3.1.19 \(\int (a+a \sin (c+d x))^2 \tan ^6(c+d x) \, dx\) [19]

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

[Out]

-9/2*a^2*x+2*a^2*cos(d*x+c)/d+6*a^2*sec(d*x+c)/d-2*a^2*sec(d*x+c)^3/d+2/5*a^2*sec(d*x+c)^5/d+9/2*a^2*tan(d*x+c
)/d-3/2*a^2*tan(d*x+c)^3/d+9/10*a^2*tan(d*x+c)^5/d-1/2*a^2*sin(d*x+c)^2*tan(d*x+c)^5/d

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Rubi [A]
time = 0.12, antiderivative size = 149, normalized size of antiderivative = 1.00, 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 = {2789, 3554, 8, 2670, 276, 2671, 294, 308, 209} \begin {gather*} \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} \end {gather*}

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 8

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

Rule 209

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

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 294

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 308

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 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 2671

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]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/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))^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 \text {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 ) \text {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 ) \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 {\left (7 a^2\right ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.57, size = 174, normalized size = 1.17 \begin {gather*} -\frac {a^2 \sec ^5(c+d x) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 (-500+10 (103+90 c+90 d x) \cos (c+d x)-544 \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/160*(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] - 544*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)] + 3
51*Sin[3*(c + d*x)] + 5*Sin[5*(c + d*x)]))/d

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Maple [A]
time = 0.24, size = 251, normalized size = 1.68

method	result	size
risch	\(-\frac {9 a^{2} x}{2}-\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {-\frac {156 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{5}-24 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+\frac {54 i a^{2}}{5}-16 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-30 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{5} d}\)	\(188\)
derivativedivides	\(\frac {a^{2} \left (\frac {\sin ^{9}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {4 \left (\sin ^{9}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}+\frac {8 \left (\sin ^{9}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )}+\frac {8 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )}{5}-\frac {7 d x}{2}-\frac {7 c}{2}\right )+2 a^{2} \left (\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(251\)
default	\(\frac {a^{2} \left (\frac {\sin ^{9}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {4 \left (\sin ^{9}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}+\frac {8 \left (\sin ^{9}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )}+\frac {8 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )}{5}-\frac {7 d x}{2}-\frac {7 c}{2}\right )+2 a^{2} \left (\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(251\)

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,method=_RETURNVERBOSE)

[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 = 0.51, size = 152, normalized size = 1.02 \begin {gather*} \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 {gather*}

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 = 0.37, size = 152, normalized size = 1.02 \begin {gather*} -\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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int 2 \sin {\left (c + d x \right )} \tan ^{6}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\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))**2*tan(d*x+c)**6,x)

[Out]

a**2*(Integral(2*sin(c + d*x)*tan(c + d*x)**6, x) + Integral(sin(c + d*x)**2*tan(c + d*x)**6, x) + Integral(ta
n(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))^2*tan(d*x+c)^6,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 10.92, size = 392, normalized size = 2.63 \begin {gather*} -\frac {9\,a^2\,x}{2}-\frac {\frac {9\,a^2\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (18\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (180\,c+180\,d\,x-422\right )}{10}\right )+\frac {174\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}-\frac {a^2\,\left (45\,c+45\,d\,x-128\right )}{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (18\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (180\,c+180\,d\,x-90\right )}{10}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (27\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (270\,c+270\,d\,x-168\right )}{10}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {63\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (315\,c+315\,d\,x-360\right )}{10}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (27\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (270\,c+270\,d\,x-600\right )}{10}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (36\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (360\,c+360\,d\,x-424\right )}{10}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {63\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (315\,c+315\,d\,x-536\right )}{10}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (36\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (360\,c+360\,d\,x-600\right )}{10}\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 )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \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))^2,x)

[Out]

- (9*a^2*x)/2 - ((9*a^2*(c + d*x))/2 - tan(c/2 + (d*x)/2)*(18*a^2*(c + d*x) - (a^2*(180*c + 180*d*x - 422))/10
) + (174*a^2*tan(c/2 + (d*x)/2)^5)/5 - (a^2*(45*c + 45*d*x - 128))/10 + tan(c/2 + (d*x)/2)^9*(18*a^2*(c + d*x)
- (a^2*(180*c + 180*d*x - 90))/10) + tan(c/2 + (d*x)/2)^4*(27*a^2*(c + d*x) - (a^2*(270*c + 270*d*x - 168))/1
0) - tan(c/2 + (d*x)/2)^8*((63*a^2*(c + d*x))/2 - (a^2*(315*c + 315*d*x - 360))/10) - tan(c/2 + (d*x)/2)^6*(27
*a^2*(c + d*x) - (a^2*(270*c + 270*d*x - 600))/10) - tan(c/2 + (d*x)/2)^3*(36*a^2*(c + d*x) - (a^2*(360*c + 36
0*d*x - 424))/10) + tan(c/2 + (d*x)/2)^2*((63*a^2*(c + d*x))/2 - (a^2*(315*c + 315*d*x - 536))/10) + tan(c/2 +
(d*x)/2)^7*(36*a^2*(c + d*x) - (a^2*(360*c + 360*d*x - 600))/10))/(d*(tan(c/2 + (d*x)/2) - 1)^5*(tan(c/2 + (d
*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)

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