\(\int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [1541]

Optimal result

Integrand size = 31, antiderivative size = 112 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac {(a-b) (a A+b (A-2 B)) \log (1+\sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2916, 833, 647, 31} \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac {(a-b) (a A+b (A-2 B)) \log (\sin (c+d x)+1)}{4 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) ((a A+b B) \sin (c+d x)+a B+A b)}{2 d} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*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, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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^3 \text {Subst}\left (\int \frac {(a+x)^2 \left (A+\frac {B x}{b}\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}-\frac {b \text {Subst}\left (\int \frac {-a^2 A+A b^2+2 a b B+2 b B x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 d} \\ & = \frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}-\frac {((a-b) (a A+b (A-2 B))) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{4 d}+\frac {((a+b) (a A-b (A+2 B))) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = -\frac {(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac {(a-b) (a A+b (A-2 B)) \log (1+\sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.55 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (a^2-b^2\right ) ((a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))-(a-b) (a A+b (A-2 B)) \log (1+\sin (c+d x)))-2 a^3 (-A b+a B) \sec ^2(c+d x)-2 \left (a^2-b^2\right ) \left (a^2 A+A b^2+2 a b B\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 A b+4 a A b^3+2 b^4 B\right ) \tan ^2(c+d x)}{4 \left (-a^2+b^2\right ) d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.67

method	result	size
derivativedivides	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B \,a^{2}}{2 \cos \left (d x +c \right )^{2}}+\frac {A a b}{\cos \left (d x +c \right )^{2}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,b^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(187\)
default	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B \,a^{2}}{2 \cos \left (d x +c \right )^{2}}+\frac {A a b}{\cos \left (d x +c \right )^{2}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,b^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(187\)
parallelrisch	\(\frac {-2 B \,b^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (-A -2 B \right ) b +a A \right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a A +b \left (A -2 B \right )\right ) \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-2 A a b -B \,a^{2}-B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (2 A \,a^{2}+2 A \,b^{2}+4 B a b \right ) \sin \left (d x +c \right )+2 A a b +B \,a^{2}+B \,b^{2}}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(198\)
risch	\(-i B \,b^{2} x -\frac {2 i B \,b^{2} c}{d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,a^{2}-A \,b^{2}+4 i A a b \,{\mathrm e}^{i \left (d x +c \right )}-2 B a b +2 i B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}\)	\(335\)
norman	\(\frac {\frac {\left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 A a b +2 B \,a^{2}+2 B \,b^{2}}{d}+\frac {4 \left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 \left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (4 A a b +2 B \,a^{2}+2 B \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (10 A a b +5 B \,a^{2}+5 B \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (10 A a b +5 B \,a^{2}+5 B \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b -2 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b +2 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}-\frac {B \,b^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(418\)

[In]

int(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.21 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - {\left (A - 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - 2 \, B a b - {\left (A + 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a^{2} + 4 \, A a b + 2 \, B b^{2} + 2 \, {\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\int \left (A + B \sin {\left (c + d x \right )}\right ) \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**3*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Integral((A + B*sin(c + d*x))*(a + b*sin(c + d*x))**2*sec(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.09 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - {\left (A - 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - 2 \, B a b - {\left (A + 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (B a^{2} + 2 \, A a b + B b^{2} + {\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.30 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - A b^{2} + 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (A a^{2} - 2 \, B a b - A b^{2} - 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (B b^{2} \sin \left (d x + c\right )^{2} + A a^{2} \sin \left (d x + c\right ) + 2 \, B a b \sin \left (d x + c\right ) + A b^{2} \sin \left (d x + c\right ) + B a^{2} + 2 \, A a b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.65 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-b\right )\,\left (A\,a+A\,b-2\,B\,b\right )}{4\,d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{2}+B\,a\,b+\frac {A\,b^2}{2}\right )+\frac {B\,a^2}{2}+\frac {B\,b^2}{2}+A\,a\,b}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (A\,b-A\,a+2\,B\,b\right )}{4\,d} \]

[In]

int(((A + B*sin(c + d*x))*(a + b*sin(c + d*x))^2)/cos(c + d*x)^3,x)

[Out]

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