3.2.93 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx\) [193]

Optimal. Leaf size=208 \[ -\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}}-\frac {(3 A+B) \cos (e+f x)}{8 a f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {(3 A+B) \cos (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {(3 A+B) \tanh ^{-1}(\sin (e+f x)) \cos (e+f x)}{8 a^2 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.33, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3051, 2822, 2820, 3855} \begin {gather*} \frac {(3 A+B) \cos (e+f x)}{8 a^2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac {(3 A+B) \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a^2 c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {(3 A+B) \cos (e+f x)}{8 a f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (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])^(5/2)*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

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

Rule 2820

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Di
st[Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[1/Cos[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 2822

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[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[(m + n + 1)/(a*(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, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

Rule 3051

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]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.69, size = 305, normalized size = 1.47 \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 (-2 A \cos ^2(e+f x)+(-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+(A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-(3 A+B) \log \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 )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+(3 A+B) \log \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 )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{8 f (a (1+\sin (e+f x)))^{5/2} (c-c \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2)),x]

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(434\) vs. \(2(184)=368\).
time = 0.30, size = 435, normalized size = 2.09

method	result	size
default	\(-\frac {\left (3 A \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-3 A \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+B \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-B \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-2 A \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+3 A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-3 A \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+B \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-B \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+A \left (\cos ^{2}\left (f x +e \right )\right )+3 B \left (\cos ^{2}\left (f x +e \right )\right )-3 A \sin \left (f x +e \right )-B \sin \left (f x +e \right )-A -3 B \right ) \cos \left (f x +e \right )}{8 f \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}}\)	\(435\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)/((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(3/2)), x)

________________________________________________________________________________________

Fricas [A]
time = 0.47, size = 449, normalized size = 2.16 \begin {gather*} \left [\frac {{\left ({\left (3 \, A + B\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + {\left (3 \, A + B\right )} \cos \left (f x + e\right )^{3}\right )} \sqrt {a c} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) - 2 \, {\left ({\left (3 \, A + B\right )} \cos \left (f x + e\right )^{2} - {\left (3 \, A + B\right )} \sin \left (f x + e\right ) - A - 3 \, B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}}, -\frac {{\left ({\left (3 \, A + B\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + {\left (3 \, A + B\right )} \cos \left (f x + e\right )^{3}\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left ({\left (3 \, A + B\right )} \cos \left (f x + e\right )^{2} - {\left (3 \, A + B\right )} \sin \left (f x + e\right ) - A - 3 \, B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{8 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\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))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(((3*A + B)*cos(f*x + e)^3*sin(f*x + e) + (3*A + B)*cos(f*x + e)^3)*sqrt(a*c)*log(-(a*c*cos(f*x + e)^3 -
2*a*c*cos(f*x + e) - 2*sqrt(a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e))/cos(f*x + e
)^3) - 2*((3*A + B)*cos(f*x + e)^2 - (3*A + B)*sin(f*x + e) - A - 3*B)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*
x + e) + c))/(a^3*c^2*f*cos(f*x + e)^3*sin(f*x + e) + a^3*c^2*f*cos(f*x + e)^3), -1/8*(((3*A + B)*cos(f*x + e)
^3*sin(f*x + e) + (3*A + B)*cos(f*x + e)^3)*sqrt(-a*c)*arctan(sqrt(-a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(
f*x + e) + c)/(a*c*cos(f*x + e)*sin(f*x + e))) + ((3*A + B)*cos(f*x + e)^2 - (3*A + B)*sin(f*x + e) - A - 3*B)
*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^2*f*cos(f*x + e)^3*sin(f*x + e) + a^3*c^2*f*cos(f*
x + e)^3)]

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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))**(5/2)/(c-c*sin(f*x+e))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

________________________________________________________________________________________

Giac [A]
time = 0.54, size = 318, normalized size = 1.53 \begin {gather*} \frac {\frac {2 \, {\left (3 \, A \sqrt {a} \sqrt {c} + B \sqrt {a} \sqrt {c}\right )} \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {4 \, {\left (3 \, A \sqrt {a} \sqrt {c} + B \sqrt {a} \sqrt {c}\right )} \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, {\left (3 \, A \sqrt {a} \sqrt {c} + B \sqrt {a} \sqrt {c}\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - {\left (3 \, A \sqrt {a} \sqrt {c} + B \sqrt {a} \sqrt {c}\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - A \sqrt {a} \sqrt {c} + B \sqrt {a} \sqrt {c}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{3} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{32 \, 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))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

1/32*(2*(3*A*sqrt(a)*sqrt(c) + B*sqrt(a)*sqrt(c))*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(a^3*c^2*sgn(cos(
-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 4*(3*A*sqrt(a)*sqrt(c) + B*sqrt(a)*sqrt(c))
*log(abs(cos(-1/4*pi + 1/2*f*x + 1/2*e)))/(a^3*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f
*x + 1/2*e))) + (2*(3*A*sqrt(a)*sqrt(c) + B*sqrt(a)*sqrt(c))*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4 - (3*A*sqrt(a)*s
qrt(c) + B*sqrt(a)*sqrt(c))*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - A*sqrt(a)*sqrt(c) + B*sqrt(a)*sqrt(c))/((cos(-1
/4*pi + 1/2*f*x + 1/2*e)^2 - 1)*a^3*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*s
gn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,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))^(5/2)*(c - c*sin(e + f*x))^(3/2)),x)

[Out]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(3/2)), x)

________________________________________________________________________________________