3.994 \(\int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=162 \[ \frac{a^7 (A+B)}{16 d (a-a \sin (c+d x))^4}+\frac{a^5 (3 A-B)}{32 d (a-a \sin (c+d x))^2}+\frac{a^4 (2 A-B)}{16 d (a-a \sin (c+d x))}-\frac{a^4 (A-B)}{32 d (a \sin (c+d x)+a)}+\frac{a^3 (5 A-3 B) \tanh ^{-1}(\sin (c+d x))}{32 d}+\frac{a^6 A}{12 d (a-a \sin (c+d x))^3} \]

[Out]

(a^3*(5*A - 3*B)*ArcTanh[Sin[c + d*x]])/(32*d) + (a^7*(A + B))/(16*d*(a - a*Sin[c + d*x])^4) + (a^6*A)/(12*d*(
a - a*Sin[c + d*x])^3) + (a^5*(3*A - B))/(32*d*(a - a*Sin[c + d*x])^2) + (a^4*(2*A - B))/(16*d*(a - a*Sin[c +
d*x])) - (a^4*(A - B))/(32*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.186258, antiderivative size = 162, 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} \[ \frac{a^7 (A+B)}{16 d (a-a \sin (c+d x))^4}+\frac{a^5 (3 A-B)}{32 d (a-a \sin (c+d x))^2}+\frac{a^4 (2 A-B)}{16 d (a-a \sin (c+d x))}-\frac{a^4 (A-B)}{32 d (a \sin (c+d x)+a)}+\frac{a^3 (5 A-3 B) \tanh ^{-1}(\sin (c+d x))}{32 d}+\frac{a^6 A}{12 d (a-a \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(5*A - 3*B)*ArcTanh[Sin[c + d*x]])/(32*d) + (a^7*(A + B))/(16*d*(a - a*Sin[c + d*x])^4) + (a^6*A)/(12*d*(
a - a*Sin[c + d*x])^3) + (a^5*(3*A - B))/(32*d*(a - a*Sin[c + d*x])^2) + (a^4*(2*A - B))/(16*d*(a - a*Sin[c +
d*x])) - (a^4*(A - B))/(32*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 ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac{a^9 \operatorname{Subst}\left (\int \frac{A+\frac{B x}{a}}{(a-x)^5 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^9 \operatorname{Subst}\left (\int \left (\frac{A+B}{4 a^2 (a-x)^5}+\frac{A}{4 a^3 (a-x)^4}+\frac{3 A-B}{16 a^4 (a-x)^3}+\frac{2 A-B}{16 a^5 (a-x)^2}+\frac{A-B}{32 a^5 (a+x)^2}+\frac{5 A-3 B}{32 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^7 (A+B)}{16 d (a-a \sin (c+d x))^4}+\frac{a^6 A}{12 d (a-a \sin (c+d x))^3}+\frac{a^5 (3 A-B)}{32 d (a-a \sin (c+d x))^2}+\frac{a^4 (2 A-B)}{16 d (a-a \sin (c+d x))}-\frac{a^4 (A-B)}{32 d (a+a \sin (c+d x))}+\frac{\left (a^4 (5 A-3 B)\right ) \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{32 d}\\ &=\frac{a^3 (5 A-3 B) \tanh ^{-1}(\sin (c+d x))}{32 d}+\frac{a^7 (A+B)}{16 d (a-a \sin (c+d x))^4}+\frac{a^6 A}{12 d (a-a \sin (c+d x))^3}+\frac{a^5 (3 A-B)}{32 d (a-a \sin (c+d x))^2}+\frac{a^4 (2 A-B)}{16 d (a-a \sin (c+d x))}-\frac{a^4 (A-B)}{32 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.631038, size = 151, normalized size = 0.93 \[ \frac{a^9 \left (\frac{2 A-B}{16 a^5 (a-a \sin (c+d x))}-\frac{A-B}{32 a^5 (a \sin (c+d x)+a)}+\frac{3 A-B}{32 a^4 (a-a \sin (c+d x))^2}+\frac{A+B}{16 a^2 (a-a \sin (c+d x))^4}+\frac{(5 A-3 B) \tanh ^{-1}(\sin (c+d x))}{32 a^6}+\frac{A}{12 a^3 (a-a \sin (c+d x))^3}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^9*(((5*A - 3*B)*ArcTanh[Sin[c + d*x]])/(32*a^6) + (A + B)/(16*a^2*(a - a*Sin[c + d*x])^4) + A/(12*a^3*(a -
a*Sin[c + d*x])^3) + (3*A - B)/(32*a^4*(a - a*Sin[c + d*x])^2) + (2*A - B)/(16*a^5*(a - a*Sin[c + d*x])) - (A
- B)/(32*a^5*(a + a*Sin[c + d*x]))))/d

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Maple [B]  time = 0.223, size = 669, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

3/8/d*a^3*A/cos(d*x+c)^8+1/8/d*B*a^3/cos(d*x+c)^8+1/16/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^6+5/16/d*a^3*A*sin(d*x+
c)^3/cos(d*x+c)^6+1/4/d*B*a^3*sin(d*x+c)^4/cos(d*x+c)^6+1/12/d*a^3*A*sin(d*x+c)^4/cos(d*x+c)^6+15/128/d*a^3*A*
sin(d*x+c)^3/cos(d*x+c)^2+15/128/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)^2+35/128/d*a^3*A*sec(d*x+c)*tan(d*x+c)+3/8/d*
B*a^3*sin(d*x+c)^4/cos(d*x+c)^8+3/8/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)^8+1/8/d*a^3*A*tan(d*x+c)*sec(d*x+c)^7+1/8/
d*a^3*A*sin(d*x+c)^4/cos(d*x+c)^8+1/8/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^8+3/8/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^8+
5/16/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)^6+7/48/d*a^3*A*tan(d*x+c)*sec(d*x+c)^5-1/128/d*B*a^3*sin(d*x+c)^3-3/32/d*
B*a^3*ln(sec(d*x+c)+tan(d*x+c))+15/128/d*a^3*A*sin(d*x+c)+5/32/d*a^3*A*ln(sec(d*x+c)+tan(d*x+c))+1/24/d*a^3*A*
sin(d*x+c)^4/cos(d*x+c)^4-1/128/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^2+1/64/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^4+15/64
/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*B*a^3*sin(d*x+c)^4/cos(d*x+c)^4+15/64/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)
^4+35/192/d*a^3*A*tan(d*x+c)*sec(d*x+c)^3+3/32*a^3*B*sin(d*x+c)/d

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Maxima [A]  time = 1.06497, size = 250, normalized size = 1.54 \begin{align*} \frac{3 \,{\left (5 \, A - 3 \, B\right )} a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (5 \, A - 3 \, B\right )} a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{4} - 9 \,{\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} + 7 \,{\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{2} + 3 \,{\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right ) - 32 \, A a^{3}\right )}}{\sin \left (d x + c\right )^{5} - 3 \, \sin \left (d x + c\right )^{4} + 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right )^{2} - 3 \, \sin \left (d x + c\right ) + 1}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/192*(3*(5*A - 3*B)*a^3*log(sin(d*x + c) + 1) - 3*(5*A - 3*B)*a^3*log(sin(d*x + c) - 1) - 2*(3*(5*A - 3*B)*a^
3*sin(d*x + c)^4 - 9*(5*A - 3*B)*a^3*sin(d*x + c)^3 + 7*(5*A - 3*B)*a^3*sin(d*x + c)^2 + 3*(5*A - 3*B)*a^3*sin
(d*x + c) - 32*A*a^3)/(sin(d*x + c)^5 - 3*sin(d*x + c)^4 + 2*sin(d*x + c)^3 + 2*sin(d*x + c)^2 - 3*sin(d*x + c
) + 1))/d

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Fricas [B]  time = 1.86546, size = 846, normalized size = 5.22 \begin{align*} \frac{6 \,{\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 26 \,{\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 12 \,{\left (3 \, A - 5 \, B\right )} a^{3} + 3 \,{\left (3 \,{\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \,{\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} -{\left ({\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \,{\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \,{\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \,{\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} -{\left ({\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \,{\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \,{\left (3 \,{\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - 2 \,{\left (5 \, A - 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{192 \,{\left (3 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} -{\left (d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/192*(6*(5*A - 3*B)*a^3*cos(d*x + c)^4 - 26*(5*A - 3*B)*a^3*cos(d*x + c)^2 + 12*(3*A - 5*B)*a^3 + 3*(3*(5*A -
3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*a^3*cos(d*x + c)^2 - ((5*A - 3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*
a^3*cos(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) - 3*(3*(5*A - 3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*
a^3*cos(d*x + c)^2 - ((5*A - 3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*a^3*cos(d*x + c)^2)*sin(d*x + c))*log(-si
n(d*x + c) + 1) + 6*(3*(5*A - 3*B)*a^3*cos(d*x + c)^2 - 2*(5*A - 3*B)*a^3)*sin(d*x + c))/(3*d*cos(d*x + c)^4 -
4*d*cos(d*x + c)^2 - (d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2)*sin(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)**9*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.50438, size = 320, normalized size = 1.98 \begin{align*} \frac{12 \,{\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 12 \,{\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{12 \,{\left (5 \, A a^{3} \sin \left (d x + c\right ) - 3 \, B a^{3} \sin \left (d x + c\right ) + 7 \, A a^{3} - 5 \, B a^{3}\right )}}{\sin \left (d x + c\right ) + 1} + \frac{125 \, A a^{3} \sin \left (d x + c\right )^{4} - 75 \, B a^{3} \sin \left (d x + c\right )^{4} - 596 \, A a^{3} \sin \left (d x + c\right )^{3} + 348 \, B a^{3} \sin \left (d x + c\right )^{3} + 1110 \, A a^{3} \sin \left (d x + c\right )^{2} - 618 \, B a^{3} \sin \left (d x + c\right )^{2} - 996 \, A a^{3} \sin \left (d x + c\right ) + 492 \, B a^{3} \sin \left (d x + c\right ) + 405 \, A a^{3} - 99 \, B a^{3}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{4}}}{768 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/768*(12*(5*A*a^3 - 3*B*a^3)*log(abs(sin(d*x + c) + 1)) - 12*(5*A*a^3 - 3*B*a^3)*log(abs(sin(d*x + c) - 1)) -
12*(5*A*a^3*sin(d*x + c) - 3*B*a^3*sin(d*x + c) + 7*A*a^3 - 5*B*a^3)/(sin(d*x + c) + 1) + (125*A*a^3*sin(d*x
+ c)^4 - 75*B*a^3*sin(d*x + c)^4 - 596*A*a^3*sin(d*x + c)^3 + 348*B*a^3*sin(d*x + c)^3 + 1110*A*a^3*sin(d*x +
c)^2 - 618*B*a^3*sin(d*x + c)^2 - 996*A*a^3*sin(d*x + c) + 492*B*a^3*sin(d*x + c) + 405*A*a^3 - 99*B*a^3)/(sin
(d*x + c) - 1)^4)/d