3.976 \(\int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=132 \[ \frac{a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac{a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac{a^3 (A-B)}{16 d (a \sin (c+d x)+a)}+\frac{a^2 (2 A-B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^4 A}{8 d (a-a \sin (c+d x))^2} \]

[Out]

(a^2*(2*A - B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^5*(A + B))/(12*d*(a - a*Sin[c + d*x])^3) + (a^4*A)/(8*d*(a -
a*Sin[c + d*x])^2) + (a^3*(3*A - B))/(16*d*(a - a*Sin[c + d*x])) - (a^3*(A - B))/(16*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.156748, antiderivative size = 132, 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 = {2836, 77, 206} \[ \frac{a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac{a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac{a^3 (A-B)}{16 d (a \sin (c+d x)+a)}+\frac{a^2 (2 A-B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^4 A}{8 d (a-a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(2*A - B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^5*(A + B))/(12*d*(a - a*Sin[c + d*x])^3) + (a^4*A)/(8*d*(a -
a*Sin[c + d*x])^2) + (a^3*(3*A - B))/(16*d*(a - a*Sin[c + d*x])) - (a^3*(A - B))/(16*d*(a + a*Sin[c + d*x]))

Rule 2836

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac{a^7 \operatorname{Subst}\left (\int \frac{A+\frac{B x}{a}}{(a-x)^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^7 \operatorname{Subst}\left (\int \left (\frac{A+B}{4 a^2 (a-x)^4}+\frac{A}{4 a^3 (a-x)^3}+\frac{3 A-B}{16 a^4 (a-x)^2}+\frac{A-B}{16 a^4 (a+x)^2}+\frac{2 A-B}{8 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac{a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac{a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac{a^3 (A-B)}{16 d (a+a \sin (c+d x))}+\frac{\left (a^3 (2 A-B)\right ) \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=\frac{a^2 (2 A-B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac{a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac{a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac{a^3 (A-B)}{16 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.656443, size = 90, normalized size = 0.68 \[ \frac{a^2 \left (\frac{3 B-9 A}{\sin (c+d x)-1}-\frac{3 (A-B)}{\sin (c+d x)+1}-\frac{4 (A+B)}{(\sin (c+d x)-1)^3}+6 (2 A-B) \tanh ^{-1}(\sin (c+d x))+\frac{6 A}{(\sin (c+d x)-1)^2}\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(6*(2*A - B)*ArcTanh[Sin[c + d*x]] - (4*(A + B))/(-1 + Sin[c + d*x])^3 + (6*A)/(-1 + Sin[c + d*x])^2 + (-
9*A + 3*B)/(-1 + Sin[c + d*x]) - (3*(A - B))/(1 + Sin[c + d*x])))/(48*d)

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Maple [B]  time = 0.115, size = 379, normalized size = 2.9 \begin{align*}{\frac{{a}^{2}A \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{a}^{2}A \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2}A \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{16\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{16\,d}}+{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{B{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{B{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{12\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2}A}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{B{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{B{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{B{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{B{a}^{2}\sin \left ( dx+c \right ) }{8\,d}}-{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{a}^{2}A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,{a}^{2}A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{B{a}^{2}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/6/d*a^2*A*sin(d*x+c)^3/cos(d*x+c)^6+1/8/d*a^2*A*sin(d*x+c)^3/cos(d*x+c)^4+1/16/d*a^2*A*sin(d*x+c)^3/cos(d*x+
c)^2+1/16/d*a^2*A*sin(d*x+c)+1/4/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+1/6/d*B*a^2*sin(d*x+c)^4/cos(d*x+c)^6+1/12/
d*B*a^2*sin(d*x+c)^4/cos(d*x+c)^4+1/3/d*a^2*A/cos(d*x+c)^6+1/3/d*B*a^2*sin(d*x+c)^3/cos(d*x+c)^6+1/4/d*B*a^2*s
in(d*x+c)^3/cos(d*x+c)^4+1/8/d*B*a^2*sin(d*x+c)^3/cos(d*x+c)^2+1/8/d*B*a^2*sin(d*x+c)-1/8/d*B*a^2*ln(sec(d*x+c
)+tan(d*x+c))+1/6/d*a^2*A*tan(d*x+c)*sec(d*x+c)^5+5/24/d*a^2*A*tan(d*x+c)*sec(d*x+c)^3+5/16/d*a^2*A*sec(d*x+c)
*tan(d*x+c)+1/6/d*B*a^2/cos(d*x+c)^6

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Maxima [A]  time = 1.02922, size = 200, normalized size = 1.52 \begin{align*} \frac{3 \,{\left (2 \, A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{3} - 6 \,{\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{2} +{\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right ) + 2 \,{\left (4 \, A + B\right )} a^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(3*(2*A - B)*a^2*log(sin(d*x + c) + 1) - 3*(2*A - B)*a^2*log(sin(d*x + c) - 1) - 2*(3*(2*A - B)*a^2*sin(d
*x + c)^3 - 6*(2*A - B)*a^2*sin(d*x + c)^2 + (2*A - B)*a^2*sin(d*x + c) + 2*(4*A + B)*a^2)/(sin(d*x + c)^4 - 2
*sin(d*x + c)^3 + 2*sin(d*x + c) - 1))/d

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Fricas [B]  time = 1.77586, size = 640, normalized size = 4.85 \begin{align*} -\frac{12 \,{\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} - 8 \,{\left (A - 2 \, B\right )} a^{2} - 3 \,{\left ({\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \,{\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \,{\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left ({\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \,{\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \,{\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (3 \,{\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} - 4 \,{\left (2 \, A - B\right )} a^{2}\right )} \sin \left (d x + c\right )}{48 \,{\left (d \cos \left (d x + c\right )^{4} + 2 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(12*(2*A - B)*a^2*cos(d*x + c)^2 - 8*(A - 2*B)*a^2 - 3*((2*A - B)*a^2*cos(d*x + c)^4 + 2*(2*A - B)*a^2*c
os(d*x + c)^2*sin(d*x + c) - 2*(2*A - B)*a^2*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + 3*((2*A - B)*a^2*cos(d*x
+ c)^4 + 2*(2*A - B)*a^2*cos(d*x + c)^2*sin(d*x + c) - 2*(2*A - B)*a^2*cos(d*x + c)^2)*log(-sin(d*x + c) + 1)
- 2*(3*(2*A - B)*a^2*cos(d*x + c)^2 - 4*(2*A - B)*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4 + 2*d*cos(d*x + c)^2*si
n(d*x + c) - 2*d*cos(d*x + c)^2)

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

[Out]

Timed out

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Giac [A]  time = 1.36955, size = 282, normalized size = 2.14 \begin{align*} \frac{6 \,{\left (2 \, A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \,{\left (2 \, A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{6 \,{\left (2 \, A a^{2} \sin \left (d x + c\right ) - B a^{2} \sin \left (d x + c\right ) + 3 \, A a^{2} - 2 \, B a^{2}\right )}}{\sin \left (d x + c\right ) + 1} + \frac{22 \, A a^{2} \sin \left (d x + c\right )^{3} - 11 \, B a^{2} \sin \left (d x + c\right )^{3} - 84 \, A a^{2} \sin \left (d x + c\right )^{2} + 39 \, B a^{2} \sin \left (d x + c\right )^{2} + 114 \, A a^{2} \sin \left (d x + c\right ) - 45 \, B a^{2} \sin \left (d x + c\right ) - 60 \, A a^{2} + 9 \, B a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/96*(6*(2*A*a^2 - B*a^2)*log(abs(sin(d*x + c) + 1)) - 6*(2*A*a^2 - B*a^2)*log(abs(sin(d*x + c) - 1)) - 6*(2*A
*a^2*sin(d*x + c) - B*a^2*sin(d*x + c) + 3*A*a^2 - 2*B*a^2)/(sin(d*x + c) + 1) + (22*A*a^2*sin(d*x + c)^3 - 11
*B*a^2*sin(d*x + c)^3 - 84*A*a^2*sin(d*x + c)^2 + 39*B*a^2*sin(d*x + c)^2 + 114*A*a^2*sin(d*x + c) - 45*B*a^2*
sin(d*x + c) - 60*A*a^2 + 9*B*a^2)/(sin(d*x + c) - 1)^3)/d