3.179 \(\int (a+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx\)

Optimal. Leaf size=175 \[ \frac{\left (13 a^2-6 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}-\frac{\left (11 a^2-18 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5}{16} x \left (a^2-6 b^2\right )-\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}-\frac{2 a b \sin ^5(c+d x)}{5 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{2 a b \sin (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \tan (c+d x)}{d} \]

[Out]

(5*(a^2 - 6*b^2)*x)/16 + (2*a*b*ArcTanh[Sin[c + d*x]])/d - (2*a*b*Sin[c + d*x])/d - ((11*a^2 - 18*b^2)*Cos[c +
d*x]*Sin[c + d*x])/(16*d) + ((13*a^2 - 6*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (a^2*Cos[c + d*x]^5*Sin[c
+ d*x])/(6*d) - (2*a*b*Sin[c + d*x]^3)/(3*d) - (2*a*b*Sin[c + d*x]^5)/(5*d) + (b^2*Tan[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.461275, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {3872, 2911, 2592, 302, 206, 455, 1814, 1157, 388, 203} \[ \frac{\left (13 a^2-6 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}-\frac{\left (11 a^2-18 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{5}{16} x \left (a^2-6 b^2\right )-\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}-\frac{2 a b \sin ^5(c+d x)}{5 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{2 a b \sin (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b^2 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(a^2 - 6*b^2)*x)/16 + (2*a*b*ArcTanh[Sin[c + d*x]])/d - (2*a*b*Sin[c + d*x])/d - ((11*a^2 - 18*b^2)*Cos[c +
d*x]*Sin[c + d*x])/(16*d) + ((13*a^2 - 6*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (a^2*Cos[c + d*x]^5*Sin[c
+ d*x])/(6*d) - (2*a*b*Sin[c + d*x]^3)/(3*d) - (2*a*b*Sin[c + d*x]^5)/(5*d) + (b^2*Tan[c + d*x])/d

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rule 2592

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

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 206

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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+b \sec (c+d x))^2 \sin ^6(c+d x) \, dx &=\int (-b-a \cos (c+d x))^2 \sin ^4(c+d x) \tan ^2(c+d x) \, dx\\ &=(2 a b) \int \sin ^5(c+d x) \tan (c+d x) \, dx+\int \left (b^2+a^2 \cos ^2(c+d x)\right ) \sin ^4(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^6 \left (a^2+b^2+b^2 x^2\right )}{\left (1+x^2\right )^4} \, dx,x,\tan (c+d x)\right )}{d}+\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x^6}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-a^2+6 a^2 x^2-6 a^2 x^4-6 b^2 x^6}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{6 d}+\frac{(2 a b) \operatorname{Subst}\left (\int \left (-1-x^2-x^4+\frac{1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{2 a b \sin (c+d x)}{d}+\frac{\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{2 a b \sin ^5(c+d x)}{5 d}+\frac{\operatorname{Subst}\left (\int \frac{-3 \left (3 a^2-2 b^2\right )+24 \left (a^2-b^2\right ) x^2+24 b^2 x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{24 d}+\frac{(2 a b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{2 a b \sin (c+d x)}{d}-\frac{\left (11 a^2-18 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{2 a b \sin ^5(c+d x)}{5 d}-\frac{\operatorname{Subst}\left (\int \frac{-3 \left (5 a^2-14 b^2\right )-48 b^2 x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{48 d}\\ &=\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{2 a b \sin (c+d x)}{d}-\frac{\left (11 a^2-18 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{2 a b \sin ^5(c+d x)}{5 d}+\frac{b^2 \tan (c+d x)}{d}+\frac{\left (5 \left (a^2-6 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=\frac{5}{16} \left (a^2-6 b^2\right ) x+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{2 a b \sin (c+d x)}{d}-\frac{\left (11 a^2-18 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (13 a^2-6 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{2 a b \sin ^5(c+d x)}{5 d}+\frac{b^2 \tan (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 1.6659, size = 193, normalized size = 1.1 \[ \frac{\tan (c+d x) \left (-5 \left (29 a^2-84 b^2\right ) \cos (2 (c+d x))+35 a^2 \cos (4 (c+d x))-5 a^2 \cos (6 (c+d x))-185 a^2+232 a b \cos (3 (c+d x))-24 a b \cos (5 (c+d x))-30 b^2 \cos (4 (c+d x))+1410 b^2\right )+60 \left (5 \left (a^2-6 b^2\right ) (c+d x)-32 a b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+32 a b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-2128 a b \sin (c+d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*(5*(a^2 - 6*b^2)*(c + d*x) - 32*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*a*b*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) - 2128*a*b*Sin[c + d*x] + (-185*a^2 + 1410*b^2 - 5*(29*a^2 - 84*b^2)*Cos[2*(c + d*x)] +
232*a*b*Cos[3*(c + d*x)] + 35*a^2*Cos[4*(c + d*x)] - 30*b^2*Cos[4*(c + d*x)] - 24*a*b*Cos[5*(c + d*x)] - 5*a^2
*Cos[6*(c + d*x)])*Tan[c + d*x])/(960*d)

________________________________________________________________________________________

Maple [A]  time = 0.044, size = 246, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d}}-{\frac{5\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24\,d}}-{\frac{5\,{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16\,d}}+{\frac{5\,{a}^{2}x}{16}}+{\frac{5\,{a}^{2}c}{16\,d}}-{\frac{2\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{2\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-2\,{\frac{ab\sin \left ( dx+c \right ) }{d}}+2\,{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{d\cos \left ( dx+c \right ) }}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}\cos \left ( dx+c \right ) }{d}}+{\frac{5\,{b}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{15\,{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}-{\frac{15\,{b}^{2}x}{8}}-{\frac{15\,{b}^{2}c}{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d*a^2*cos(d*x+c)*sin(d*x+c)^5-5/24/d*a^2*cos(d*x+c)*sin(d*x+c)^3-5/16*a^2*cos(d*x+c)*sin(d*x+c)/d+5/16*a^
2*x+5/16/d*a^2*c-2/5*a*b*sin(d*x+c)^5/d-2/3*a*b*sin(d*x+c)^3/d-2*a*b*sin(d*x+c)/d+2/d*a*b*ln(sec(d*x+c)+tan(d*
x+c))+1/d*b^2*sin(d*x+c)^7/cos(d*x+c)+1/d*b^2*sin(d*x+c)^5*cos(d*x+c)+5/4/d*b^2*cos(d*x+c)*sin(d*x+c)^3+15/8/d
*b^2*cos(d*x+c)*sin(d*x+c)-15/8*b^2*x-15/8/d*b^2*c

________________________________________________________________________________________

Maxima [A]  time = 1.46326, size = 234, normalized size = 1.34 \begin{align*} \frac{5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 64 \,{\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a b - 120 \,{\left (15 \, d x + 15 \, c - \frac{9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} b^{2}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(5*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^2 - 64*(6*sin(d*x
+ c)^5 + 10*sin(d*x + c)^3 - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1) + 30*sin(d*x + c))*a*b - 120
*(15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1) - 8*tan(d*x + c)
)*b^2)/d

________________________________________________________________________________________

Fricas [A]  time = 1.95119, size = 468, normalized size = 2.67 \begin{align*} \frac{75 \,{\left (a^{2} - 6 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 240 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 240 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) -{\left (40 \, a^{2} \cos \left (d x + c\right )^{6} + 96 \, a b \cos \left (d x + c\right )^{5} - 352 \, a b \cos \left (d x + c\right )^{3} - 10 \,{\left (13 \, a^{2} - 6 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 736 \, a b \cos \left (d x + c\right ) + 15 \,{\left (11 \, a^{2} - 18 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 240 \, b^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(75*(a^2 - 6*b^2)*d*x*cos(d*x + c) + 240*a*b*cos(d*x + c)*log(sin(d*x + c) + 1) - 240*a*b*cos(d*x + c)*l
og(-sin(d*x + c) + 1) - (40*a^2*cos(d*x + c)^6 + 96*a*b*cos(d*x + c)^5 - 352*a*b*cos(d*x + c)^3 - 10*(13*a^2 -
6*b^2)*cos(d*x + c)^4 + 736*a*b*cos(d*x + c) + 15*(11*a^2 - 18*b^2)*cos(d*x + c)^2 - 240*b^2)*sin(d*x + c))/(
d*cos(d*x + c))

________________________________________________________________________________________

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+b*sec(d*x+c))**2*sin(d*x+c)**6,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.36594, size = 512, normalized size = 2.93 \begin{align*} \frac{480 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 480 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 75 \,{\left (a^{2} - 6 \, b^{2}\right )}{\left (d x + c\right )} - \frac{480 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + \frac{2 \,{\left (75 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 480 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 210 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 425 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 3040 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 870 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 990 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 8256 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 660 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 990 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 8256 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 660 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 425 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3040 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 870 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 75 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 480 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 210 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^2*sin(d*x+c)^6,x, algorithm="giac")

[Out]

1/240*(480*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 480*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 75*(a^2 - 6*b
^2)*(d*x + c) - 480*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(75*a^2*tan(1/2*d*x + 1/2*c)^11
- 480*a*b*tan(1/2*d*x + 1/2*c)^11 - 210*b^2*tan(1/2*d*x + 1/2*c)^11 + 425*a^2*tan(1/2*d*x + 1/2*c)^9 - 3040*a*
b*tan(1/2*d*x + 1/2*c)^9 - 870*b^2*tan(1/2*d*x + 1/2*c)^9 + 990*a^2*tan(1/2*d*x + 1/2*c)^7 - 8256*a*b*tan(1/2*
d*x + 1/2*c)^7 - 660*b^2*tan(1/2*d*x + 1/2*c)^7 - 990*a^2*tan(1/2*d*x + 1/2*c)^5 - 8256*a*b*tan(1/2*d*x + 1/2*
c)^5 + 660*b^2*tan(1/2*d*x + 1/2*c)^5 - 425*a^2*tan(1/2*d*x + 1/2*c)^3 - 3040*a*b*tan(1/2*d*x + 1/2*c)^3 + 870
*b^2*tan(1/2*d*x + 1/2*c)^3 - 75*a^2*tan(1/2*d*x + 1/2*c) - 480*a*b*tan(1/2*d*x + 1/2*c) + 210*b^2*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d