\(\int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\) [1557]

Optimal result

Integrand size = 31, antiderivative size = 228 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {(a A+3 A b+2 b B) \log (1-\sin (c+d x))}{4 (a+b)^3 d}+\frac {(a A-3 A b+2 b B) \log (1+\sin (c+d x))}{4 (a-b)^3 d}+\frac {b^2 \left (4 a A b-3 a^2 B-b^2 B\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {b \left (a^2 A+3 A b^2-4 a b B\right )}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec ^2(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]

[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-B*a-(A*a-B*b)*sin(d*x+c))/(a^2-b^2)/d/(a+b*sin(d*x+c))

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2916, 837, 815} \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {b \left (a^2 A-4 a b B+3 A b^2\right )}{2 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {b^2 \left (-3 a^2 B+4 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {\sec ^2(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {(a A+3 A b+2 b B) \log (1-\sin (c+d x))}{4 d (a+b)^3}+\frac {(a A-3 A b+2 b B) \log (\sin (c+d x)+1)}{4 d (a-b)^3} \]

[In]

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

[Out]

-1/4*((a*A + 3*A*b + 2*b*B)*Log[1 - Sin[c + d*x]])/((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 815

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 837

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] +
Dist[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 2916

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/b)*x)^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*} \text {integral}& = \frac {b^3 \text {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 \text {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 \text {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*}

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {(a A-b B) ((a-b) \log (1-\sin (c+d x))-(a+b) \log (1+\sin (c+d x))+2 b \log (a+b \sin (c+d x)))}{(a-b) (a+b)}+\frac {\sec ^2(c+d x) (A b-a B+(-a A+b B) \sin (c+d x))}{a+b \sin (c+d x)}+b \left (a^2 A+3 A b^2-4 a b B\right ) \left (-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 b}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{2 \left (-a^2+b^2\right ) d} \]

[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)

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {A +B}{4 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a A -3 A b -2 B b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{3}}-\frac {A -B}{4 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a A -3 A b +2 B b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{3}}-\frac {b^{2} \left (A b -B a \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {b^{2} \left (4 A a b -3 B \,a^{2}-B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\)	\(191\)
default	\(\frac {-\frac {A +B}{4 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a A -3 A b -2 B b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{3}}-\frac {A -B}{4 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a A -3 A b +2 B b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{3}}-\frac {b^{2} \left (A b -B a \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {b^{2} \left (4 A a b -3 B \,a^{2}-B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\)	\(191\)
parallelrisch	\(\frac {8 \left (b \sin \left (3 d x +3 c \right )+2 a \cos \left (2 d x +2 c \right )+b \sin \left (d x +c \right )+2 a \right ) b^{2} \left (A a b -\frac {3}{4} B \,a^{2}-\frac {1}{4} B \,b^{2}\right ) a \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\left (b \sin \left (3 d x +3 c \right )+2 a \cos \left (2 d x +2 c \right )+b \sin \left (d x +c \right )+2 a \right ) a \left (a A +3 b \left (A +\frac {2 B}{3}\right )\right ) \left (a -b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\left (b \sin \left (3 d x +3 c \right )+2 a \cos \left (2 d x +2 c \right )+b \sin \left (d x +c \right )+2 a \right ) \left (a A -3 b \left (A -\frac {2 B}{3}\right )\right ) a \left (a +b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \left (3 a \left (a -b \right ) \left (a +b \right ) \left (A b -B a \right ) \cos \left (2 d x +2 c \right )+b \left (A \,a^{2} b -3 A \,b^{3}-\frac {3}{2} B \,a^{3}+\frac {7}{2} B a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (-2 A \,a^{4}+3 A \,a^{2} b^{2}-3 A \,b^{4}+\frac {1}{2} B \,a^{3} b +\frac {3}{2} B a \,b^{3}\right ) \sin \left (d x +c \right )+5 a \left (a -b \right ) \left (a +b \right ) \left (A b -B a \right )\right ) \left (a -b \right )\right ) \left (a +b \right )}{2 \left (a -b \right )^{3} \left (a +b \right )^{3} a d \left (b \sin \left (3 d x +3 c \right )+2 a \cos \left (2 d x +2 c \right )+b \sin \left (d x +c \right )+2 a \right )}\)	\(419\)
norman	\(\frac {\frac {\left (3 A \,a^{4}-5 A \,a^{2} b^{2}-2 A \,b^{4}-2 B \,a^{3} b +6 B a \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (3 A \,a^{4}-5 A \,a^{2} b^{2}-2 A \,b^{4}-2 B \,a^{3} b +6 B a \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (A \,a^{4}+A \,a^{2} b^{2}+2 A \,b^{4}-2 B \,a^{3} b -2 B a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (A \,a^{4}+A \,a^{2} b^{2}+2 A \,b^{4}-2 B \,a^{3} b -2 B a \,b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 \left (A b -B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}-\frac {2 \left (A b -B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}-\frac {2 \left (2 A b -2 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {b^{2} \left (4 A a b -3 B \,a^{2}-B \,b^{2}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {\left (a A -3 A b +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\left (a A +3 A b +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\)	\(629\)
risch	\(\text {Expression too large to display}\)	\(1389\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 598 vs. \(2 (219) = 438\).

Time = 0.76 (sec) , antiderivative size = 598, normalized size of antiderivative = 2.62 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\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 )}} \]

[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)

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\left (A + B \sin {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.52 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (A a - {\left (3 \, A - 2 \, B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (A a + {\left (3 \, A + 2 \, B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left (B a^{3} - 2 \, A a^{2} b + 3 \, B a b^{2} - 2 \, A b^{3} + {\left (A a^{2} b - 4 \, B a b^{2} + 3 \, A b^{3}\right )} \sin \left (d x + c\right )^{2} + {\left (A a^{3} - B a^{2} b - A a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )\right )}}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}}{4 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, B a^{2} b^{3} - 4 \, A a b^{4} + B b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (A a - 3 \, A b + 2 \, B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (A a + 3 \, A b + 2 \, B b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (A a^{2} b \sin \left (d x + c\right )^{2} - 4 \, B a b^{2} \sin \left (d x + c\right )^{2} + 3 \, A b^{3} \sin \left (d x + c\right )^{2} + A a^{3} \sin \left (d x + c\right ) - B a^{2} b \sin \left (d x + c\right ) - A a b^{2} \sin \left (d x + c\right ) + B b^{3} \sin \left (d x + c\right ) + B a^{3} - 2 \, A a^{2} b + 3 \, B a b^{2} - 2 \, A b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\right )}}}{4 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 12.16 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.43 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^2\,\left (A\,a^2\,b-4\,B\,a\,b^2+3\,A\,b^3\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {-B\,a^3+2\,A\,a^2\,b-3\,B\,a\,b^2+2\,A\,b^3}{2\,{\left (a^2-b^2\right )}^2}+\frac {\sin \left (c+d\,x\right )\,\left (A\,a-B\,b\right )}{2\,\left (a^2-b^2\right )}}{d\,\left (-b\,{\sin \left (c+d\,x\right )}^3-a\,{\sin \left (c+d\,x\right )}^2+b\,\sin \left (c+d\,x\right )+a\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (A\,a+b\,\left (3\,A+2\,B\right )\right )}{d\,\left (4\,a^3+12\,a^2\,b+12\,a\,b^2+4\,b^3\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (A\,a-b\,\left (3\,A-2\,B\right )\right )}{d\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (3\,B\,a^2\,b^2-4\,A\,a\,b^3+B\,b^4\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]

[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))