3.2.14 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx\) [114]

Optimal. Leaf size=180 \[ \frac {3 (5 A-3 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}+\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

3/32*(5*A-3*B)*cos(f*x+e)/a/c/f/(c-c*sin(f*x+e))^(3/2)+1/4*(A+B)*sec(f*x+e)/a/c/f/(c-c*sin(f*x+e))^(3/2)+3/64*
(5*A-3*B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a/c^(5/2)/f*2^(1/2)-1/8*(5*A-3*B)*sec
(f*x+e)/a/c^2/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3046, 2938, 2766, 2729, 2728, 212} \begin {gather*} \frac {3 (5 A-3 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(5*A - 3*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(32*Sqrt[2]*a*c^(5/2)*f) +
(3*(5*A - 3*B)*Cos[e + f*x])/(32*a*c*f*(c - c*Sin[e + f*x])^(3/2)) + ((A + B)*Sec[e + f*x])/(4*a*c*f*(c - c*Si
n[e + f*x])^(3/2)) - ((5*A - 3*B)*Sec[e + f*x])/(8*a*c^2*f*Sqrt[c - c*Sin[e + f*x]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2766

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*((
g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[a*((2*p + 1)/(2*g^2*(p + 1))), In
t[(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0
] && LtQ[p, -1] && IntegerQ[2*p]

Rule 2938

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p
+ 1))), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 3046

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

Rubi steps

\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx &=\frac {\int \frac {\sec ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx}{a c}\\ &=\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(5 A-3 B) \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{8 a c^2}\\ &=\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {(3 (5 A-3 B)) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{16 a c}\\ &=\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {(3 (5 A-3 B)) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{64 a c^2}\\ &=\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(3 (5 A-3 B)) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{32 a c^2 f}\\ &=\frac {3 (5 A-3 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}+\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.58, size = 404, normalized size = 2.24 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (8 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+4 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+(7 A-B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-(3+3 i) \sqrt [4]{-1} (5 A-3 B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+8 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (7 A-B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{32 a f (1+\sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(-A + B)*(Cos[(e + f*x)/2] - S
in[(e + f*x)/2])^4 + 4*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (
7*A - B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - (3 + 3*I)*(-1)^(1/4)*
(5*A - 3*B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2]) + 8*(A + B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(7*A
- B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))/(32*a*f*
(1 + Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs. \(2(157)=314\).
time = 8.98, size = 350, normalized size = 1.94

method	result	size
default	\(-\frac {\sin \left (f x +e \right ) \left (-30 A \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+40 A \,c^{\frac {5}{2}}+18 B \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-24 B \,c^{\frac {5}{2}}\right )+\left (-15 A \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+30 A \,c^{\frac {5}{2}}+9 B \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-18 B \,c^{\frac {5}{2}}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+30 A \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-24 A \,c^{\frac {5}{2}}-18 B \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+40 B \,c^{\frac {5}{2}}}{64 c^{\frac {9}{2}} a \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(350\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64/c^(9/2)/a*(sin(f*x+e)*(-30*A*(c+c*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c
^(1/2))*c^2+40*A*c^(5/2)+18*B*(c+c*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2
))*c^2-24*B*c^(5/2))+(-15*A*(c+c*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))
*c^2+30*A*c^(5/2)+9*B*(c+c*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2-1
8*B*c^(5/2))*cos(f*x+e)^2+30*A*(c+c*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/
2))*c^2-24*A*c^(5/2)-18*B*(c+c*sin(f*x+e))^(1/2)*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c
^2+40*B*c^(5/2))/(sin(f*x+e)-1)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 304, normalized size = 1.69 \begin {gather*} -\frac {3 \, \sqrt {2} {\left ({\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (5 \, A - 3 \, B\right )} \sin \left (f x + e\right ) - 12 \, A + 20 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{128 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(3*sqrt(2)*((5*A - 3*B)*cos(f*x + e)^3 + 2*(5*A - 3*B)*cos(f*x + e)*sin(f*x + e) - 2*(5*A - 3*B)*cos(f*
x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x +
e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*
sin(f*x + e) - cos(f*x + e) - 2)) + 4*(3*(5*A - 3*B)*cos(f*x + e)^2 + 4*(5*A - 3*B)*sin(f*x + e) - 12*A + 20*B
)*sqrt(-c*sin(f*x + e) + c))/(a*c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*cos(f*x + e)*sin(f*x + e) - 2*a*c^3*f*cos(f*x
+ e))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 528 vs. \(2 (165) = 330\).
time = 0.63, size = 528, normalized size = 2.93 \begin {gather*} \frac {\frac {12 \, \sqrt {2} {\left (5 \, A \sqrt {c} - 3 \, B \sqrt {c}\right )} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{a c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (A \sqrt {c} + B \sqrt {c} - \frac {16 \, A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {90 \, A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - \frac {54 \, B \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}{a c^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {128 \, \sqrt {2} {\left (A \sqrt {c} - B \sqrt {c}\right )}}{a c^{3} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\frac {16 \, \sqrt {2} A a c^{\frac {7}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {\sqrt {2} A a c^{\frac {7}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - \frac {\sqrt {2} B a c^{\frac {7}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}}{a^{2} c^{6}}}{512 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/512*(12*sqrt(2)*(5*A*sqrt(c) - 3*B*sqrt(c))*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x
+ 1/2*e) + 1))/(a*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2)*(A*sqrt(c) + B*sqrt(c) - 16*A*sqrt(c)*(c
os(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 90*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x
+ 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 54*B*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(
cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2/(a*c^3*(cos(-1/4*pi + 1/2*f*x +
1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 128*sqrt(2)*(A*sqrt(c) - B*sqrt(c))/(a*c^3*((cos(-1/4*pi
+ 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - (16*s
qrt(2)*A*a*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2
*f*x + 1/2*e) + 1) - sqrt(2)*A*a*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/
2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - sqrt(2)*B*a*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(
sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(a^2*c^6))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,\sin \left (e+f\,x\right )}{\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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