3.1557 \(\int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=228 \[ -\frac {b \left (a^2 A-4 a b B+3 A b^2\right )}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {b^2 \left (-3 a^2 B+4 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\sec ^2(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {(a A+3 A b+2 b B) \log (1-\sin (c+d x))}{4 d (a+b)^3}+\frac {(a A-3 A b+2 b B) \log (\sin (c+d x)+1)}{4 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 228, normalized size of antiderivative = 1.00, 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 = {2837, 823, 801} \[ -\frac {b \left (a^2 A-4 a b B+3 A b^2\right )}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {b^2 \left (-3 a^2 B+4 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\sec ^2(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {(a A+3 A b+2 b B) \log (1-\sin (c+d x))}{4 d (a+b)^3}+\frac {(a A-3 A b+2 b B) \log (\sin (c+d x)+1)}{4 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 1.72, size = 246, normalized size = 1.08 \[ \frac {b \left (a^2 A-4 a b B+3 A b^2\right ) \left (\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (\sin (c+d x)+1)}{2 b (a-b)^2}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}\right )+\frac {(a A-b B) ((a-b) \log (1-\sin (c+d x))-(a+b) \log (\sin (c+d x)+1)+2 b \log (a+b \sin (c+d x)))}{(a-b) (a+b)}+\frac {\sec ^2(c+d x) ((b B-a A) \sin (c+d x)-a B+A b)}{a+b \sin (c+d x)}}{2 d \left (b^2-a^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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fricas [B]  time = 1.35, size = 598, normalized size = 2.62 \[ \frac {2 \, B a^{5} - 2 \, A a^{4} b - 4 \, B a^{3} b^{2} + 4 \, A a^{2} b^{3} + 2 \, B a b^{4} - 2 \, A b^{5} - 2 \, {\left (A a^{4} b - 4 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + 4 \, B a b^{4} - 3 \, A b^{5}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (3 \, B a^{2} b^{3} - 4 \, A a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (3 \, B a^{3} b^{2} - 4 \, A a^{2} b^{3} + B a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left ({\left (A a^{4} b + 2 \, B a^{3} b^{2} - 6 \, {\left (A - B\right )} a^{2} b^{3} - 2 \, {\left (4 \, A - 3 \, B\right )} a b^{4} - {\left (3 \, A - 2 \, B\right )} b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (A a^{5} + 2 \, B a^{4} b - 6 \, {\left (A - B\right )} a^{3} b^{2} - 2 \, {\left (4 \, A - 3 \, B\right )} a^{2} b^{3} - {\left (3 \, A - 2 \, B\right )} a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A a^{4} b + 2 \, B a^{3} b^{2} - 6 \, {\left (A + B\right )} a^{2} b^{3} + 2 \, {\left (4 \, A + 3 \, B\right )} a b^{4} - {\left (3 \, A + 2 \, B\right )} b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (A a^{5} + 2 \, B a^{4} b - 6 \, {\left (A + B\right )} a^{3} b^{2} + 2 \, {\left (4 \, A + 3 \, B\right )} a^{2} b^{3} - {\left (3 \, A + 2 \, B\right )} a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{5} - B a^{4} b - 2 \, A a^{3} b^{2} + 2 \, B a^{2} b^{3} + A a b^{4} - B b^{5}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 335, normalized size = 1.47 \[ -\frac {\frac {4 \, {\left (3 \, B a^{2} b^{3} - 4 \, A a b^{4} + B b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (A a - 3 \, A b + 2 \, B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (A a + 3 \, A b + 2 \, B b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (A a^{2} b \sin \left (d x + c\right )^{2} - 4 \, B a b^{2} \sin \left (d x + c\right )^{2} + 3 \, A b^{3} \sin \left (d x + c\right )^{2} + A a^{3} \sin \left (d x + c\right ) - B a^{2} b \sin \left (d x + c\right ) - A a b^{2} \sin \left (d x + c\right ) + B b^{3} \sin \left (d x + c\right ) + B a^{3} - 2 \, A a^{2} b + 3 \, B a b^{2} - 2 \, A b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.80, size = 388, normalized size = 1.70 \[ -\frac {A}{4 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {B}{4 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a A}{4 d \left (a +b \right )^{3}}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) A b}{4 d \left (a +b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) B b}{2 d \left (a +b \right )^{3}}-\frac {b^{3} A}{d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {b^{2} a B}{d \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {4 b^{3} \ln \left (a +b \sin \left (d x +c \right )\right ) A a}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {3 b^{2} \ln \left (a +b \sin \left (d x +c \right )\right ) B \,a^{2}}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right ) B}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {A}{4 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {B}{4 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a A}{4 d \left (a -b \right )^{3}}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) A b}{4 d \left (a -b \right )^{3}}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) B b}{2 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x)

[Out]

-1/4/d/(a+b)^2/(sin(d*x+c)-1)*A-1/4/d/(a+b)^2/(sin(d*x+c)-1)*B-1/4/d/(a+b)^3*ln(sin(d*x+c)-1)*a*A-3/4/d/(a+b)^
3*ln(sin(d*x+c)-1)*A*b-1/2/d/(a+b)^3*ln(sin(d*x+c)-1)*B*b-1/d*b^3/(a+b)^2/(a-b)^2/(a+b*sin(d*x+c))*A+1/d*b^2/(
a+b)^2/(a-b)^2/(a+b*sin(d*x+c))*a*B+4/d*b^3/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))*A*a-3/d*b^2/(a+b)^3/(a-b)^3*ln(
a+b*sin(d*x+c))*B*a^2-1/d*b^4/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))*B-1/4/d/(a-b)^2/(1+sin(d*x+c))*A+1/4/d/(a-b)^
2/(1+sin(d*x+c))*B+1/4/d/(a-b)^3*ln(1+sin(d*x+c))*a*A-3/4/d/(a-b)^3*ln(1+sin(d*x+c))*A*b+1/2/d/(a-b)^3*ln(1+si
n(d*x+c))*B*b

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maxima [A]  time = 0.52, size = 346, normalized size = 1.52 \[ -\frac {\frac {4 \, {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (A a - {\left (3 \, A - 2 \, B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (A a + {\left (3 \, A + 2 \, B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left (B a^{3} - 2 \, A a^{2} b + 3 \, B a b^{2} - 2 \, A b^{3} + {\left (A a^{2} b - 4 \, B a b^{2} + 3 \, A b^{3}\right )} \sin \left (d x + c\right )^{2} + {\left (A a^{3} - B a^{2} b - A a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )\right )}}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.66, size = 327, normalized size = 1.43 \[ \frac {\frac {{\sin \left (c+d\,x\right )}^2\,\left (A\,a^2\,b-4\,B\,a\,b^2+3\,A\,b^3\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {-B\,a^3+2\,A\,a^2\,b-3\,B\,a\,b^2+2\,A\,b^3}{2\,{\left (a^2-b^2\right )}^2}+\frac {\sin \left (c+d\,x\right )\,\left (A\,a-B\,b\right )}{2\,\left (a^2-b^2\right )}}{d\,\left (-b\,{\sin \left (c+d\,x\right )}^3-a\,{\sin \left (c+d\,x\right )}^2+b\,\sin \left (c+d\,x\right )+a\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (A\,a+b\,\left (3\,A+2\,B\right )\right )}{d\,\left (4\,a^3+12\,a^2\,b+12\,a\,b^2+4\,b^3\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (A\,a-b\,\left (3\,A-2\,B\right )\right )}{d\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (3\,B\,a^2\,b^2-4\,A\,a\,b^3+B\,b^4\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(c + d*x))/(cos(c + d*x)^3*(a + b*sin(c + d*x))^2),x)

[Out]

((sin(c + d*x)^2*(3*A*b^3 + A*a^2*b - 4*B*a*b^2))/(2*(a^4 + b^4 - 2*a^2*b^2)) - (2*A*b^3 - B*a^3 + 2*A*a^2*b -
3*B*a*b^2)/(2*(a^2 - b^2)^2) + (sin(c + d*x)*(A*a - B*b))/(2*(a^2 - b^2)))/(d*(a + b*sin(c + d*x) - a*sin(c +
d*x)^2 - b*sin(c + d*x)^3)) - (log(sin(c + d*x) - 1)*(A*a + b*(3*A + 2*B)))/(d*(12*a*b^2 + 12*a^2*b + 4*a^3 +
4*b^3)) + (log(sin(c + d*x) + 1)*(A*a - b*(3*A - 2*B)))/(d*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3)) - (log(a +
b*sin(c + d*x))*(B*b^4 + 3*B*a^2*b^2 - 4*A*a*b^3))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sin {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c))**2,x)

[Out]

Integral((A + B*sin(c + d*x))*sec(c + d*x)**3/(a + b*sin(c + d*x))**2, x)

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