3.982 \(\int \sec ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=104 \[ \frac{a^2 (3 A-2 B) \tan ^3(c+d x)}{15 d}+\frac{a^2 (3 A-2 B) \tan (c+d x)}{5 d}+\frac{a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac{(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.124831, antiderivative size = 104, 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} \[ \frac{a^2 (3 A-2 B) \tan ^3(c+d x)}{15 d}+\frac{a^2 (3 A-2 B) \tan (c+d x)}{5 d}+\frac{a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac{(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(3*A - 2*B)*Sec[c + d*x]^3)/(15*d) + ((A + B)*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2)/(5*d) + (a^2*(3*A -
2*B)*Tan[c + d*x])/(5*d) + (a^2*(3*A - 2*B)*Tan[c + d*x]^3)/(15*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 ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac{1}{5} (a (3 A-2 B)) \int \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=\frac{a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac{1}{5} \left (a^2 (3 A-2 B)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac{\left (a^2 (3 A-2 B)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{a^2 (3 A-2 B) \sec ^3(c+d x)}{15 d}+\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac{a^2 (3 A-2 B) \tan (c+d x)}{5 d}+\frac{a^2 (3 A-2 B) \tan ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.0200992, size = 178, normalized size = 1.71 \[ \frac{2 a^2 A \tan ^5(c+d x)}{5 d}+\frac{2 a^2 A \sec ^5(c+d x)}{5 d}-\frac{a^2 A \tan ^3(c+d x) \sec ^2(c+d x)}{d}+\frac{a^2 A \tan (c+d x) \sec ^4(c+d x)}{d}-\frac{4 a^2 B \tan ^5(c+d x)}{15 d}+\frac{a^2 B \sec ^5(c+d x)}{15 d}+\frac{a^2 B \tan ^2(c+d x) \sec ^3(c+d x)}{3 d}+\frac{2 a^2 B \tan ^3(c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(2*a^2*A*Sec[c + d*x]^5)/(5*d) + (a^2*B*Sec[c + d*x]^5)/(15*d) + (a^2*A*Sec[c + d*x]^4*Tan[c + d*x])/d + (a^2*
B*Sec[c + d*x]^3*Tan[c + d*x]^2)/(3*d) - (a^2*A*Sec[c + d*x]^2*Tan[c + d*x]^3)/d + (2*a^2*B*Sec[c + d*x]^2*Tan
[c + d*x]^3)/(3*d) + (2*a^2*A*Tan[c + d*x]^5)/(5*d) - (4*a^2*B*Tan[c + d*x]^5)/(15*d)

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Maple [B]  time = 0.106, size = 231, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ({a}^{2}A \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +B{a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{15}} \right ) +{\frac{2\,{a}^{2}A}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2\,B{a}^{2} \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) -{a}^{2}A \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) +{\frac{B{a}^{2}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.02389, size = 198, normalized size = 1.9 \begin{align*} \frac{{\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^{2} +{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} A a^{2} + 2 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} B a^{2} - \frac{{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} B a^{2}}{\cos \left (d x + c\right )^{5}} + \frac{6 \, A a^{2}}{\cos \left (d x + c\right )^{5}} + \frac{3 \, B a^{2}}{\cos \left (d x + c\right )^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/15*((3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + (3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*A
*a^2 + 2*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*B*a^2 - (5*cos(d*x + c)^2 - 3)*B*a^2/cos(d*x + c)^5 + 6*A*a^2/c
os(d*x + c)^5 + 3*B*a^2/cos(d*x + c)^5)/d

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Fricas [A]  time = 1.83644, size = 273, normalized size = 2.62 \begin{align*} -\frac{4 \,{\left (3 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 3 \,{\left (2 \, A - 3 \, B\right )} a^{2} -{\left (2 \,{\left (3 \, A - 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 3 \,{\left (3 \, A - 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.24262, size = 259, normalized size = 2.49 \begin{align*} -\frac{\frac{15 \,{\left (A a^{2} - B a^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1} + \frac{105 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 15 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 270 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 360 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 40 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 210 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 50 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 63 \, A a^{2} - 7 \, B a^{2}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(15*(A*a^2 - B*a^2)/(tan(1/2*d*x + 1/2*c) + 1) + (105*A*a^2*tan(1/2*d*x + 1/2*c)^4 + 15*B*a^2*tan(1/2*d*
x + 1/2*c)^4 - 270*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 360*A*a^2*tan(1/2*d*x + 1/
2*c)^2 - 40*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 210*A*a^2*tan(1/2*d*x + 1/2*c) + 50*B*a^2*tan(1/2*d*x + 1/2*c) + 63
*A*a^2 - 7*B*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^5)/d