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]

(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)

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Rubi [A]  time = 0.133749, antiderivative size = 129, normalized size of antiderivative = 1., 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 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 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 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*}

Mathematica [A]  time = 0.345351, 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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Maple [B]  time = 0.119, size = 295, normalized size = 2.3 \begin{align*}{\frac{1}{d} \left ({a}^{2}A \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +B{a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{35}} \right ) +{\frac{2\,{a}^{2}A}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+2\,B{a}^{2} \left ( 1/7\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) -{a}^{2}A \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{35}} \right ) \tan \left ( dx+c \right ) +{\frac{B{a}^{2}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}} \right ) } \end{align*}

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 = 1.03039, size = 240, normalized size = 1.86 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.61766, size = 377, normalized size = 2.92 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [B]  time = 1.29088, size = 439, normalized size = 3.4 \begin{align*} -\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} \end{align*}

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