3.1550 \(\int \frac{\sec ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=263 \[ -\frac{b^4 (A b-a B) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac{\left (3 a^2 A+a b (9 A+B)+b^2 (8 A+3 B)\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac{\left (3 a^2 A-a b (9 A-B)+b^2 (8 A-3 B)\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac{\sec ^4(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{4 d \left (a^2-b^2\right )}+\frac{\sec ^2(c+d x) \left (\left (3 a^3 A+a^2 b B-7 a A b^2+3 b^3 B\right ) \sin (c+d x)+4 b^2 (A b-a B)\right )}{8 d \left (a^2-b^2\right )^2} \]

[Out]

-((3*a^2*A + a*b*(9*A + B) + b^2*(8*A + 3*B))*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) + ((3*a^2*A + b^2*(8*A -
3*B) - a*b*(9*A - B))*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*d) - (b^4*(A*b - a*B)*Log[a + b*Sin[c + d*x]])/((a
^2 - b^2)^3*d) - (Sec[c + d*x]^4*(A*b - a*B - (a*A - b*B)*Sin[c + d*x]))/(4*(a^2 - b^2)*d) + (Sec[c + d*x]^2*(
4*b^2*(A*b - a*B) + (3*a^3*A - 7*a*A*b^2 + a^2*b*B + 3*b^3*B)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.447782, antiderivative size = 263, 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} \[ -\frac{b^4 (A b-a B) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac{\left (3 a^2 A+a b (9 A+B)+b^2 (8 A+3 B)\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac{\left (3 a^2 A-a b (9 A-B)+b^2 (8 A-3 B)\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac{\sec ^4(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{4 d \left (a^2-b^2\right )}+\frac{\sec ^2(c+d x) \left (\left (3 a^3 A+a^2 b B-7 a A b^2+3 b^3 B\right ) \sin (c+d x)+4 b^2 (A b-a B)\right )}{8 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a^2*A + a*b*(9*A + B) + b^2*(8*A + 3*B))*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) + ((3*a^2*A + b^2*(8*A -
3*B) - a*b*(9*A - B))*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*d) - (b^4*(A*b - a*B)*Log[a + b*Sin[c + d*x]])/((a
^2 - b^2)^3*d) - (Sec[c + d*x]^4*(A*b - a*B - (a*A - b*B)*Sin[c + d*x]))/(4*(a^2 - b^2)*d) + (Sec[c + d*x]^2*(
4*b^2*(A*b - a*B) + (3*a^3*A - 7*a*A*b^2 + a^2*b*B + 3*b^3*B)*Sin[c + d*x]))/(8*(a^2 - b^2)^2*d)

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)} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{A+\frac{B x}{b}}{(a+x) \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}-\frac{b^3 \operatorname{Subst}\left (\int \frac{-3 a^2 A+4 A b^2-a b B-3 (a A-b B) x}{(a+x) \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}+\frac{\sec ^2(c+d x) \left (4 b^2 (A b-a B)+\left (3 a^3 A-7 a A b^2+a^2 b B+3 b^3 B\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac{b \operatorname{Subst}\left (\int \frac{3 a^4 A-7 a^2 A b^2+8 A b^4+a^3 b B-5 a b^3 B+\left (3 a^3 A-7 a A b^2+a^2 b B+3 b^3 B\right ) x}{(a+x) \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}+\frac{\sec ^2(c+d x) \left (4 b^2 (A b-a B)+\left (3 a^3 A-7 a A b^2+a^2 b B+3 b^3 B\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac{b \operatorname{Subst}\left (\int \left (\frac{(a-b)^2 \left (3 a^2 A+a b (9 A+B)+b^2 (8 A+3 B)\right )}{2 b (a+b) (b-x)}+\frac{8 b^3 (-A b+a B)}{(a-b) (a+b) (a+x)}+\frac{(a+b)^2 \left (3 a^2 A+b^2 (8 A-3 B)-a b (9 A-B)\right )}{2 (a-b) 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+a b (9 A+B)+b^2 (8 A+3 B)\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac{\left (3 a^2 A+b^2 (8 A-3 B)-a b (9 A-B)\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac{b^4 (A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 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}+\frac{\sec ^2(c+d x) \left (4 b^2 (A b-a B)+\left (3 a^3 A-7 a A b^2+a^2 b B+3 b^3 B\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ \end{align*}

Mathematica [A]  time = 1.27897, size = 321, normalized size = 1.22 \[ \frac{\frac{16 b^4 (A b-a B) \log (a+b \sin (c+d x))}{\left (b^2-a^2\right )^3}-\frac{2 \left (3 a^2 A+a b (9 A+B)+b^2 (8 A+3 B)\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{(a+b)^3}+\frac{2 \left (3 a^2 A+a b (B-9 A)+b^2 (8 A-3 B)\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{(a-b)^3}+\frac{3 a A+a B+5 A b+3 b B}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{-3 a A+a B+5 A b-3 b B}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{A+B}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}+\frac{B-A}{(a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}}{16 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*(3*a^2*A + a*b*(9*A + B) + b^2*(8*A + 3*B))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(a + b)^3 + (2*(3*a
^2*A + b^2*(8*A - 3*B) + a*b*(-9*A + B))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(a - b)^3 + (16*b^4*(A*b -
a*B)*Log[a + b*Sin[c + d*x]])/(-a^2 + b^2)^3 + (A + B)/((a + b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + (3*
a*A + 5*A*b + a*B + 3*b*B)/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (-A + B)/((a - b)*(Cos[(c + d
*x)/2] + Sin[(c + d*x)/2])^4) + (-3*a*A + 5*A*b + a*B - 3*b*B)/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
)^2))/(16*d)

________________________________________________________________________________________

Maple [B]  time = 0.106, size = 586, normalized size = 2.2 \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)),x)

[Out]

-1/d*b^5/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))*A+1/d*b^4/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))*a*B+1/2/d/(8*a+8*b)/(
sin(d*x+c)-1)^2*A+1/2/d/(8*a+8*b)/(sin(d*x+c)-1)^2*B-3/16/d/(a+b)^2/(sin(d*x+c)-1)*a*A-5/16/d/(a+b)^2/(sin(d*x
+c)-1)*A*b-1/16/d/(a+b)^2/(sin(d*x+c)-1)*a*B-3/16/d/(a+b)^2/(sin(d*x+c)-1)*B*b-3/16/d/(a+b)^3*ln(sin(d*x+c)-1)
*a^2*A-9/16/d/(a+b)^3*ln(sin(d*x+c)-1)*A*a*b-1/2/d/(a+b)^3*ln(sin(d*x+c)-1)*A*b^2-1/16/d/(a+b)^3*ln(sin(d*x+c)
-1)*B*a*b-3/16/d/(a+b)^3*ln(sin(d*x+c)-1)*B*b^2-1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2*A+1/2/d/(8*a-8*b)/(1+sin(d*x+
c))^2*B-3/16/d/(a-b)^2/(1+sin(d*x+c))*a*A+5/16/d/(a-b)^2/(1+sin(d*x+c))*A*b+1/16/d/(a-b)^2/(1+sin(d*x+c))*a*B-
3/16/d/(a-b)^2/(1+sin(d*x+c))*B*b+3/16/d/(a-b)^3*ln(1+sin(d*x+c))*a^2*A-9/16/d/(a-b)^3*ln(1+sin(d*x+c))*A*a*b+
1/2/d/(a-b)^3*ln(1+sin(d*x+c))*A*b^2+1/16/d/(a-b)^3*ln(1+sin(d*x+c))*B*a*b-3/16/d/(a-b)^3*ln(1+sin(d*x+c))*B*b
^2

________________________________________________________________________________________

Maxima [A]  time = 1.01621, size = 495, normalized size = 1.88 \begin{align*} \frac{\frac{16 \,{\left (B a b^{4} - A b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac{{\left (3 \, A a^{2} -{\left (9 \, A - B\right )} a b +{\left (8 \, A - 3 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{{\left (3 \, A a^{2} +{\left (9 \, A + B\right )} a b +{\left (8 \, A + 3 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{2 \,{\left (2 \, B a^{3} - 2 \, A a^{2} b - 6 \, B a b^{2} + 6 \, A b^{3} -{\left (3 \, A a^{3} + B a^{2} b - 7 \, A a b^{2} + 3 \, B b^{3}\right )} \sin \left (d x + c\right )^{3} + 4 \,{\left (B a b^{2} - A b^{3}\right )} \sin \left (d x + c\right )^{2} +{\left (5 \, A a^{3} - B a^{2} b - 9 \, A a b^{2} + 5 \, B b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{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)),x, algorithm="maxima")

[Out]

1/16*(16*(B*a*b^4 - A*b^5)*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (3*A*a^2 - (9*A - B)*
a*b + (8*A - 3*B)*b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (3*A*a^2 + (9*A + B)*a*b + (8*A
+ 3*B)*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(2*B*a^3 - 2*A*a^2*b - 6*B*a*b^2 + 6*A*
b^3 - (3*A*a^3 + B*a^2*b - 7*A*a*b^2 + 3*B*b^3)*sin(d*x + c)^3 + 4*(B*a*b^2 - A*b^3)*sin(d*x + c)^2 + (5*A*a^3
- B*a^2*b - 9*A*a*b^2 + 5*B*b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^
4 - 2*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2))/d

________________________________________________________________________________________

Fricas [A]  time = 9.24453, size = 929, normalized size = 3.53 \begin{align*} \frac{4 \, B a^{5} - 4 \, A a^{4} b - 8 \, B a^{3} b^{2} + 8 \, A a^{2} b^{3} + 4 \, B a b^{4} - 4 \, A b^{5} + 16 \,{\left (B a b^{4} - A b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) +{\left (3 \, A a^{5} + B a^{4} b - 10 \, A a^{3} b^{2} - 6 \, B a^{2} b^{3} +{\left (15 \, A - 8 \, B\right )} a b^{4} +{\left (8 \, A - 3 \, B\right )} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, A a^{5} + B a^{4} b - 10 \, A a^{3} b^{2} - 6 \, B a^{2} b^{3} +{\left (15 \, A + 8 \, B\right )} a b^{4} -{\left (8 \, A + 3 \, B\right )} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \,{\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (2 \, A a^{5} - 2 \, B a^{4} b - 4 \, A a^{3} b^{2} + 4 \, B a^{2} b^{3} + 2 \, A a b^{4} - 2 \, B b^{5} +{\left (3 \, A a^{5} + B a^{4} b - 10 \, A a^{3} b^{2} + 2 \, B a^{2} b^{3} + 7 \, A a b^{4} - 3 \, B b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \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)),x, algorithm="fricas")

[Out]

1/16*(4*B*a^5 - 4*A*a^4*b - 8*B*a^3*b^2 + 8*A*a^2*b^3 + 4*B*a*b^4 - 4*A*b^5 + 16*(B*a*b^4 - A*b^5)*cos(d*x + c
)^4*log(b*sin(d*x + c) + a) + (3*A*a^5 + B*a^4*b - 10*A*a^3*b^2 - 6*B*a^2*b^3 + (15*A - 8*B)*a*b^4 + (8*A - 3*
B)*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*A*a^5 + B*a^4*b - 10*A*a^3*b^2 - 6*B*a^2*b^3 + (15*A + 8*B)*
a*b^4 - (8*A + 3*B)*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 8*(B*a^3*b^2 - A*a^2*b^3 - B*a*b^4 + A*b^5)*c
os(d*x + c)^2 + 2*(2*A*a^5 - 2*B*a^4*b - 4*A*a^3*b^2 + 4*B*a^2*b^3 + 2*A*a*b^4 - 2*B*b^5 + (3*A*a^5 + B*a^4*b
- 10*A*a^3*b^2 + 2*B*a^2*b^3 + 7*A*a*b^4 - 3*B*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^
4 - b^6)*d*cos(d*x + c)^4)

________________________________________________________________________________________

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)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.29397, size = 728, normalized size = 2.77 \begin{align*} \frac{\frac{16 \,{\left (B a b^{5} - A b^{6}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac{{\left (3 \, A a^{2} + 9 \, A a b + B a b + 8 \, A b^{2} + 3 \, B b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{{\left (3 \, A a^{2} - 9 \, A a b + B a b + 8 \, A b^{2} - 3 \, B b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{2 \,{\left (6 \, B a b^{4} \sin \left (d x + c\right )^{4} - 6 \, A b^{5} \sin \left (d x + c\right )^{4} - 3 \, A a^{5} \sin \left (d x + c\right )^{3} - B a^{4} b \sin \left (d x + c\right )^{3} + 10 \, A a^{3} b^{2} \sin \left (d x + c\right )^{3} - 2 \, B a^{2} b^{3} \sin \left (d x + c\right )^{3} - 7 \, A a b^{4} \sin \left (d x + c\right )^{3} + 3 \, B b^{5} \sin \left (d x + c\right )^{3} + 4 \, B a^{3} b^{2} \sin \left (d x + c\right )^{2} - 4 \, A a^{2} b^{3} \sin \left (d x + c\right )^{2} - 16 \, B a b^{4} \sin \left (d x + c\right )^{2} + 16 \, A b^{5} \sin \left (d x + c\right )^{2} + 5 \, A a^{5} \sin \left (d x + c\right ) - B a^{4} b \sin \left (d x + c\right ) - 14 \, A a^{3} b^{2} \sin \left (d x + c\right ) + 6 \, B a^{2} b^{3} \sin \left (d x + c\right ) + 9 \, A a b^{4} \sin \left (d x + c\right ) - 5 \, B b^{5} \sin \left (d x + c\right ) + 2 \, B a^{5} - 2 \, A a^{4} b - 8 \, B a^{3} b^{2} + 8 \, A a^{2} b^{3} + 12 \, B a b^{4} - 12 \, A b^{5}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{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)),x, algorithm="giac")

[Out]

1/16*(16*(B*a*b^5 - A*b^6)*log(abs(b*sin(d*x + c) + a))/(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (3*A*a^2 + 9*A
*a*b + B*a*b + 8*A*b^2 + 3*B*b^2)*log(abs(-sin(d*x + c) + 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + (3*A*a^2 - 9*A
*a*b + B*a*b + 8*A*b^2 - 3*B*b^2)*log(abs(-sin(d*x + c) - 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 2*(6*B*a*b^4*s
in(d*x + c)^4 - 6*A*b^5*sin(d*x + c)^4 - 3*A*a^5*sin(d*x + c)^3 - B*a^4*b*sin(d*x + c)^3 + 10*A*a^3*b^2*sin(d*
x + c)^3 - 2*B*a^2*b^3*sin(d*x + c)^3 - 7*A*a*b^4*sin(d*x + c)^3 + 3*B*b^5*sin(d*x + c)^3 + 4*B*a^3*b^2*sin(d*
x + c)^2 - 4*A*a^2*b^3*sin(d*x + c)^2 - 16*B*a*b^4*sin(d*x + c)^2 + 16*A*b^5*sin(d*x + c)^2 + 5*A*a^5*sin(d*x
+ c) - B*a^4*b*sin(d*x + c) - 14*A*a^3*b^2*sin(d*x + c) + 6*B*a^2*b^3*sin(d*x + c) + 9*A*a*b^4*sin(d*x + c) -
5*B*b^5*sin(d*x + c) + 2*B*a^5 - 2*A*a^4*b - 8*B*a^3*b^2 + 8*A*a^2*b^3 + 12*B*a*b^4 - 12*A*b^5)/((a^6 - 3*a^4*
b^2 + 3*a^2*b^4 - b^6)*(sin(d*x + c)^2 - 1)^2))/d