3.187 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=323 \[ \frac {8 c^5 (3 A-7 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {4 c^4 (3 A-7 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {c^3 (3 A-7 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {c^2 (3 A-7 B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {c (3 A-7 B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a f (a \sin (e+f x)+a)^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a \sin (e+f x)+a)^{5/2}} \]

[Out]

1/4*(3*A-7*B)*c*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/a/f/(a+a*sin(f*x+e))^(3/2)-1/4*(A-B)*cos(f*x+e)*(c-c*sin(f*x
+e))^(9/2)/f/(a+a*sin(f*x+e))^(5/2)+(3*A-7*B)*c^3*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f/(a+a*sin(f*x+e))^(1/
2)+1/3*(3*A-7*B)*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a^2/f/(a+a*sin(f*x+e))^(1/2)+8*(3*A-7*B)*c^5*cos(f*x+e)
*ln(1+sin(f*x+e))/a^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)+4*(3*A-7*B)*c^4*cos(f*x+e)*(c-c*sin(f*x+
e))^(1/2)/a^2/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.71, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2972, 2739, 2740, 2737, 2667, 31} \[ \frac {4 c^4 (3 A-7 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {c^3 (3 A-7 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {c^2 (3 A-7 B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {8 c^5 (3 A-7 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c (3 A-7 B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a f (a \sin (e+f x)+a)^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a \sin (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(8*(3*A - 7*B)*c^5*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(a^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]
]) + (4*(3*A - 7*B)*c^4*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a^2*f*Sqrt[a + a*Sin[e + f*x]]) + ((3*A - 7*B)
*c^3*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(a^2*f*Sqrt[a + a*Sin[e + f*x]]) + ((3*A - 7*B)*c^2*Cos[e + f*x]
*(c - c*Sin[e + f*x])^(5/2))/(3*a^2*f*Sqrt[a + a*Sin[e + f*x]]) + ((3*A - 7*B)*c*Cos[e + f*x]*(c - c*Sin[e + f
*x])^(7/2))/(4*a*f*(a + a*Sin[e + f*x])^(3/2)) - ((A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(4*f*(a + a
*Sin[e + f*x])^(5/2))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2737

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(
a*c*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[Cos[e + f*x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)), x] - Dist[(b*(2*m - 1)
)/(d*(2*n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[2*m + n + 1, 0])

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2972

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] + Dist[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 1, 0]

Rubi steps

\begin {align*} \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A-7 B) \int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a}\\ &=\frac {(3 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {((3 A-7 B) c) \int \frac {(c-c \sin (e+f x))^{7/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=\frac {(3 A-7 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (2 (3 A-7 B) c^2\right ) \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=\frac {(3 A-7 B) c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (4 (3 A-7 B) c^3\right ) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=\frac {4 (3 A-7 B) c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (8 (3 A-7 B) c^4\right ) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=\frac {4 (3 A-7 B) c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (8 (3 A-7 B) c^5 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{a \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {4 (3 A-7 B) c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (8 (3 A-7 B) c^5 \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {8 (3 A-7 B) c^5 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {4 (3 A-7 B) c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{4 f (a+a \sin (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 7.01, size = 573, normalized size = 1.77 \[ -\frac {(28 A-97 B) \sin (e+f x) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{4 f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}-\frac {(A-7 B) \cos (2 (e+f x)) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{4 f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}+\frac {16 (2 A-3 B) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3}{f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}-\frac {8 (A-B) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}+\frac {16 (3 A-7 B) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}-\frac {B \sin (3 (e+f x)) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{12 f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(-8*(A - B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^(9/2))/(f*(Cos[(e + f*x)/2] - Sin[(e +
f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(5/2)) + (16*(2*A - 3*B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(c - c*Sin[
e + f*x])^(9/2))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(5/2)) - ((A - 7*B)*Cos[2*(
e + f*x)]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(c - c*Sin[e + f*x])^(9/2))/(4*f*(Cos[(e + f*x)/2] - Sin[(e
+ f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(5/2)) + (16*(3*A - 7*B)*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*(Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2])^5*(c - c*Sin[e + f*x])^(9/2))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a*(1 +
Sin[e + f*x]))^(5/2)) - ((28*A - 97*B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*Sin[e + f*x]*(c - c*Sin[e + f*
x])^(9/2))/(4*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(5/2)) - (B*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^5*(c - c*Sin[e + f*x])^(9/2)*Sin[3*(e + f*x)])/(12*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^
9*(a*(1 + Sin[e + f*x]))^(5/2))

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fricas [F]  time = 45.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left ({\left (A - 4 \, B\right )} c^{4} \cos \left (f x + e\right )^{4} - 4 \, {\left (2 \, A - 3 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} + 8 \, {\left (A - B\right )} c^{4} + {\left (B c^{4} \cos \left (f x + e\right )^{4} + 4 \, {\left (A - 2 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} - 8 \, {\left (A - B\right )} c^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(-((A - 4*B)*c^4*cos(f*x + e)^4 - 4*(2*A - 3*B)*c^4*cos(f*x + e)^2 + 8*(A - B)*c^4 + (B*c^4*cos(f*x +
e)^4 + 4*(A - 2*B)*c^4*cos(f*x + e)^2 - 8*(A - B)*c^4)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x
+ e) + c)/(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e)), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.68, size = 1287, normalized size = 3.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x)

[Out]

-1/6/f*(-108*B*cos(f*x+e)^3*sin(f*x+e)-1152*A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2688*B*ln((1-cos(f*x+e)
+sin(f*x+e))/sin(f*x+e))+396*A*sin(f*x+e)-174*A*sin(f*x+e)*cos(f*x+e)+36*A*cos(f*x+e)^3*sin(f*x+e)-3*A*cos(f*x
+e)^4*sin(f*x+e)+15*B*sin(f*x+e)*cos(f*x+e)^4+1008*B*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-255*A*cos(f*x+e)^2*sin(
f*x+e)+581*B*cos(f*x+e)^2*sin(f*x+e)-93*B*cos(f*x+e)^4+396*A-932*B+219*A*cos(f*x+e)^3-336*B*ln(2/(cos(f*x+e)+1
))*cos(f*x+e)^3-432*A*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+576*A*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-288*A*cos(f*x+e)
*ln(2/(cos(f*x+e)+1))+442*B*sin(f*x+e)*cos(f*x+e)-1344*B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+672*B*cos(f*x+e)*ln(2
/(cos(f*x+e)+1))+864*A*cos(f*x+e)^2*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-2016*B*cos(f*x+e)^2*ln((1-cos(f*x
+e)+sin(f*x+e))/sin(f*x+e))-288*A*cos(f*x+e)*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+3*A*cos(f*x+e)^5-17*B*cos(f*x+e)^
5+2*B*cos(f*x+e)^6+336*B*ln(2/(cos(f*x+e)+1))*cos(f*x+e)^2*sin(f*x+e)+672*B*ln(2/(cos(f*x+e)+1))*sin(f*x+e)*co
s(f*x+e)+2*B*cos(f*x+e)^5*sin(f*x+e)+576*A*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-1344*B*cos(f*x+
e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-1152*A*sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2688*B*
sin(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+1023*B*cos(f*x+e)^2+144*A*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1)
)+33*A*cos(f*x+e)^4-144*A*cos(f*x+e)^2*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-288*A*cos(f*x+e)^3*ln((1-cos(f*x+e)+sin
(f*x+e))/sin(f*x+e))+672*B*cos(f*x+e)^3*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-473*B*cos(f*x+e)^3-932*B*sin(
f*x+e)+576*A*sin(f*x+e)*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-1344*B*sin(f*x+e)*cos(f*x+e)*ln((1
-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+288*A*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*sin(f*x+e)*cos(f*x+e)^2-672
*B*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*sin(f*x+e)*cos(f*x+e)^2-222*A*cos(f*x+e)+490*B*cos(f*x+e)-429*A*co
s(f*x+e)^2+576*A*ln(2/(cos(f*x+e)+1))-1344*B*ln(2/(cos(f*x+e)+1)))*(-c*(sin(f*x+e)-1))^(9/2)/(sin(f*x+e)*cos(f
*x+e)^4+cos(f*x+e)^5+4*sin(f*x+e)*cos(f*x+e)^3-5*cos(f*x+e)^4-12*cos(f*x+e)^2*sin(f*x+e)-8*cos(f*x+e)^3-8*sin(
f*x+e)*cos(f*x+e)+20*cos(f*x+e)^2+16*sin(f*x+e)+8*cos(f*x+e)-16)/(a*(1+sin(f*x+e)))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(-c*sin(f*x + e) + c)^(9/2)/(a*sin(f*x + e) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x))^(5/2),x)

[Out]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(9/2)/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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