\(\int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [1542]

Optimal result

Integrand size = 31, antiderivative size = 122 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (3 a^2 A-A b^2-2 a b B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{4 d}+\frac {\sec ^2(c+d x) \left (2 b (2 a A-b B)+\left (3 a^2 A+A b^2-2 a b B\right ) \sin (c+d x)\right )}{8 d} \]

[Out]

1/8*(3*A*a^2-A*b^2-2*B*a*b)*arctanh(sin(d*x+c))/d+1/4*sec(d*x+c)^4*(B+A*sin(d*x+c))*(a+b*sin(d*x+c))^2/d+1/8*s
ec(d*x+c)^2*(2*b*(2*A*a-B*b)+(3*A*a^2+A*b^2-2*B*a*b)*sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2916, 835, 792, 212} \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (3 a^2 A-2 a b B-A b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\sec ^2(c+d x) \left (\left (3 a^2 A-2 a b B+A b^2\right ) \sin (c+d x)+2 b (2 a A-b B)\right )}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2 (A \sin (c+d x)+B)}{4 d} \]

[In]

Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

((3*a^2*A - A*b^2 - 2*a*b*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (Sec[c + d*x]^4*(B + A*Sin[c + d*x])*(a + b*Sin[c
+ d*x])^2)/(4*d) + (Sec[c + d*x]^2*(2*b*(2*a*A - b*B) + (3*a^2*A + A*b^2 - 2*a*b*B)*Sin[c + d*x]))/(8*d)

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^5 \text {Subst}\left (\int \frac {(a+x)^2 \left (A+\frac {B x}{b}\right )}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^4(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{4 d}-\frac {b^3 \text {Subst}\left (\int \frac {(a+x) (-3 a A+2 b B-A x)}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {\sec ^4(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{4 d}+\frac {\sec ^2(c+d x) \left (2 b (2 a A-b B)+\left (3 a^2 A+A b^2-2 a b B\right ) \sin (c+d x)\right )}{8 d}+\frac {\left (b \left (3 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 )}{8 d} \\ & = \frac {\left (3 a^2 A-A b^2-2 a b B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{4 d}+\frac {\sec ^2(c+d x) \left (2 b (2 a A-b B)+\left (3 a^2 A+A b^2-2 a b B\right ) \sin (c+d x)\right )}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.52 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {4 \left (-a^2+b^2\right ) \sec ^4(c+d x) (a+b \sin (c+d x))^3 (A b-a B+(-a A+b B) \sin (c+d x))+\left (-3 a^2 A+A b^2+2 a b B\right ) \left (\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (1+\sin (c+d x)))+2 a^3 b \sec ^2(c+d x)-2 \left (a^4-b^4\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 b+4 a b^3\right ) \tan ^2(c+d x)\right )}{16 \left (a^2-b^2\right )^2 d} \]

[In]

Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(4*(-a^2 + b^2)*Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3*(A*b - a*B + (-(a*A) + b*B)*Sin[c + d*x]) + (-3*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^2 - b^2)^2*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(235\) vs. \(2(116)=232\).

Time = 0.95 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.93

method	result	size
derivativedivides	\(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,a^{2}}{4 \cos \left (d x +c \right )^{4}}+\frac {A a b}{2 \cos \left (d x +c \right )^{4}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,b^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(236\)
default	\(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,a^{2}}{4 \cos \left (d x +c \right )^{4}}+\frac {A a b}{2 \cos \left (d x +c \right )^{4}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,b^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(236\)
parallelrisch	\(\frac {-6 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A \,a^{2}-\frac {1}{3} A \,b^{2}-\frac {2}{3} B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A \,a^{2}-\frac {1}{3} A \,b^{2}-\frac {2}{3} B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (-2 A a b -B \,a^{2}-B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-2 A a b -B \,a^{2}+B \,b^{2}\right ) \cos \left (4 d x +4 c \right )+\left (3 A \,a^{2}-A \,b^{2}-2 B a b \right ) \sin \left (3 d x +3 c \right )+\left (11 A \,a^{2}+7 A \,b^{2}+14 B a b \right ) \sin \left (d x +c \right )+10 A a b +5 B \,a^{2}+3 B \,b^{2}}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(263\)
risch	\(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (3 A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-2 B a b \,{\mathrm e}^{6 i \left (d x +c \right )}+11 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+7 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+14 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+32 i A a b \,{\mathrm e}^{3 i \left (d x +c \right )}-11 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-7 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-8 i B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-14 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}+16 i B \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-3 A \,a^{2}+A \,b^{2}+2 B a b -8 i B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2}}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{4 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2}}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{4 d}\)	\(398\)
norman	\(\frac {\frac {\left (7 A \,a^{2}+11 A \,b^{2}+22 B a b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 A a b +2 B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 A a b +2 B \,a^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (12 A a b +6 B \,a^{2}+4 B \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (12 A a b +6 B \,a^{2}+4 B \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (16 A a b +8 B \,a^{2}+12 B \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (16 A a b +8 B \,a^{2}+12 B \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (5 A \,a^{2}+A \,b^{2}+2 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (5 A \,a^{2}+A \,b^{2}+2 B a b \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (9 A \,a^{2}+5 A \,b^{2}+10 B a b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (9 A \,a^{2}+5 A \,b^{2}+10 B a b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (27 A \,a^{2}+31 A \,b^{2}+62 B a b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (27 A \,a^{2}+31 A \,b^{2}+62 B a b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {\left (3 A \,a^{2}-A \,b^{2}-2 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 A \,a^{2}-A \,b^{2}-2 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(523\)

[In]

int(sec(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(A*a^2*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+1/4*B*a^2/cos(d*x+c)
^4+1/2*A*a*b/cos(d*x+c)^4+2*B*a*b*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-
1/8*ln(sec(d*x+c)+tan(d*x+c)))+A*b^2*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+
c)-1/8*ln(sec(d*x+c)+tan(d*x+c)))+1/4*B*b^2*sin(d*x+c)^4/cos(d*x+c)^4)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.42 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, B b^{2} \cos \left (d x + c\right )^{2} + 4 \, B a^{2} + 8 \, A a b + 4 \, B b^{2} + 2 \, {\left (2 \, A a^{2} + 4 \, B a b + 2 \, A b^{2} + {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.40 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (4 \, B b^{2} \sin \left (d x + c\right )^{2} - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} + 2 \, B a^{2} + 4 \, A a b - 2 \, B b^{2} + {\left (5 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.53 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \sin \left (d x + c\right )^{3} - 2 \, B a b \sin \left (d x + c\right )^{3} - A b^{2} \sin \left (d x + c\right )^{3} - 4 \, B b^{2} \sin \left (d x + c\right )^{2} - 5 \, A a^{2} \sin \left (d x + c\right ) - 2 \, B a b \sin \left (d x + c\right ) - A b^{2} \sin \left (d x + c\right ) - 2 \, B a^{2} - 4 \, A a b + 2 \, B b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*((3*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x + c) + 1)) - (3*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x + c
) - 1)) - 2*(3*A*a^2*sin(d*x + c)^3 - 2*B*a*b*sin(d*x + c)^3 - A*b^2*sin(d*x + c)^3 - 4*B*b^2*sin(d*x + c)^2 -
5*A*a^2*sin(d*x + c) - 2*B*a*b*sin(d*x + c) - A*b^2*sin(d*x + c) - 2*B*a^2 - 4*A*a*b + 2*B*b^2)/(sin(d*x + c)
^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 12.18 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.48 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\sin \left (c+d\,x\right )\,\left (\frac {5\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}\right )+\frac {B\,a^2}{4}-\frac {B\,b^2}{4}+{\sin \left (c+d\,x\right )}^3\,\left (-\frac {3\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}\right )+\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {A\,a\,b}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\frac {4\,\sin \left (c+d\,x\right )\,\left (-\frac {3\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )}{-\frac {3\,A\,a^2}{4}+\frac {B\,a\,b}{2}+\frac {A\,b^2}{4}}\right )\,\left (-\frac {3\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}\right )}{d} \]

[In]

int(((A + B*sin(c + d*x))*(a + b*sin(c + d*x))^2)/cos(c + d*x)^5,x)

[Out]

(sin(c + d*x)*((5*A*a^2)/8 + (A*b^2)/8 + (B*a*b)/4) + (B*a^2)/4 - (B*b^2)/4 + sin(c + d*x)^3*((A*b^2)/8 - (3*A
*a^2)/8 + (B*a*b)/4) + (B*b^2*sin(c + d*x)^2)/2 + (A*a*b)/2)/(d*(sin(c + d*x)^4 - 2*sin(c + d*x)^2 + 1)) - (at
anh((4*sin(c + d*x)*((A*b^2)/16 - (3*A*a^2)/16 + (B*a*b)/8))/((A*b^2)/4 - (3*A*a^2)/4 + (B*a*b)/2))*((A*b^2)/8
- (3*A*a^2)/8 + (B*a*b)/4))/d