\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx\) [194]

Optimal result

Integrand size = 40, antiderivative size = 245 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}-\frac {A \cos (e+f x)}{2 a f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 A \cos (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {3 A \cos (e+f x)}{8 a^2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {3 A \text {arctanh}(\sin (e+f x)) \cos (e+f x)}{8 a^2 c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3051, 2822, 2820, 3855} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 A \cos (e+f x) \text {arctanh}(\sin (e+f x))}{8 a^2 c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {3 A \cos (e+f x)}{8 a^2 c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac {3 A \cos (e+f x)}{8 a^2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac {(A-B) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}-\frac {A \cos (e+f x)}{2 a f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}} \]

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2)),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])^(5/2)) - (A*Cos[e + f*x])/(2*a*
f*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2)) + (3*A*Cos[e + f*x])/(8*a^2*f*Sqrt[a + a*Sin[e + f*x]
]*(c - c*Sin[e + f*x])^(5/2)) + (3*A*Cos[e + f*x])/(8*a^2*c*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3
/2)) + (3*A*ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(8*a^2*c^2*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*} \text {integral}& = -\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}+\frac {A \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx}{a} \\ & = -\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}-\frac {A \cos (e+f x)}{2 a f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {(3 A) \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx}{2 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))^{5/2}}-\frac {A \cos (e+f x)}{2 a f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 A \cos (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {(3 A) \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx}{4 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))^{5/2}}-\frac {A \cos (e+f x)}{2 a f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 A \cos (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {3 A \cos (e+f x)}{8 a^2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {(3 A) \int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx}{8 a^2 c^2} \\ & = -\frac {(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}-\frac {A \cos (e+f x)}{2 a f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 A \cos (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {3 A \cos (e+f x)}{8 a^2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {(3 A \cos (e+f x)) \int \sec (e+f x) \, dx}{8 a^2 c^2 \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))^{5/2}}-\frac {A \cos (e+f x)}{2 a f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 A \cos (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {3 A \cos (e+f x)}{8 a^2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {3 A \text {arctanh}(\sin (e+f x)) \cos (e+f x)}{8 a^2 c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.58 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.78 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sec ^3(e+f x) \left (16 B-9 A \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-12 A \cos (2 (e+f x)) \left (\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )-3 A \cos (4 (e+f x)) \left (\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )+9 A \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )+22 A \sin (e+f x)+6 A \sin (3 (e+f x))\right )}{64 a^2 c^2 f \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(Sec[e + f*x]^3*(16*B - 9*A*Log[1 - Tan[(e + f*x)/2]] - 12*A*Cos[2*(e + f*x)]*(Log[1 - Tan[(e + f*x)/2]] - Log
[1 + Tan[(e + f*x)/2]]) - 3*A*Cos[4*(e + f*x)]*(Log[1 - Tan[(e + f*x)/2]] - Log[1 + Tan[(e + f*x)/2]]) + 9*A*L
og[1 + Tan[(e + f*x)/2]] + 22*A*Sin[e + f*x] + 6*A*Sin[3*(e + f*x)]))/(64*a^2*c^2*f*Sqrt[a*(1 + Sin[e + f*x])]
*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 3.04 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.56

method	result	size
default	\(\frac {3 A \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-3 A \cos \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-2 \cos \left (f x +e \right ) B +3 \tan \left (f x +e \right ) A +2 \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right ) A +2 \left (\sec ^{3}\left (f x +e \right )\right ) B}{8 a^{2} c^{2} f \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(136\)
parts	\(\frac {A \left (3 \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-3 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \cos \left (f x +e \right )+3 \tan \left (f x +e \right )+2 \left (\sec ^{2}\left (f x +e \right )\right ) \tan \left (f x +e \right )\right )}{8 f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2} c^{2}}+\frac {B \left (\tan ^{2}\left (f x +e \right )\right ) \sec \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )+1\right )}{4 f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2} c^{2}}\)	\(175\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.25 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\left [\frac {3 \, \sqrt {a c} A \cos \left (f x + e\right )^{5} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, {\left ({\left (3 \, A \cos \left (f x + e\right )^{2} + 2 \, A\right )} \sin \left (f x + e\right ) + 2 \, B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}, -\frac {3 \, \sqrt {-a c} A \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - {\left ({\left (3 \, A \cos \left (f x + e\right )^{2} + 2 \, A\right )} \sin \left (f x + e\right ) + 2 \, B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{8 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}\right ] \]

[In]

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

[Out]

[1/16*(3*sqrt(a*c)*A*cos(f*x + e)^5*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*cos(f*x + e)^2 + 2*A)*sin(f*x + e
) + 2*B)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^3*f*cos(f*x + e)^5), -1/8*(3*sqrt(-a*c)*A*
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)))*cos(f*x
+ e)^5 - ((3*A*cos(f*x + e)^2 + 2*A)*sin(f*x + e) + 2*B)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(
a^3*c^3*f*cos(f*x + e)^5)]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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