3.983 \(\int \sec ^8(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=129 \[ \frac {a^2 (5 A-2 B) \tan ^5(c+d x)}{35 d}+\frac {2 a^2 (5 A-2 B) \tan ^3(c+d x)}{21 d}+\frac {a^2 (5 A-2 B) \tan (c+d x)}{7 d}+\frac {a^2 (5 A-2 B) \sec ^5(c+d x)}{35 d}+\frac {(A+B) \sec ^7(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

[Out]

1/35*a^2*(5*A-2*B)*sec(d*x+c)^5/d+1/7*(A+B)*sec(d*x+c)^7*(a+a*sin(d*x+c))^2/d+1/7*a^2*(5*A-2*B)*tan(d*x+c)/d+2
/21*a^2*(5*A-2*B)*tan(d*x+c)^3/d+1/35*a^2*(5*A-2*B)*tan(d*x+c)^5/d

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Rubi [A]  time = 0.13, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2855, 2669, 3767} \[ \frac {a^2 (5 A-2 B) \tan ^5(c+d x)}{35 d}+\frac {2 a^2 (5 A-2 B) \tan ^3(c+d x)}{21 d}+\frac {a^2 (5 A-2 B) \tan (c+d x)}{7 d}+\frac {a^2 (5 A-2 B) \sec ^5(c+d x)}{35 d}+\frac {(A+B) \sec ^7(c+d x) (a \sin (c+d x)+a)^2}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(5*A - 2*B)*Sec[c + d*x]^5)/(35*d) + ((A + B)*Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2)/(7*d) + (a^2*(5*A -
2*B)*Tan[c + d*x])/(7*d) + (2*a^2*(5*A - 2*B)*Tan[c + d*x]^3)/(21*d) + (a^2*(5*A - 2*B)*Tan[c + d*x]^5)/(35*d)

Rule 2669

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \sec ^8(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac {(A+B) \sec ^7(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac {1}{7} (a (5 A-2 B)) \int \sec ^6(c+d x) (a+a \sin (c+d x)) \, dx\\ &=\frac {a^2 (5 A-2 B) \sec ^5(c+d x)}{35 d}+\frac {(A+B) \sec ^7(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac {1}{7} \left (a^2 (5 A-2 B)\right ) \int \sec ^6(c+d x) \, dx\\ &=\frac {a^2 (5 A-2 B) \sec ^5(c+d x)}{35 d}+\frac {(A+B) \sec ^7(c+d x) (a+a \sin (c+d x))^2}{7 d}-\frac {\left (a^2 (5 A-2 B)\right ) \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{7 d}\\ &=\frac {a^2 (5 A-2 B) \sec ^5(c+d x)}{35 d}+\frac {(A+B) \sec ^7(c+d x) (a+a \sin (c+d x))^2}{7 d}+\frac {a^2 (5 A-2 B) \tan (c+d x)}{7 d}+\frac {2 a^2 (5 A-2 B) \tan ^3(c+d x)}{21 d}+\frac {a^2 (5 A-2 B) \tan ^5(c+d x)}{35 d}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 130, normalized size = 1.01 \[ \frac {a^2 \left (8 (2 B-5 A) \tan ^7(c+d x)+(30 A+9 B) \sec ^7(c+d x)-35 (5 A-2 B) \tan ^3(c+d x) \sec ^4(c+d x)+28 (5 A-2 B) \tan ^5(c+d x) \sec ^2(c+d x)+105 A \tan (c+d x) \sec ^6(c+d x)+21 B \tan ^2(c+d x) \sec ^5(c+d x)\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*((30*A + 9*B)*Sec[c + d*x]^7 + 105*A*Sec[c + d*x]^6*Tan[c + d*x] + 21*B*Sec[c + d*x]^5*Tan[c + d*x]^2 - 3
5*(5*A - 2*B)*Sec[c + d*x]^4*Tan[c + d*x]^3 + 28*(5*A - 2*B)*Sec[c + d*x]^2*Tan[c + d*x]^5 + 8*(-5*A + 2*B)*Ta
n[c + d*x]^7))/(105*d)

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fricas [A]  time = 0.68, size = 157, normalized size = 1.22 \[ -\frac {16 \, {\left (5 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 8 \, {\left (5 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 5 \, {\left (2 \, A - 5 \, B\right )} a^{2} - {\left (8 \, {\left (5 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 12 \, {\left (5 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 5 \, {\left (5 \, A - 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{5} + 2 \, d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/105*(16*(5*A - 2*B)*a^2*cos(d*x + c)^4 - 8*(5*A - 2*B)*a^2*cos(d*x + c)^2 - 5*(2*A - 5*B)*a^2 - (8*(5*A - 2
*B)*a^2*cos(d*x + c)^4 - 12*(5*A - 2*B)*a^2*cos(d*x + c)^2 - 5*(5*A - 2*B)*a^2)*sin(d*x + c))/(d*cos(d*x + c)^
5 + 2*d*cos(d*x + c)^3*sin(d*x + c) - 2*d*cos(d*x + c)^3)

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giac [B]  time = 0.23, size = 325, normalized size = 2.52 \[ -\frac {\frac {35 \, {\left (9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A a^{2} - 5 \, B a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}} + \frac {1365 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 210 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 5775 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12250 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 175 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 14350 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 910 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10185 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 756 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3955 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 427 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 760 \, A a^{2} - 31 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{7}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/840*(35*(9*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 6*B*a^2*tan(1/2*d*x + 1/2*c)^2 + 15*A*a^2*tan(1/2*d*x + 1/2*c) -
9*B*a^2*tan(1/2*d*x + 1/2*c) + 8*A*a^2 - 5*B*a^2)/(tan(1/2*d*x + 1/2*c) + 1)^3 + (1365*A*a^2*tan(1/2*d*x + 1/2
*c)^6 + 210*B*a^2*tan(1/2*d*x + 1/2*c)^6 - 5775*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 105*B*a^2*tan(1/2*d*x + 1/2*c)^
5 + 12250*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 175*B*a^2*tan(1/2*d*x + 1/2*c)^4 - 14350*A*a^2*tan(1/2*d*x + 1/2*c)^3
+ 910*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 10185*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 756*B*a^2*tan(1/2*d*x + 1/2*c)^2 -
3955*A*a^2*tan(1/2*d*x + 1/2*c) + 427*B*a^2*tan(1/2*d*x + 1/2*c) + 760*A*a^2 - 31*B*a^2)/(tan(1/2*d*x + 1/2*c)
- 1)^7)/d

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maple [B]  time = 0.71, size = 295, normalized size = 2.29 \[ \frac {a^{2} A \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+B \,a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{35 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{35 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{35}\right )+\frac {2 a^{2} A}{7 \cos \left (d x +c \right )^{7}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )-a^{2} A \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )+\frac {B \,a^{2}}{7 \cos \left (d x +c \right )^{7}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^2*A*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)+B*a^
2*(1/7*sin(d*x+c)^4/cos(d*x+c)^7+3/35*sin(d*x+c)^4/cos(d*x+c)^5+1/35*sin(d*x+c)^4/cos(d*x+c)^3-1/35*sin(d*x+c)
^4/cos(d*x+c)-1/35*(2+sin(d*x+c)^2)*cos(d*x+c))+2/7*a^2*A/cos(d*x+c)^7+2*B*a^2*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+
4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)-a^2*A*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)
^4-8/35*sec(d*x+c)^2)*tan(d*x+c)+1/7*B*a^2/cos(d*x+c)^7)

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maxima [A]  time = 0.34, size = 178, normalized size = 1.38 \[ \frac {{\left (15 \, \tan \left (d x + c\right )^{7} + 42 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3}\right )} A a^{2} + 3 \, {\left (5 \, \tan \left (d x + c\right )^{7} + 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} A a^{2} + 2 \, {\left (15 \, \tan \left (d x + c\right )^{7} + 42 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3}\right )} B a^{2} - \frac {3 \, {\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} B a^{2}}{\cos \left (d x + c\right )^{7}} + \frac {30 \, A a^{2}}{\cos \left (d x + c\right )^{7}} + \frac {15 \, B a^{2}}{\cos \left (d x + c\right )^{7}}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/105*((15*tan(d*x + c)^7 + 42*tan(d*x + c)^5 + 35*tan(d*x + c)^3)*A*a^2 + 3*(5*tan(d*x + c)^7 + 21*tan(d*x +
c)^5 + 35*tan(d*x + c)^3 + 35*tan(d*x + c))*A*a^2 + 2*(15*tan(d*x + c)^7 + 42*tan(d*x + c)^5 + 35*tan(d*x + c)
^3)*B*a^2 - 3*(7*cos(d*x + c)^2 - 5)*B*a^2/cos(d*x + c)^7 + 30*A*a^2/cos(d*x + c)^7 + 15*B*a^2/cos(d*x + c)^7)
/d

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mupad [B]  time = 12.23, size = 274, normalized size = 2.12 \[ -\frac {a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {25\,A\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{4}-\frac {105\,A\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{4}-\frac {95\,A\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{8}+\frac {15\,A\,\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{8}-21\,B\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {105\,B\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{8}-\frac {41\,B\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{8}+\frac {55\,B\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{16}+\frac {9\,B\,\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{16}-\frac {125\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {55\,A\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{2}-\frac {25\,A\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{2}+5\,A\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )+\frac {5\,A\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{2}+\frac {37\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {19\,B\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{4}-\frac {B\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{4}+\frac {13\,B\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{4}-B\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\right )}{1680\,d\,{\cos \left (\frac {c}{2}-\frac {\pi }{4}+\frac {d\,x}{2}\right )}^3\,{\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d\,x}{2}\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^8,x)

[Out]

-(a^2*cos(c/2 + (d*x)/2)*((25*A*cos((5*c)/2 + (5*d*x)/2))/4 - (105*A*cos((3*c)/2 + (3*d*x)/2))/4 - (95*A*cos((
7*c)/2 + (7*d*x)/2))/8 + (15*A*cos((9*c)/2 + (9*d*x)/2))/8 - 21*B*cos(c/2 + (d*x)/2) + (105*B*cos((3*c)/2 + (3
*d*x)/2))/8 - (41*B*cos((5*c)/2 + (5*d*x)/2))/8 + (55*B*cos((7*c)/2 + (7*d*x)/2))/16 + (9*B*cos((9*c)/2 + (9*d
*x)/2))/16 - (125*A*sin(c/2 + (d*x)/2))/2 + (55*A*sin((3*c)/2 + (3*d*x)/2))/2 - (25*A*sin((5*c)/2 + (5*d*x)/2)
)/2 + 5*A*sin((7*c)/2 + (7*d*x)/2) + (5*A*sin((9*c)/2 + (9*d*x)/2))/2 + (37*B*sin(c/2 + (d*x)/2))/4 + (19*B*si
n((3*c)/2 + (3*d*x)/2))/4 - (B*sin((5*c)/2 + (5*d*x)/2))/4 + (13*B*sin((7*c)/2 + (7*d*x)/2))/4 - B*sin((9*c)/2
+ (9*d*x)/2)))/(1680*d*cos(c/2 - pi/4 + (d*x)/2)^3*cos(c/2 + pi/4 + (d*x)/2)^7)

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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)**8*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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