\(\int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [1543]

Optimal result

Integrand size = 31, antiderivative size = 160 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (5 a^2 A-A b^2-2 a b B\right ) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac {\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac {\left (5 a^2 A-A b^2-2 a b B\right ) \sec (c+d x) \tan (c+d x)}{16 d} \]

[Out]

1/16*(5*A*a^2-A*b^2-2*B*a*b)*arctanh(sin(d*x+c))/d+1/6*sec(d*x+c)^6*(B+A*sin(d*x+c))*(a+b*sin(d*x+c))^2/d+1/24
*sec(d*x+c)^4*(2*b*(4*A*a-B*b)+(5*A*a^2+3*A*b^2-2*B*a*b)*sin(d*x+c))/d+1/16*(5*A*a^2-A*b^2-2*B*a*b)*sec(d*x+c)
*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2916, 835, 792, 205, 212} \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (5 a^2 A-2 a b B-A b^2\right ) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\sec ^4(c+d x) \left (\left (5 a^2 A-2 a b B+3 A b^2\right ) \sin (c+d x)+2 b (4 a A-b B)\right )}{24 d}+\frac {\left (5 a^2 A-2 a b B-A b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {\sec ^6(c+d x) (a+b \sin (c+d x))^2 (A \sin (c+d x)+B)}{6 d} \]

[In]

Int[Sec[c + d*x]^7*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

((5*a^2*A - A*b^2 - 2*a*b*B)*ArcTanh[Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^6*(B + A*Sin[c + d*x])*(a + b*Sin[c
+ d*x])^2)/(6*d) + (Sec[c + d*x]^4*(2*b*(4*a*A - b*B) + (5*a^2*A + 3*A*b^2 - 2*a*b*B)*Sin[c + d*x]))/(24*d) +
((5*a^2*A - A*b^2 - 2*a*b*B)*Sec[c + d*x]*Tan[c + d*x])/(16*d)

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 212

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

Rule 792

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a*(e*f + d*g) - (
c*d*f - a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^m*(
a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 2916

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

Rubi steps \begin{align*} \text {integral}& = \frac {b^7 \text {Subst}\left (\int \frac {(a+x)^2 \left (A+\frac {B x}{b}\right )}{\left (b^2-x^2\right )^4} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}-\frac {b^5 \text {Subst}\left (\int \frac {(a+x) (-5 a A+2 b B-3 A x)}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{6 d} \\ & = \frac {\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac {\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac {\left (b^3 \left (5 a^2 A-A b^2-2 a b B\right )\right ) \text {Subst}\left (\int \frac {1}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac {\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac {\left (5 a^2 A-A b^2-2 a b B\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {\left (b \left (5 a^2 A-A b^2-2 a b B\right )\right ) \text {Subst}\left (\int \frac {1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{16 d} \\ & = \frac {\left (5 a^2 A-A b^2-2 a b B\right ) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac {\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac {\left (5 a^2 A-A b^2-2 a b B\right ) \sec (c+d x) \tan (c+d x)}{16 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.51 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {b \sec ^6(c+d x) (a+b \sin (c+d x))^3 (A b-a B+(-a A+b B) \sin (c+d x))+\frac {1}{4} b \sec ^4(c+d x) (a+b \sin (c+d x))^3 (3 A b+(-5 a A+2 b B) \sin (c+d x))-\frac {3 b \left (-5 a^2 A+A b^2+2 a b B\right ) \left (\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (1+\sin (c+d x)))+2 a^3 b \sec ^2(c+d x)-2 \left (a^4-b^4\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 b+4 a b^3\right ) \tan ^2(c+d x)\right )}{16 (a-b) (a+b)}}{6 b \left (-a^2+b^2\right ) d} \]

[In]

Integrate[Sec[c + d*x]^7*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(b*Sec[c + d*x]^6*(a + b*Sin[c + d*x])^3*(A*b - a*B + (-(a*A) + b*B)*Sin[c + d*x]) + (b*Sec[c + d*x]^4*(a + b*
Sin[c + d*x])^3*(3*A*b + (-5*a*A + 2*b*B)*Sin[c + d*x]))/4 - (3*b*(-5*a^2*A + A*b^2 + 2*a*b*B)*((a^2 - b^2)^2*
(Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + 2*a^3*b*Sec[c + d*x]^2 - 2*(a^4 - b^4)*Sec[c + d*x]*Tan[c +
d*x] + (-6*a^3*b + 4*a*b^3)*Tan[c + d*x]^2))/(16*(a - b)*(a + b)))/(6*b*(-a^2 + b^2)*d)

Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.89

method	result	size
derivativedivides	\(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{2}}{6 \cos \left (d x +c \right )^{6}}+\frac {A a b}{3 \cos \left (d x +c \right )^{6}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+B \,b^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )}{d}\)	\(302\)
default	\(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{2}}{6 \cos \left (d x +c \right )^{6}}+\frac {A a b}{3 \cos \left (d x +c \right )^{6}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+B \,b^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )}{d}\)	\(302\)
parallelrisch	\(\frac {-15 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (A \,a^{2}-\frac {1}{5} A \,b^{2}-\frac {2}{5} B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (A \,a^{2}-\frac {1}{5} A \,b^{2}-\frac {2}{5} B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-240 A a b -120 B \,a^{2}-132 B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-96 A a b -48 B \,a^{2}+24 B \,b^{2}\right ) \cos \left (4 d x +4 c \right )+\left (-16 A a b -8 B \,a^{2}+4 B \,b^{2}\right ) \cos \left (6 d x +6 c \right )+\left (170 A \,a^{2}-34 A \,b^{2}-68 B a b \right ) \sin \left (3 d x +3 c \right )+\left (30 A \,a^{2}-6 A \,b^{2}-12 B a b \right ) \sin \left (5 d x +5 c \right )+\left (396 A \,a^{2}+228 A \,b^{2}+456 B a b \right ) \sin \left (d x +c \right )+352 A a b +176 B \,a^{2}+104 B \,b^{2}}{48 d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\)	\(352\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (15 A \,a^{2} {\mathrm e}^{10 i \left (d x +c \right )}-3 A \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}-6 B a b \,{\mathrm e}^{10 i \left (d x +c \right )}+85 A \,a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-17 A \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-34 B a b \,{\mathrm e}^{8 i \left (d x +c \right )}+198 A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+114 A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+228 B a b \,{\mathrm e}^{6 i \left (d x +c \right )}+256 i B \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-198 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-114 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-96 i B \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-228 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+512 i A a b \,{\mathrm e}^{5 i \left (d x +c \right )}-96 i B \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-85 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+17 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+34 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}+64 i B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-15 A \,a^{2}+3 A \,b^{2}+6 B a b \right )}{24 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2}}{16 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{16 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{8 d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2}}{16 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{16 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{8 d}\)	\(519\)

[In]

int(sec(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(A*a^2*(-(-1/6*sec(d*x+c)^5-5/24*sec(d*x+c)^3-5/16*sec(d*x+c))*tan(d*x+c)+5/16*ln(sec(d*x+c)+tan(d*x+c)))+
1/6*B*a^2/cos(d*x+c)^6+1/3*A*a*b/cos(d*x+c)^6+2*B*a*b*(1/6*sin(d*x+c)^3/cos(d*x+c)^6+1/8*sin(d*x+c)^3/cos(d*x+
c)^4+1/16*sin(d*x+c)^3/cos(d*x+c)^2+1/16*sin(d*x+c)-1/16*ln(sec(d*x+c)+tan(d*x+c)))+A*b^2*(1/6*sin(d*x+c)^3/co
s(d*x+c)^6+1/8*sin(d*x+c)^3/cos(d*x+c)^4+1/16*sin(d*x+c)^3/cos(d*x+c)^2+1/16*sin(d*x+c)-1/16*ln(sec(d*x+c)+tan
(d*x+c)))+B*b^2*(1/6*sin(d*x+c)^4/cos(d*x+c)^6+1/12*sin(d*x+c)^4/cos(d*x+c)^4))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.27 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 24 \, B b^{2} \cos \left (d x + c\right )^{2} + 16 \, B a^{2} + 32 \, A a b + 16 \, B b^{2} + 2 \, {\left (3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, A a^{2} + 16 \, B a b + 8 \, A b^{2} + 2 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \]

[In]

integrate(sec(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d
*x + c)^6*log(-sin(d*x + c) + 1) - 24*B*b^2*cos(d*x + c)^2 + 16*B*a^2 + 32*A*a*b + 16*B*b^2 + 2*(3*(5*A*a^2 -
2*B*a*b - A*b^2)*cos(d*x + c)^4 + 8*A*a^2 + 16*B*a*b + 8*A*b^2 + 2*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^2)
*sin(d*x + c))/(d*cos(d*x + c)^6)

Sympy [F(-1)]

Timed out. \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**7*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.32 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{5} + 12 \, B b^{2} \sin \left (d x + c\right )^{2} - 8 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} + 8 \, B a^{2} + 16 \, A a b - 4 \, B b^{2} + 3 \, {\left (11 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(sin(d*x + c) + 1) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(sin(d*x + c) - 1
) - 2*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^5 + 12*B*b^2*sin(d*x + c)^2 - 8*(5*A*a^2 - 2*B*a*b - A*b^2)*
sin(d*x + c)^3 + 8*B*a^2 + 16*A*a*b - 4*B*b^2 + 3*(11*A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c))/(sin(d*x + c)^6 -
3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1))/d

Giac [A] (verification not implemented)

none

Time = 0.64 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.43 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{2} \sin \left (d x + c\right )^{5} - 6 \, B a b \sin \left (d x + c\right )^{5} - 3 \, A b^{2} \sin \left (d x + c\right )^{5} - 40 \, A a^{2} \sin \left (d x + c\right )^{3} + 16 \, B a b \sin \left (d x + c\right )^{3} + 8 \, A b^{2} \sin \left (d x + c\right )^{3} + 12 \, B b^{2} \sin \left (d x + c\right )^{2} + 33 \, A a^{2} \sin \left (d x + c\right ) + 6 \, B a b \sin \left (d x + c\right ) + 3 \, A b^{2} \sin \left (d x + c\right ) + 8 \, B a^{2} + 16 \, A a b - 4 \, B b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \]

[In]

integrate(sec(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x + c) + 1)) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x
+ c) - 1)) - 2*(15*A*a^2*sin(d*x + c)^5 - 6*B*a*b*sin(d*x + c)^5 - 3*A*b^2*sin(d*x + c)^5 - 40*A*a^2*sin(d*x
+ c)^3 + 16*B*a*b*sin(d*x + c)^3 + 8*A*b^2*sin(d*x + c)^3 + 12*B*b^2*sin(d*x + c)^2 + 33*A*a^2*sin(d*x + c) +
6*B*a*b*sin(d*x + c) + 3*A*b^2*sin(d*x + c) + 8*B*a^2 + 16*A*a*b - 4*B*b^2)/(sin(d*x + c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 12.36 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.38 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\mathrm {atanh}\left (\frac {4\,\sin \left (c+d\,x\right )\,\left (-\frac {5\,A\,a^2}{32}+\frac {B\,a\,b}{16}+\frac {A\,b^2}{32}\right )}{-\frac {5\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}}\right )\,\left (-\frac {5\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )}{d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {11\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )+\frac {B\,a^2}{6}-\frac {B\,b^2}{12}+{\sin \left (c+d\,x\right )}^3\,\left (-\frac {5\,A\,a^2}{6}+\frac {B\,a\,b}{3}+\frac {A\,b^2}{6}\right )-{\sin \left (c+d\,x\right )}^5\,\left (-\frac {5\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )+\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^2}{4}+\frac {A\,a\,b}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^6-3\,{\sin \left (c+d\,x\right )}^4+3\,{\sin \left (c+d\,x\right )}^2-1\right )} \]

[In]

int(((A + B*sin(c + d*x))*(a + b*sin(c + d*x))^2)/cos(c + d*x)^7,x)

[Out]

- (atanh((4*sin(c + d*x)*((A*b^2)/32 - (5*A*a^2)/32 + (B*a*b)/16))/((A*b^2)/8 - (5*A*a^2)/8 + (B*a*b)/4))*((A*
b^2)/16 - (5*A*a^2)/16 + (B*a*b)/8))/d - (sin(c + d*x)*((11*A*a^2)/16 + (A*b^2)/16 + (B*a*b)/8) + (B*a^2)/6 -
(B*b^2)/12 + sin(c + d*x)^3*((A*b^2)/6 - (5*A*a^2)/6 + (B*a*b)/3) - sin(c + d*x)^5*((A*b^2)/16 - (5*A*a^2)/16
+ (B*a*b)/8) + (B*b^2*sin(c + d*x)^2)/4 + (A*a*b)/3)/(d*(3*sin(c + d*x)^2 - 3*sin(c + d*x)^4 + sin(c + d*x)^6
- 1))