3.1000 \(\int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=107 \[ \frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^3}{5 d} \]

[Out]

1/15*a^5*(2*A-3*B)*cos(d*x+c)/d/(a-a*sin(d*x+c))^2+1/5*(A+B)*sec(d*x+c)^5*(a+a*sin(d*x+c))^3/d+1/15*a^5*(2*A-3
*B)*cos(d*x+c)/d/(a^2-a^2*sin(d*x+c))

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Rubi [A]  time = 0.15, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2855, 2670, 2650, 2648} \[ \frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac {(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^3}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^5*(2*A - 3*B)*Cos[c + d*x])/(15*d*(a - a*Sin[c + d*x])^2) + ((A + B)*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3)
/(5*d) + (a^5*(2*A - 3*B)*Cos[c + d*x])/(15*d*(a^2 - a^2*Sin[c + d*x]))

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2670

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 2855

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*} \int \sec ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac {1}{5} (a (2 A-3 B)) \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\\ &=\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac {1}{5} \left (a^5 (2 A-3 B)\right ) \int \frac {1}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}+\frac {1}{15} \left (a^4 (2 A-3 B)\right ) \int \frac {1}{a-a \sin (c+d x)} \, dx\\ &=\frac {a^5 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac {a^4 (2 A-3 B) \cos (c+d x)}{15 d (a-a \sin (c+d x))}+\frac {(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^3}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 94, normalized size = 0.88 \[ -\frac {a^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) (6 (2 A-3 B) \sin (c+d x)+(2 A-3 B) \cos (2 (c+d x))-16 A+9 B)}{30 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

-1/30*(a^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-16*A + 9*B + (2*A - 3*B)*Cos[2*(c + d*x)] + 6*(2*A - 3*B)*S
in[c + d*x]))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5)

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fricas [A]  time = 0.69, size = 188, normalized size = 1.76 \[ \frac {{\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} - 2 \, {\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, A - 2 \, B\right )} a^{3} \cos \left (d x + c\right ) - 3 \, {\left (A + B\right )} a^{3} + {\left ({\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right ) - 3 \, {\left (A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + 3 \, d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/15*((2*A - 3*B)*a^3*cos(d*x + c)^3 - 2*(2*A - 3*B)*a^3*cos(d*x + c)^2 - 3*(3*A - 2*B)*a^3*cos(d*x + c) - 3*(
A + B)*a^3 + ((2*A - 3*B)*a^3*cos(d*x + c)^2 + 3*(2*A - 3*B)*a^3*cos(d*x + c) - 3*(A + B)*a^3)*sin(d*x + c))/(
d*cos(d*x + c)^3 + 3*d*cos(d*x + c)^2 - 2*d*cos(d*x + c) - (d*cos(d*x + c)^2 - 2*d*cos(d*x + c) - 4*d)*sin(d*x
+ c) - 4*d)

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giac [A]  time = 0.25, size = 146, normalized size = 1.36 \[ -\frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 20 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, A a^{3} - 3 \, B a^{3}\right )}}{15 \, d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-2/15*(15*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 30*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 4
0*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 15*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 20*A*a^3*tan(1/2*d*x + 1/2*c) + 15*B*a^3*ta
n(1/2*d*x + 1/2*c) + 7*A*a^3 - 3*B*a^3)/(d*(tan(1/2*d*x + 1/2*c) - 1)^5)

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maple [B]  time = 0.66, size = 333, normalized size = 3.11 \[ \frac {a^{3} A \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )^{5}}+3 a^{3} A \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )+\frac {3 a^{3} A}{5 \cos \left (d x +c \right )^{5}}+3 B \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )-a^{3} A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {B \,a^{3}}{5 \cos \left (d x +c \right )^{5}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(1/5*sin(d*x+c)^4/cos(d*x+c)^5+1/15*sin(d*x+c)^4/cos(d*x+c)^3-1/15*sin(d*x+c)^4/cos(d*x+c)-1/15*(2+
sin(d*x+c)^2)*cos(d*x+c))+1/5*B*a^3*sin(d*x+c)^5/cos(d*x+c)^5+3*a^3*A*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(
d*x+c)^3/cos(d*x+c)^3)+3*B*a^3*(1/5*sin(d*x+c)^4/cos(d*x+c)^5+1/15*sin(d*x+c)^4/cos(d*x+c)^3-1/15*sin(d*x+c)^4
/cos(d*x+c)-1/15*(2+sin(d*x+c)^2)*cos(d*x+c))+3/5*a^3*A/cos(d*x+c)^5+3*B*a^3*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/
15*sin(d*x+c)^3/cos(d*x+c)^3)-a^3*A*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+1/5*B*a^3/cos(d*x+c)
^5)

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maxima [A]  time = 0.33, size = 188, normalized size = 1.76 \[ \frac {3 \, B a^{3} \tan \left (d x + c\right )^{5} + {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 3 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} A a^{3} + 3 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} B a^{3} - \frac {{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} A a^{3}}{\cos \left (d x + c\right )^{5}} - \frac {3 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} B a^{3}}{\cos \left (d x + c\right )^{5}} + \frac {9 \, A a^{3}}{\cos \left (d x + c\right )^{5}} + \frac {3 \, B a^{3}}{\cos \left (d x + c\right )^{5}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/15*(3*B*a^3*tan(d*x + c)^5 + (3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 3*(3*tan(d*x +
c)^5 + 5*tan(d*x + c)^3)*A*a^3 + 3*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*B*a^3 - (5*cos(d*x + c)^2 - 3)*A*a^3
/cos(d*x + c)^5 - 3*(5*cos(d*x + c)^2 - 3)*B*a^3/cos(d*x + c)^5 + 9*A*a^3/cos(d*x + c)^5 + 3*B*a^3/cos(d*x + c
)^5)/d

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mupad [B]  time = 11.31, size = 113, normalized size = 1.06 \[ -\frac {\sqrt {2}\,a^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,B-\frac {53\,A}{4}+4\,A\,\cos \left (c+d\,x\right )+\frac {3\,B\,\cos \left (c+d\,x\right )}{2}+\frac {25\,A\,\sin \left (c+d\,x\right )}{2}-\frac {15\,B\,\sin \left (c+d\,x\right )}{2}+\frac {9\,A\,\cos \left (2\,c+2\,d\,x\right )}{4}-\frac {3\,B\,\cos \left (2\,c+2\,d\,x\right )}{2}-\frac {5\,A\,\sin \left (2\,c+2\,d\,x\right )}{4}\right )}{60\,d\,{\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^6,x)

[Out]

-(2^(1/2)*a^3*cos(c/2 + (d*x)/2)*(3*B - (53*A)/4 + 4*A*cos(c + d*x) + (3*B*cos(c + d*x))/2 + (25*A*sin(c + d*x
))/2 - (15*B*sin(c + d*x))/2 + (9*A*cos(2*c + 2*d*x))/4 - (3*B*cos(2*c + 2*d*x))/2 - (5*A*sin(2*c + 2*d*x))/4)
)/(60*d*cos(c/2 + pi/4 + (d*x)/2)^5)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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