\(\int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [999]

Optimal result

Integrand size = 31, antiderivative size = 69 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=a^3 B x+\frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\frac {2 a^5 B \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )} \]

[Out]

a^3*B*x+1/3*(A+B)*sec(d*x+c)^3*(a+a*sin(d*x+c))^3/d-2*a^5*B*cos(d*x+c)/d/(a^2-a^2*sin(d*x+c))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 69, 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 = {2934, 2749, 2759, 8} \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=a^3 B x-\frac {2 a^5 B \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

a^3*B*x + ((A + B)*Sec[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(3*d) - (2*a^5*B*Cos[c + d*x])/(d*(a^2 - a^2*Sin[c +
d*x]))

Rule 8

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

Rule 2749

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a/g)^
(2*m), Int[(g*Cos[e + f*x])^(2*m + p)/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 -
b^2, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]

Rule 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2934

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

Rubi steps \begin{align*} \text {integral}& = \frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-(a B) \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx \\ & = \frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\left (a^5 B\right ) \int \frac {\cos ^2(c+d x)}{(a-a \sin (c+d x))^2} \, dx \\ & = \frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\frac {2 a^5 B \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )}+\left (a^3 B\right ) \int 1 \, dx \\ & = a^3 B x+\frac {(A+B) \sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\frac {2 a^5 B \cos (c+d x)}{d \left (a^2-a^2 \sin (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.90 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.75 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {a^3 \left (-3 (2 A+3 B (2+c+d x)) \cos \left (\frac {1}{2} (c+d x)\right )+(2 A+B (14+3 c+3 d x)) \cos \left (\frac {3}{2} (c+d x)\right )+6 B (2 (2+c+d x)+(c+d x) \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

-1/6*(a^3*(-3*(2*A + 3*B*(2 + c + d*x))*Cos[(c + d*x)/2] + (2*A + B*(14 + 3*c + 3*d*x))*Cos[(3*(c + d*x))/2] +
6*B*(2*(2 + c + d*x) + (c + d*x)*Cos[c + d*x])*Sin[(c + d*x)/2]))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25

method	result	size
risch	\(a^{3} B x -\frac {2 \left (3 A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-12 i B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}+9 B \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-A \,a^{3}-7 B \,a^{3}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\)	\(86\)
parallelrisch	\(-\frac {2 a^{3} \left (-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) d x}{2}+\left (\frac {3}{2} d x B +A -B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+B \left (-\frac {3 d x}{2}+4\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {d x B}{2}+\frac {A}{3}-\frac {5 B}{3}\right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(89\)
derivativedivides	\(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+B \,a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+\frac {A \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {A \,a^{3}}{\cos \left (d x +c \right )^{3}}+\frac {B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}-A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {B \,a^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(248\)
default	\(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+B \,a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+\frac {A \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {A \,a^{3}}{\cos \left (d x +c \right )^{3}}+\frac {B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}-A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {B \,a^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(248\)
norman	\(\frac {\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+a^{3} B x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{3} B x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {20 A \,a^{3}-4 B \,a^{3}}{3 d}-a^{3} B x -\frac {\left (8 A \,a^{3}+24 B \,a^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (38 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (46 A \,a^{3}+26 B \,a^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (98 A \,a^{3}+182 B \,a^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (212 A \,a^{3}+188 B \,a^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-a^{3} B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{3} B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{3} B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{3} B x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{3} B x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 a^{3} \left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \left (A -B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{3} \left (7 A +9 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{3} \left (11 A +5 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a^{3} \left (11 A +5 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{3} \left (61 A +67 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{3} \left (61 A +67 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(531\)

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

a^3*B*x-2/3*(3*A*a^3*exp(2*I*(d*x+c))-12*I*B*a^3*exp(I*(d*x+c))+9*B*a^3*exp(2*I*(d*x+c))-A*a^3-7*B*a^3)/(exp(I
*(d*x+c))-I)^3/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (68) = 136\).

Time = 0.30 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.42 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {6 \, B a^{3} d x + 2 \, {\left (A + B\right )} a^{3} - {\left (3 \, B a^{3} d x + {\left (A + 7 \, B\right )} a^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B a^{3} d x + {\left (A - 5 \, B\right )} a^{3}\right )} \cos \left (d x + c\right ) - {\left (6 \, B a^{3} d x - 2 \, {\left (A + B\right )} a^{3} + {\left (3 \, B a^{3} d x - {\left (A + 7 \, B\right )} a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/3*(6*B*a^3*d*x + 2*(A + B)*a^3 - (3*B*a^3*d*x + (A + 7*B)*a^3)*cos(d*x + c)^2 + (3*B*a^3*d*x + (A - 5*B)*a^
3)*cos(d*x + c) - (6*B*a^3*d*x - 2*(A + B)*a^3 + (3*B*a^3*d*x - (A + 7*B)*a^3)*cos(d*x + c))*sin(d*x + c))/(d*
cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**4*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (68) = 136\).

Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.38 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {3 \, A a^{3} \tan \left (d x + c\right )^{3} + 3 \, B a^{3} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} B a^{3} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} A a^{3}}{\cos \left (d x + c\right )^{3}} - \frac {3 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} B a^{3}}{\cos \left (d x + c\right )^{3}} + \frac {3 \, A a^{3}}{\cos \left (d x + c\right )^{3}} + \frac {B a^{3}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.35 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (d x + c\right )} B a^{3} - \frac {2 \, {\left (3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a^{3} - 5 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*B*a^3 - 2*(3*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 3*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*B*a^3*tan(1/2
*d*x + 1/2*c) + A*a^3 - 5*B*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 9.98 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.03 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=B\,a^3\,x-\frac {\frac {a^3\,\left (2\,A-10\,B+3\,B\,\left (c+d\,x\right )\right )}{3}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (6\,A-6\,B+9\,B\,\left (c+d\,x\right )\right )}{3}-3\,B\,a^3\,\left (c+d\,x\right )\right )+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3\,\left (24\,B-9\,B\,\left (c+d\,x\right )\right )}{3}+3\,B\,a^3\,\left (c+d\,x\right )\right )-B\,a^3\,\left (c+d\,x\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \]

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^4,x)

[Out]

B*a^3*x - ((a^3*(2*A - 10*B + 3*B*(c + d*x)))/3 + tan(c/2 + (d*x)/2)^2*((a^3*(6*A - 6*B + 9*B*(c + d*x)))/3 -
3*B*a^3*(c + d*x)) + tan(c/2 + (d*x)/2)*((a^3*(24*B - 9*B*(c + d*x)))/3 + 3*B*a^3*(c + d*x)) - B*a^3*(c + d*x)
)/(d*(tan(c/2 + (d*x)/2) - 1)^3)