3.1558 \(\int \frac{\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=372 \[ -\frac{b \left (-12 a^2 A b^2+3 a^4 A+2 a^3 b B+22 a b^3 B-15 A b^4\right )}{8 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{b^4 \left (-5 a^2 B+6 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}-\frac{\left (3 a^2 A+2 a b (6 A+B)+b^2 (15 A+8 B)\right ) \log (1-\sin (c+d x))}{16 d (a+b)^4}+\frac{\left (3 a^2 A-2 a b (6 A-B)+b^2 (15 A-8 B)\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^4}-\frac{\sec ^4(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{\sec ^2(c+d x) \left (\left (3 a^3 A+2 a^2 b B-9 a A b^2+4 b^3 B\right ) \sin (c+d x)+b \left (a^2 A-6 a b B+5 A b^2\right )\right )}{8 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))} \]

[Out]

-((3*a^2*A + 2*a*b*(6*A + B) + b^2*(15*A + 8*B))*Log[1 - Sin[c + d*x]])/(16*(a + b)^4*d) + ((3*a^2*A + b^2*(15
*A - 8*B) - 2*a*b*(6*A - B))*Log[1 + Sin[c + d*x]])/(16*(a - b)^4*d) - (b^4*(6*a*A*b - 5*a^2*B - b^2*B)*Log[a
+ b*Sin[c + d*x]])/((a^2 - b^2)^4*d) - (b*(3*a^4*A - 12*a^2*A*b^2 - 15*A*b^4 + 2*a^3*b*B + 22*a*b^3*B))/(8*(a^
2 - b^2)^3*d*(a + b*Sin[c + d*x])) - (Sec[c + d*x]^4*(A*b - a*B - (a*A - b*B)*Sin[c + d*x]))/(4*(a^2 - b^2)*d*
(a + b*Sin[c + d*x])) + (Sec[c + d*x]^2*(b*(a^2*A + 5*A*b^2 - 6*a*b*B) + (3*a^3*A - 9*a*A*b^2 + 2*a^2*b*B + 4*
b^3*B)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.564163, antiderivative size = 372, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {2837, 823, 801} \[ -\frac{b \left (-12 a^2 A b^2+3 a^4 A+2 a^3 b B+22 a b^3 B-15 A b^4\right )}{8 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac{b^4 \left (-5 a^2 B+6 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}-\frac{\left (3 a^2 A+2 a b (6 A+B)+b^2 (15 A+8 B)\right ) \log (1-\sin (c+d x))}{16 d (a+b)^4}+\frac{\left (3 a^2 A-2 a b (6 A-B)+b^2 (15 A-8 B)\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^4}-\frac{\sec ^4(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{\sec ^2(c+d x) \left (\left (3 a^3 A+2 a^2 b B-9 a A b^2+4 b^3 B\right ) \sin (c+d x)+b \left (a^2 A-6 a b B+5 A b^2\right )\right )}{8 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x])^2,x]

[Out]

-((3*a^2*A + 2*a*b*(6*A + B) + b^2*(15*A + 8*B))*Log[1 - Sin[c + d*x]])/(16*(a + b)^4*d) + ((3*a^2*A + b^2*(15
*A - 8*B) - 2*a*b*(6*A - B))*Log[1 + Sin[c + d*x]])/(16*(a - b)^4*d) - (b^4*(6*a*A*b - 5*a^2*B - b^2*B)*Log[a
+ b*Sin[c + d*x]])/((a^2 - b^2)^4*d) - (b*(3*a^4*A - 12*a^2*A*b^2 - 15*A*b^4 + 2*a^3*b*B + 22*a*b^3*B))/(8*(a^
2 - b^2)^3*d*(a + b*Sin[c + d*x])) - (Sec[c + d*x]^4*(A*b - a*B - (a*A - b*B)*Sin[c + d*x]))/(4*(a^2 - b^2)*d*
(a + b*Sin[c + d*x])) + (Sec[c + d*x]^2*(b*(a^2*A + 5*A*b^2 - 6*a*b*B) + (3*a^3*A - 9*a*A*b^2 + 2*a^2*b*B + 4*
b^3*B)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]))

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{A+\frac{B x}{b}}{(a+x)^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\sec ^4(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{b^3 \operatorname{Subst}\left (\int \frac{-3 a^2 A+5 A b^2-2 a b B-4 (a A-b B) x}{(a+x)^2 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^4(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\sec ^2(c+d x) \left (b \left (a^2 A+5 A b^2-6 a b B\right )+\left (3 a^3 A-9 a A b^2+2 a^2 b B+4 b^3 B\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{b \operatorname{Subst}\left (\int \frac{3 a^4 A-6 a^2 A b^2+15 A b^4+2 a^3 b B-14 a b^3 B+2 \left (3 a^3 A-9 a A b^2+2 a^2 b B+4 b^3 B\right ) x}{(a+x)^2 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac{\sec ^4(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\sec ^2(c+d x) \left (b \left (a^2 A+5 A b^2-6 a b B\right )+\left (3 a^3 A-9 a A b^2+2 a^2 b B+4 b^3 B\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{b \operatorname{Subst}\left (\int \left (\frac{(a-b)^2 \left (3 a^2 A+2 a b (6 A+B)+b^2 (15 A+8 B)\right )}{2 b (a+b)^2 (b-x)}+\frac{3 a^4 A-12 a^2 A b^2-15 A b^4+2 a^3 b B+22 a b^3 B}{\left (a^2-b^2\right ) (a+x)^2}+\frac{8 b^3 \left (-6 a A b+5 a^2 B+b^2 B\right )}{\left (-a^2+b^2\right )^2 (a+x)}+\frac{(a+b)^2 \left (3 a^2 A+b^2 (15 A-8 B)-2 a b (6 A-B)\right )}{2 (a-b)^2 b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac{\left (3 a^2 A+2 a b (6 A+B)+b^2 (15 A+8 B)\right ) \log (1-\sin (c+d x))}{16 (a+b)^4 d}+\frac{\left (3 a^2 A+b^2 (15 A-8 B)-2 a b (6 A-B)\right ) \log (1+\sin (c+d x))}{16 (a-b)^4 d}-\frac{b^4 \left (6 a A b-5 a^2 B-b^2 B\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}-\frac{b \left (3 a^4 A-12 a^2 A b^2-15 A b^4+2 a^3 b B+22 a b^3 B\right )}{8 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac{\sec ^4(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\sec ^2(c+d x) \left (b \left (a^2 A+5 A b^2-6 a b B\right )+\left (3 a^3 A-9 a A b^2+2 a^2 b B+4 b^3 B\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 4.12523, size = 370, normalized size = 0.99 \[ \frac{-\frac{\left (3 a^3 A+2 a^2 b B-9 a A b^2+4 b^3 B\right ) ((a-b) \log (1-\sin (c+d x))-(a+b) \log (\sin (c+d x)+1)+2 b \log (a+b \sin (c+d x)))}{(a-b) (a+b)}+b \left (12 a^2 A b^2-3 a^4 A-2 a^3 b B-22 a b^3 B+15 A b^4\right ) \left (\frac{1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac{\log (\sin (c+d x)+1)}{2 b (a-b)^2}-\frac{2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}\right )+\frac{2 \left (b^2-a^2\right ) \sec ^4(c+d x) ((b B-a A) \sin (c+d x)-a B+A b)}{a+b \sin (c+d x)}+\frac{\sec ^2(c+d x) \left (\left (3 a^3 A+2 a^2 b B-9 a A b^2+4 b^3 B\right ) \sin (c+d x)+b \left (a^2 A-6 a b B+5 A b^2\right )\right )}{a+b \sin (c+d x)}}{8 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x])^2,x]

[Out]

(-(((3*a^3*A - 9*a*A*b^2 + 2*a^2*b*B + 4*b^3*B)*((a - b)*Log[1 - Sin[c + d*x]] - (a + b)*Log[1 + Sin[c + d*x]]
+ 2*b*Log[a + b*Sin[c + d*x]]))/((a - b)*(a + b))) + (2*(-a^2 + b^2)*Sec[c + d*x]^4*(A*b - a*B + (-(a*A) + b*
B)*Sin[c + d*x]))/(a + b*Sin[c + d*x]) + (Sec[c + d*x]^2*(b*(a^2*A + 5*A*b^2 - 6*a*b*B) + (3*a^3*A - 9*a*A*b^2
+ 2*a^2*b*B + 4*b^3*B)*Sin[c + d*x]))/(a + b*Sin[c + d*x]) + b*(-3*a^4*A + 12*a^2*A*b^2 + 15*A*b^4 - 2*a^3*b*
B - 22*a*b^3*B)*(-Log[1 - Sin[c + d*x]]/(2*b*(a + b)^2) + Log[1 + Sin[c + d*x]]/(2*(a - b)^2*b) - (2*a*Log[a +
b*Sin[c + d*x]])/((a - b)^2*(a + b)^2) + 1/((a^2 - b^2)*(a + b*Sin[c + d*x]))))/(8*(a^2 - b^2)^2*d)

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Maple [A]  time = 0.175, size = 675, normalized size = 1.8 \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)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x)

[Out]

-6/d*b^5/(a+b)^4/(a-b)^4*ln(a+b*sin(d*x+c))*A*a+5/d*b^4/(a+b)^4/(a-b)^4*ln(a+b*sin(d*x+c))*B*a^2-1/d*b^4/(a+b)
^3/(a-b)^3/(a+b*sin(d*x+c))*a*B+1/d*b^5/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))*A-3/4/d/(a+b)^4*ln(sin(d*x+c)-1)*A*a*
b-1/8/d/(a+b)^4*ln(sin(d*x+c)-1)*B*a*b-3/4/d/(a-b)^4*ln(1+sin(d*x+c))*A*a*b+1/8/d/(a-b)^4*ln(1+sin(d*x+c))*B*a
*b+1/16/d/(a-b)^2/(1+sin(d*x+c))^2*B-1/16/d/(a-b)^2/(1+sin(d*x+c))^2*A+1/16/d/(a+b)^2/(sin(d*x+c)-1)^2*A+1/16/
d/(a+b)^2/(sin(d*x+c)-1)^2*B-1/16/d/(a+b)^3/(sin(d*x+c)-1)*a*B-5/16/d/(a+b)^3/(sin(d*x+c)-1)*B*b-3/16/d/(a+b)^
4*ln(sin(d*x+c)-1)*a^2*A-15/16/d/(a+b)^4*ln(sin(d*x+c)-1)*A*b^2-1/2/d/(a+b)^4*ln(sin(d*x+c)-1)*B*b^2-3/16/d/(a
-b)^3/(1+sin(d*x+c))*a*A+7/16/d/(a-b)^3/(1+sin(d*x+c))*A*b+1/16/d/(a-b)^3/(1+sin(d*x+c))*a*B-5/16/d/(a-b)^3/(1
+sin(d*x+c))*B*b+3/16/d/(a-b)^4*ln(1+sin(d*x+c))*a^2*A+15/16/d/(a-b)^4*ln(1+sin(d*x+c))*A*b^2-1/2/d/(a-b)^4*ln
(1+sin(d*x+c))*B*b^2-7/16/d/(a+b)^3/(sin(d*x+c)-1)*A*b+1/d*b^6/(a+b)^4/(a-b)^4*ln(a+b*sin(d*x+c))*B-3/16/d/(a+
b)^3/(sin(d*x+c)-1)*a*A

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Maxima [A]  time = 1.07899, size = 890, normalized size = 2.39 \begin{align*} \frac{\frac{16 \,{\left (5 \, B a^{2} b^{4} - 6 \, A a b^{5} + B b^{6}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} + \frac{{\left (3 \, A a^{2} - 2 \,{\left (6 \, A - B\right )} a b +{\left (15 \, A - 8 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac{{\left (3 \, A a^{2} + 2 \,{\left (6 \, A + B\right )} a b +{\left (15 \, A + 8 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac{2 \,{\left (2 \, B a^{5} - 4 \, A a^{4} b - 12 \, B a^{3} b^{2} + 20 \, A a^{2} b^{3} - 14 \, B a b^{4} + 8 \, A b^{5} -{\left (3 \, A a^{4} b + 2 \, B a^{3} b^{2} - 12 \, A a^{2} b^{3} + 22 \, B a b^{4} - 15 \, A b^{5}\right )} \sin \left (d x + c\right )^{4} -{\left (3 \, A a^{5} + 2 \, B a^{4} b - 12 \, A a^{3} b^{2} + 2 \, B a^{2} b^{3} + 9 \, A a b^{4} - 4 \, B b^{5}\right )} \sin \left (d x + c\right )^{3} +{\left (5 \, A a^{4} b + 10 \, B a^{3} b^{2} - 28 \, A a^{2} b^{3} + 38 \, B a b^{4} - 25 \, A b^{5}\right )} \sin \left (d x + c\right )^{2} +{\left (5 \, A a^{5} - 16 \, A a^{3} b^{2} + 6 \, B a^{2} b^{3} + 11 \, A a b^{4} - 6 \, B b^{5}\right )} \sin \left (d x + c\right )\right )}}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} +{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{5} +{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{4} - 2 \,{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{3} - 2 \,{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{2} +{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/16*(16*(5*B*a^2*b^4 - 6*A*a*b^5 + B*b^6)*log(b*sin(d*x + c) + a)/(a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 +
b^8) + (3*A*a^2 - 2*(6*A - B)*a*b + (15*A - 8*B)*b^2)*log(sin(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b
^3 + b^4) - (3*A*a^2 + 2*(6*A + B)*a*b + (15*A + 8*B)*b^2)*log(sin(d*x + c) - 1)/(a^4 + 4*a^3*b + 6*a^2*b^2 +
4*a*b^3 + b^4) + 2*(2*B*a^5 - 4*A*a^4*b - 12*B*a^3*b^2 + 20*A*a^2*b^3 - 14*B*a*b^4 + 8*A*b^5 - (3*A*a^4*b + 2*
B*a^3*b^2 - 12*A*a^2*b^3 + 22*B*a*b^4 - 15*A*b^5)*sin(d*x + c)^4 - (3*A*a^5 + 2*B*a^4*b - 12*A*a^3*b^2 + 2*B*a
^2*b^3 + 9*A*a*b^4 - 4*B*b^5)*sin(d*x + c)^3 + (5*A*a^4*b + 10*B*a^3*b^2 - 28*A*a^2*b^3 + 38*B*a*b^4 - 25*A*b^
5)*sin(d*x + c)^2 + (5*A*a^5 - 16*A*a^3*b^2 + 6*B*a^2*b^3 + 11*A*a*b^4 - 6*B*b^5)*sin(d*x + c))/(a^7 - 3*a^5*b
^2 + 3*a^3*b^4 - a*b^6 + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*sin(d*x + c)^5 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 -
a*b^6)*sin(d*x + c)^4 - 2*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*sin(d*x + c)^3 - 2*(a^7 - 3*a^5*b^2 + 3*a^3*b
^4 - a*b^6)*sin(d*x + c)^2 + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*sin(d*x + c)))/d

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Fricas [B]  time = 21.0228, size = 1997, normalized size = 5.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(4*B*a^7 - 4*A*a^6*b - 12*B*a^5*b^2 + 12*A*a^4*b^3 + 12*B*a^3*b^4 - 12*A*a^2*b^5 - 4*B*a*b^6 + 4*A*b^7 -
2*(3*A*a^6*b + 2*B*a^5*b^2 - 15*A*a^4*b^3 + 20*B*a^3*b^4 - 3*A*a^2*b^5 - 22*B*a*b^6 + 15*A*b^7)*cos(d*x + c)^4
+ 2*(A*a^6*b - 6*B*a^5*b^2 + 3*A*a^4*b^3 + 12*B*a^3*b^4 - 9*A*a^2*b^5 - 6*B*a*b^6 + 5*A*b^7)*cos(d*x + c)^2 +
16*((5*B*a^2*b^5 - 6*A*a*b^6 + B*b^7)*cos(d*x + c)^4*sin(d*x + c) + (5*B*a^3*b^4 - 6*A*a^2*b^5 + B*a*b^6)*cos
(d*x + c)^4)*log(b*sin(d*x + c) + a) + ((3*A*a^6*b + 2*B*a^5*b^2 - 15*A*a^4*b^3 - 20*B*a^3*b^4 + 5*(9*A - 8*B)
*a^2*b^5 + 6*(8*A - 5*B)*a*b^6 + (15*A - 8*B)*b^7)*cos(d*x + c)^4*sin(d*x + c) + (3*A*a^7 + 2*B*a^6*b - 15*A*a
^5*b^2 - 20*B*a^4*b^3 + 5*(9*A - 8*B)*a^3*b^4 + 6*(8*A - 5*B)*a^2*b^5 + (15*A - 8*B)*a*b^6)*cos(d*x + c)^4)*lo
g(sin(d*x + c) + 1) - ((3*A*a^6*b + 2*B*a^5*b^2 - 15*A*a^4*b^3 - 20*B*a^3*b^4 + 5*(9*A + 8*B)*a^2*b^5 - 6*(8*A
+ 5*B)*a*b^6 + (15*A + 8*B)*b^7)*cos(d*x + c)^4*sin(d*x + c) + (3*A*a^7 + 2*B*a^6*b - 15*A*a^5*b^2 - 20*B*a^4
*b^3 + 5*(9*A + 8*B)*a^3*b^4 - 6*(8*A + 5*B)*a^2*b^5 + (15*A + 8*B)*a*b^6)*cos(d*x + c)^4)*log(-sin(d*x + c) +
1) + 2*(2*A*a^7 - 2*B*a^6*b - 6*A*a^5*b^2 + 6*B*a^4*b^3 + 6*A*a^3*b^4 - 6*B*a^2*b^5 - 2*A*a*b^6 + 2*B*b^7 + (
3*A*a^7 + 2*B*a^6*b - 15*A*a^5*b^2 + 21*A*a^3*b^4 - 6*B*a^2*b^5 - 9*A*a*b^6 + 4*B*b^7)*cos(d*x + c)^2)*sin(d*x
+ c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*cos(d*x + c)^4*sin(d*x + c) + (a^9 - 4*a^7*b^2 + 6
*a^5*b^4 - 4*a^3*b^6 + a*b^8)*d*cos(d*x + c)^4)

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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)**5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.35944, size = 1027, normalized size = 2.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(16*(5*B*a^2*b^5 - 6*A*a*b^6 + B*b^7)*log(abs(b*sin(d*x + c) + a))/(a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2
*b^7 + b^9) - (3*A*a^2 + 12*A*a*b + 2*B*a*b + 15*A*b^2 + 8*B*b^2)*log(abs(-sin(d*x + c) + 1))/(a^4 + 4*a^3*b +
6*a^2*b^2 + 4*a*b^3 + b^4) + (3*A*a^2 - 12*A*a*b + 2*B*a*b + 15*A*b^2 - 8*B*b^2)*log(abs(-sin(d*x + c) - 1))/
(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - 16*(5*B*a^2*b^5*sin(d*x + c) - 6*A*a*b^6*sin(d*x + c) + B*b^7*si
n(d*x + c) + 6*B*a^3*b^4 - 7*A*a^2*b^5 + A*b^7)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(b*sin(d*x +
c) + a)) + 2*(30*B*a^2*b^4*sin(d*x + c)^4 - 36*A*a*b^5*sin(d*x + c)^4 + 6*B*b^6*sin(d*x + c)^4 - 3*A*a^6*sin(d
*x + c)^3 - 2*B*a^5*b*sin(d*x + c)^3 + 15*A*a^4*b^2*sin(d*x + c)^3 - 12*B*a^3*b^3*sin(d*x + c)^3 - 5*A*a^2*b^4
*sin(d*x + c)^3 + 14*B*a*b^5*sin(d*x + c)^3 - 7*A*b^6*sin(d*x + c)^3 + 12*B*a^4*b^2*sin(d*x + c)^2 - 16*A*a^3*
b^3*sin(d*x + c)^2 - 68*B*a^2*b^4*sin(d*x + c)^2 + 88*A*a*b^5*sin(d*x + c)^2 - 16*B*b^6*sin(d*x + c)^2 + 5*A*a
^6*sin(d*x + c) - 2*B*a^5*b*sin(d*x + c) - 17*A*a^4*b^2*sin(d*x + c) + 20*B*a^3*b^3*sin(d*x + c) + 3*A*a^2*b^4
*sin(d*x + c) - 18*B*a*b^5*sin(d*x + c) + 9*A*b^6*sin(d*x + c) + 2*B*a^6 - 4*A*a^5*b - 14*B*a^4*b^2 + 24*A*a^3
*b^3 + 36*B*a^2*b^4 - 56*A*a*b^5 + 12*B*b^6)/((a^8 - 4*a^6*b^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8)*(sin(d*x + c)^2
- 1)^2))/d