3.19 \(\int (a+a \sin (c+d x))^2 \tan ^6(c+d x) \, dx\)

Optimal. Leaf size=149 \[ \frac{2 a^2 \cos (c+d x)}{d}+\frac{9 a^2 \tan ^5(c+d x)}{10 d}-\frac{3 a^2 \tan ^3(c+d x)}{2 d}+\frac{9 a^2 \tan (c+d x)}{2 d}+\frac{2 a^2 \sec ^5(c+d x)}{5 d}-\frac{2 a^2 \sec ^3(c+d x)}{d}+\frac{6 a^2 \sec (c+d x)}{d}-\frac{a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}-\frac{9 a^2 x}{2} \]

[Out]

(-9*a^2*x)/2 + (2*a^2*Cos[c + d*x])/d + (6*a^2*Sec[c + d*x])/d - (2*a^2*Sec[c + d*x]^3)/d + (2*a^2*Sec[c + d*x
]^5)/(5*d) + (9*a^2*Tan[c + d*x])/(2*d) - (3*a^2*Tan[c + d*x]^3)/(2*d) + (9*a^2*Tan[c + d*x]^5)/(10*d) - (a^2*
Sin[c + d*x]^2*Tan[c + d*x]^5)/(2*d)

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Rubi [A]  time = 0.165139, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2710, 3473, 8, 2590, 270, 2591, 288, 302, 203} \[ \frac{2 a^2 \cos (c+d x)}{d}+\frac{9 a^2 \tan ^5(c+d x)}{10 d}-\frac{3 a^2 \tan ^3(c+d x)}{2 d}+\frac{9 a^2 \tan (c+d x)}{2 d}+\frac{2 a^2 \sec ^5(c+d x)}{5 d}-\frac{2 a^2 \sec ^3(c+d x)}{d}+\frac{6 a^2 \sec (c+d x)}{d}-\frac{a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}-\frac{9 a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^6,x]

[Out]

(-9*a^2*x)/2 + (2*a^2*Cos[c + d*x])/d + (6*a^2*Sec[c + d*x])/d - (2*a^2*Sec[c + d*x]^3)/d + (2*a^2*Sec[c + d*x
]^5)/(5*d) + (9*a^2*Tan[c + d*x])/(2*d) - (3*a^2*Tan[c + d*x]^3)/(2*d) + (9*a^2*Tan[c + d*x]^5)/(10*d) - (a^2*
Sin[c + d*x]^2*Tan[c + d*x]^5)/(2*d)

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]

Rule 8

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

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 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 2591

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int (a+a \sin (c+d x))^2 \tan ^6(c+d x) \, dx &=\int \left (a^2 \tan ^6(c+d x)+2 a^2 \sin (c+d x) \tan ^6(c+d x)+a^2 \sin ^2(c+d x) \tan ^6(c+d x)\right ) \, dx\\ &=a^2 \int \tan ^6(c+d x) \, dx+a^2 \int \sin ^2(c+d x) \tan ^6(c+d x) \, dx+\left (2 a^2\right ) \int \sin (c+d x) \tan ^6(c+d x) \, dx\\ &=\frac{a^2 \tan ^5(c+d x)}{5 d}-a^2 \int \tan ^4(c+d x) \, dx+\frac{a^2 \operatorname{Subst}\left (\int \frac{x^8}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^6} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^2 \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan ^5(c+d x)}{5 d}-\frac{a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}+a^2 \int \tan ^2(c+d x) \, dx-\frac{\left (2 a^2\right ) \operatorname{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{\left (7 a^2\right ) \operatorname{Subst}\left (\int \frac{x^6}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{2 a^2 \cos (c+d x)}{d}+\frac{6 a^2 \sec (c+d x)}{d}-\frac{2 a^2 \sec ^3(c+d x)}{d}+\frac{2 a^2 \sec ^5(c+d x)}{5 d}+\frac{a^2 \tan (c+d x)}{d}-\frac{a^2 \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan ^5(c+d x)}{5 d}-\frac{a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}-a^2 \int 1 \, dx+\frac{\left (7 a^2\right ) \operatorname{Subst}\left (\int \left (1-x^2+x^4-\frac{1}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-a^2 x+\frac{2 a^2 \cos (c+d x)}{d}+\frac{6 a^2 \sec (c+d x)}{d}-\frac{2 a^2 \sec ^3(c+d x)}{d}+\frac{2 a^2 \sec ^5(c+d x)}{5 d}+\frac{9 a^2 \tan (c+d x)}{2 d}-\frac{3 a^2 \tan ^3(c+d x)}{2 d}+\frac{9 a^2 \tan ^5(c+d x)}{10 d}-\frac{a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}-\frac{\left (7 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac{9 a^2 x}{2}+\frac{2 a^2 \cos (c+d x)}{d}+\frac{6 a^2 \sec (c+d x)}{d}-\frac{2 a^2 \sec ^3(c+d x)}{d}+\frac{2 a^2 \sec ^5(c+d x)}{5 d}+\frac{9 a^2 \tan (c+d x)}{2 d}-\frac{3 a^2 \tan ^3(c+d x)}{2 d}+\frac{9 a^2 \tan ^5(c+d x)}{10 d}-\frac{a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.837593, size = 174, normalized size = 1.17 \[ -\frac{a^2 \sec ^5(c+d x) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 (250 \sin (c+d x)-720 c \sin (2 (c+d x))-720 d x \sin (2 (c+d x))-824 \sin (2 (c+d x))+351 \sin (3 (c+d x))+5 \sin (5 (c+d x))+10 (90 c+90 d x+103) \cos (c+d x)-544 \cos (2 (c+d x))-180 c \cos (3 (c+d x))-180 d x \cos (3 (c+d x))-206 \cos (3 (c+d x))+20 \cos (4 (c+d x))-500)}{160 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^6,x]

[Out]

-(a^2*Sec[c + d*x]^5*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(-500 + 10*(103 + 90*c + 90*d*x)*Cos[c + d*x] - 5
44*Cos[2*(c + d*x)] - 206*Cos[3*(c + d*x)] - 180*c*Cos[3*(c + d*x)] - 180*d*x*Cos[3*(c + d*x)] + 20*Cos[4*(c +
d*x)] + 250*Sin[c + d*x] - 824*Sin[2*(c + d*x)] - 720*c*Sin[2*(c + d*x)] - 720*d*x*Sin[2*(c + d*x)] + 351*Sin
[3*(c + d*x)] + 5*Sin[5*(c + d*x)]))/(160*d)

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Maple [A]  time = 0.078, size = 251, normalized size = 1.7 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{5\,\cos \left ( dx+c \right ) }}+{\frac{8\,\cos \left ( dx+c \right ) }{5} \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{7}+{\frac{7\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{35\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{35\,\sin \left ( dx+c \right ) }{16}} \right ) }-{\frac{7\,dx}{2}}-{\frac{7\,c}{2}} \right ) +2\,{a}^{2} \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{\cos \left ( dx+c \right ) }}+ \left ({\frac{16}{5}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{6}+6/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+8/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{a}^{2} \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}+\tan \left ( dx+c \right ) -dx-c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^2*tan(d*x+c)^6,x)

[Out]

1/d*(a^2*(1/5*sin(d*x+c)^9/cos(d*x+c)^5-4/15*sin(d*x+c)^9/cos(d*x+c)^3+8/5*sin(d*x+c)^9/cos(d*x+c)+8/5*(sin(d*
x+c)^7+7/6*sin(d*x+c)^5+35/24*sin(d*x+c)^3+35/16*sin(d*x+c))*cos(d*x+c)-7/2*d*x-7/2*c)+2*a^2*(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*si
n(d*x+c)^2)*cos(d*x+c))+a^2*(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 = 1.60183, size = 205, normalized size = 1.38 \begin{align*} \frac{{\left (6 \, \tan \left (d x + c\right )^{5} - 20 \, \tan \left (d x + c\right )^{3} - 105 \, d x - 105 \, c + \frac{15 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} + 90 \, \tan \left (d x + c\right )\right )} a^{2} + 2 \,{\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^{2} + 12 \, a^{2}{\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 )}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^6,x, algorithm="maxima")

[Out]

1/30*((6*tan(d*x + c)^5 - 20*tan(d*x + c)^3 - 105*d*x - 105*c + 15*tan(d*x + c)/(tan(d*x + c)^2 + 1) + 90*tan(
d*x + c))*a^2 + 2*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*a^2 + 12*a^2*((15*co
s(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 = 1.70126, size = 382, normalized size = 2.56 \begin{align*} -\frac{45 \, a^{2} d x \cos \left (d x + c\right )^{3} - 10 \, a^{2} \cos \left (d x + c\right )^{4} - 90 \, a^{2} d x \cos \left (d x + c\right ) + 78 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2} -{\left (5 \, a^{2} \cos \left (d x + c\right )^{4} - 90 \, a^{2} d x \cos \left (d x + c\right ) + 84 \, a^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2}\right )} \sin \left (d x + c\right )}{10 \,{\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^6,x, algorithm="fricas")

[Out]

-1/10*(45*a^2*d*x*cos(d*x + c)^3 - 10*a^2*cos(d*x + c)^4 - 90*a^2*d*x*cos(d*x + c) + 78*a^2*cos(d*x + c)^2 - 4
*a^2 - (5*a^2*cos(d*x + c)^4 - 90*a^2*d*x*cos(d*x + c) + 84*a^2*cos(d*x + c)^2 - 6*a^2)*sin(d*x + c))/(d*cos(d
*x + c)^3 + 2*d*cos(d*x + c)*sin(d*x + c) - 2*d*cos(d*x + c))

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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+a*sin(d*x+c))**2*tan(d*x+c)**6,x)

[Out]

Timed out

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Giac [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+a*sin(d*x+c))^2*tan(d*x+c)^6,x, algorithm="giac")

[Out]

Timed out