x)/2)^9*((a*(30*c + 30*d*x
- 30))/15 - 2*a*(c + d*x)) - tan(c/2 + (d*x)/2)^4*((a*(30*c + 30*d*x - 52))/15 - 2*a*(c + d*x)) - tan(c/2 + (d
*x)/2)^7*((a*(60*c + 60*d*x - 40))/15 - 4*a*(c + d*x)) - tan(c/2 + (d*x)/2)^2*((a*(15*c + 15*d*x - 156))/15 -
a*(c + d*x)) + tan(c/2 + (d*x)/2)^6*((a*(30*c + 30*d*x - 140))/15 - 2*a*(c + d*x)) + tan(c/2 + (d*x)/2)^3*((a*
(60*c + 60*d*x - 344))/15 - 4*a*(c + d*x)) + (44*a*tan(c/2 + (d*x)/2)^5)/15)/(d*(tan(c/2 + (d*x)/2) - 1)^5*(ta
n(c/2 + (d*x)/2) + 1)^3*(tan(c/2 + (d*x)/2)^2 + 1)) - a*x

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