\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx\) [325]

Optimal result

Integrand size = 37, antiderivative size = 261 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=-\frac {\left (B \left (5 c^2-34 c d-3 d^2\right )+A \left (3 c^2-14 c d+43 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} (c-d)^3 f}-\frac {2 d^{3/2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{5/2} (c-d)^3 \sqrt {c+d} f}-\frac {(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}} \]

[Out]

-1/4*(A-B)*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^(5/2)-1/16*(3*A*c-11*A*d+5*B*c+3*B*d)*cos(f*x+e)/a/(c-d)^2/f/(a
+a*sin(f*x+e))^(3/2)-1/32*(B*(5*c^2-34*c*d-3*d^2)+A*(3*c^2-14*c*d+43*d^2))*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1
/2)/(a+a*sin(f*x+e))^(1/2))/a^(5/2)/(c-d)^3/f*2^(1/2)-2*d^(3/2)*(-A*d+B*c)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/
(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(5/2)/(c-d)^3/f/(c+d)^(1/2)

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {3057, 3064, 2728, 212, 2852, 214} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=-\frac {\left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f (c-d)^3}-\frac {2 d^{3/2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{a^{5/2} f (c-d)^3 \sqrt {c+d}}-\frac {(3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{16 a f (c-d)^2 (a \sin (e+f x)+a)^{3/2}}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}} \]

[In]

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

[Out]

-1/16*((B*(5*c^2 - 34*c*d - 3*d^2) + A*(3*c^2 - 14*c*d + 43*d^2))*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt
[a + a*Sin[e + f*x]])])/(Sqrt[2]*a^(5/2)*(c - d)^3*f) - (2*d^(3/2)*(B*c - A*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e
+ f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(a^(5/2)*(c - d)^3*Sqrt[c + d]*f) - ((A - B)*Cos[e + f*x])/(4
*(c - d)*f*(a + a*Sin[e + f*x])^(5/2)) - ((3*A*c + 5*B*c - 11*A*d + 3*B*d)*Cos[e + f*x])/(16*a*(c - d)^2*f*(a
+ a*Sin[e + f*x])^(3/2))

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 214

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

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 2852

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

Rule 3057

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

Rule 3064

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {\int \frac {-\frac {1}{2} a (3 A c+5 B c-8 A d)-\frac {3}{2} a (A-B) d \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx}{4 a^2 (c-d)} \\ & = -\frac {(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {\frac {1}{4} a^2 \left (B c (5 c-29 d)+A \left (3 c^2-11 c d+32 d^2\right )\right )+\frac {1}{4} a^2 d (3 A c+5 B c-11 A d+3 B d) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{8 a^4 (c-d)^2} \\ & = -\frac {(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}+\frac {\left (d^2 (B c-A d)\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{a^3 (c-d)^3}+\frac {\left (B \left (5 c^2-34 c d-3 d^2\right )+A \left (3 c^2-14 c d+43 d^2\right )\right ) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2 (c-d)^3} \\ & = -\frac {(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (2 d^2 (B c-A d)\right ) \text {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{a^2 (c-d)^3 f}-\frac {\left (B \left (5 c^2-34 c d-3 d^2\right )+A \left (3 c^2-14 c d+43 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{16 a^2 (c-d)^3 f} \\ & = -\frac {\left (B \left (5 c^2-34 c d-3 d^2\right )+A \left (3 c^2-14 c d+43 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} (c-d)^3 f}-\frac {2 d^{3/2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{5/2} (c-d)^3 \sqrt {c+d} f}-\frac {(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 6.74 (sec) , antiderivative size = 912, normalized size of antiderivative = 3.49 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (8 (A-B) (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )+4 (-A+B) (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (c-d) (3 A c+5 B c-11 A d+3 B d) \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-(c-d) (3 A c+5 B c-11 A d+3 B d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+(1+i) (-1)^{3/4} \left (B \left (5 c^2-34 c d-3 d^2\right )+A \left (3 c^2-14 c d+43 d^2\right )\right ) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \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+\frac {8 d^{3/2} (-B c+A d) \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+\text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4}{\sqrt {c+d}}+\frac {8 d^{3/2} (B c-A d) \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+\text {RootSum}\left [c+4 d \text {$\#$1}+2 c \text {$\#$1}^2-4 d \text {$\#$1}^3+c \text {$\#$1}^4\&,\frac {-d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right )-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}-2 \sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}+3 d \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2+\sqrt {d} \sqrt {c+d} \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^2-c \log \left (-\text {$\#$1}+\tan \left (\frac {1}{4} (e+f x)\right )\right ) \text {$\#$1}^3}{-d-c \text {$\#$1}+3 d \text {$\#$1}^2-c \text {$\#$1}^3}\&\right ]\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4}{\sqrt {c+d}}\right )}{16 (c-d)^3 f (a (1+\sin (e+f x)))^{5/2}} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(A - B)*(c - d)^2*Sin[(e + f*x)/2] + 4*(-A + B)*(c - d)^2*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2]) + 2*(c - d)*(3*A*c + 5*B*c - 11*A*d + 3*B*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^2 - (c - d)*(3*A*c + 5*B*c - 11*A*d + 3*B*d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (1 +
I)*(-1)^(3/4)*(B*(5*c^2 - 34*c*d - 3*d^2) + A*(3*c^2 - 14*c*d + 43*d^2))*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 +
Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 + (8*d^(3/2)*(-(B*c) + A*d)*(e + f*x - 2*Log[Sec[(e
+ f*x)/4]^2] + RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) + Sqrt
[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 2*Sqrt[d]*Sqrt[c + d]*Log[-#1
+ Tan[(e + f*x)/4]]*#1 + 3*d*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4
]]*#1^2 - c*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ])*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^4)/Sqrt[c + d] + (8*d^(3/2)*(B*c - A*d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + RootSum[c + 4*d*#1 +
2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) - Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f
*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 3*d*Log[-#
1 + Tan[(e + f*x)/4]]*#1^2 + Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Log[-#1 + Tan[(e + f*x)/
4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)/Sqrt[c + d]))/(16*(c -
d)^3*f*(a*(1 + Sin[e + f*x]))^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1417\) vs. \(2(228)=456\).

Time = 1.32 (sec) , antiderivative size = 1418, normalized size of antiderivative = 5.43

method	result	size
default	\(\text {Expression too large to display}\)	\(1418\)

[In]

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

[Out]

-1/32*((64*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*d^3-64*B*arctanh((a-a*sin(f*x+e))^(
1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c*d^2-3*A*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(
1/2)/a^(1/2))*a^2*c^2+14*A*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c
*d-43*A*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2-5*B*(a*(c+d)*d)^
(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+34*B*(a*(c+d)*d)^(1/2)*2^(1/2)*arcta
nh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+3*B*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+
e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2)*cos(f*x+e)^2+2*sin(f*x+e)*(-64*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a
*d^2)^(1/2))*a^(5/2)*d^3+64*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c*d^2+3*A*(a*(c+d)
*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2-14*A*(a*(c+d)*d)^(1/2)*2^(1/2)*a
rctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+43*A*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin
(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2+5*B*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/
2)/a^(1/2))*a^2*c^2-34*B*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d
-3*B*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2)-128*A*arctanh((a-a
*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*d^3+128*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/
2))*a^(5/2)*c*d^2+6*A*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2-28
*A*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+86*A*(a*(c+d)*d)^(1/2
)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2+20*A*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^
(1/2)*a^(3/2)*c^2-72*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^(3/2)*c*d+52*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+
d)*d)^(1/2)*a^(3/2)*d^2-6*A*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^2+28*A*(a*(c+d)*d)^(1/2)*(a-a*s
in(f*x+e))^(3/2)*a^(1/2)*c*d-22*A*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d^2+10*B*(a*(c+d)*d)^(1/2)*
2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2-68*B*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2
*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d-6*B*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1
/2)*2^(1/2)/a^(1/2))*a^2*d^2+12*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a^(3/2)*c^2+8*B*(a-a*sin(f*x+e))^(1
/2)*(a*(c+d)*d)^(1/2)*a^(3/2)*c*d-20*B*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^2-10*B*(a*(c+d)*d)^(
1/2)*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^2+4*B*(a*(c+d)*d)^(1/2)*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d+6*B*(a*(c+d)*
d)^(1/2)*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d^2)*(-a*(sin(f*x+e)-1))^(1/2)/a^(9/2)/(1+sin(f*x+e))/(a*(c+d)*d)^(1/2
)/(c-d)^3/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1146 vs. \(2 (228) = 456\).

Time = 3.69 (sec) , antiderivative size = 2577, normalized size of antiderivative = 9.87 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/64*(sqrt(2)*(((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e)^3 - 4*(3*A + 5*B)*c^2 +
8*(7*A + 17*B)*c*d - 4*(43*A - 3*B)*d^2 + 3*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x
+ e)^2 - 2*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e) - (4*(3*A + 5*B)*c^2 - 8*(7
*A + 17*B)*c*d + 4*(43*A - 3*B)*d^2 - ((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e)^2
+ 2*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos
(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e)
- (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e)
- 2)) - 32*(4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d - A*a*d^2)*cos(f*x + e)^3 - 3*(B*a*c*d - A*a*d^2)*cos(f*x + e)^2
+ 2*(B*a*c*d - A*a*d^2)*cos(f*x + e) + (4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d - A*a*d^2)*cos(f*x + e)^2 + 2*(B*a*c*
d - A*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 - (6*c*d + 7*d^2)*cos(f*x
+ e)^2 - c^2 - 2*c*d - d^2 - 4*((c*d + d^2)*cos(f*x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)*cos(
f*x + e) + (c^2 + 4*c*d + 3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d/(a*c
+ a*d)) - (c^2 + 8*c*d + 9*d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*d^2)*co
s(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)
*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(4*(A - B)*c^
2 - 8*(A - B)*c*d + 4*(A - B)*d^2 + ((3*A + 5*B)*c^2 - 2*(7*A + B)*c*d + (11*A - 3*B)*d^2)*cos(f*x + e)^2 + ((
7*A + B)*c^2 - 2*(11*A - 3*B)*c*d + (15*A - 7*B)*d^2)*cos(f*x + e) - (4*(A - B)*c^2 - 8*(A - B)*c*d + 4*(A - B
)*d^2 - ((3*A + 5*B)*c^2 - 2*(7*A + B)*c*d + (11*A - 3*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e)
+ a))/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^3 + 3*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*
d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^
3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f + ((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2
- 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 -
a^3*d^3)*f)*sin(f*x + e)), 1/64*(sqrt(2)*(((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x +
e)^3 - 4*(3*A + 5*B)*c^2 + 8*(7*A + 17*B)*c*d - 4*(43*A - 3*B)*d^2 + 3*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d
+ (43*A - 3*B)*d^2)*cos(f*x + e)^2 - 2*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e)
- (4*(3*A + 5*B)*c^2 - 8*(7*A + 17*B)*c*d + 4*(43*A - 3*B)*d^2 - ((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A
- 3*B)*d^2)*cos(f*x + e)^2 + 2*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e))*sin(f*
x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x +
e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*s
in(f*x + e) - cos(f*x + e) - 2)) + 64*(4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d - A*a*d^2)*cos(f*x + e)^3 - 3*(B*a*c*d
- A*a*d^2)*cos(f*x + e)^2 + 2*(B*a*c*d - A*a*d^2)*cos(f*x + e) + (4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d - A*a*d^2)
*cos(f*x + e)^2 + 2*(B*a*c*d - A*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-d/(a*c + a*d))*arctan(1/2*sqrt(a*sin
(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-d/(a*c + a*d))/(d*cos(f*x + e))) + 4*(4*(A - B)*c^2 - 8*(A - B
)*c*d + 4*(A - B)*d^2 + ((3*A + 5*B)*c^2 - 2*(7*A + B)*c*d + (11*A - 3*B)*d^2)*cos(f*x + e)^2 + ((7*A + B)*c^2
- 2*(11*A - 3*B)*c*d + (15*A - 7*B)*d^2)*cos(f*x + e) - (4*(A - B)*c^2 - 8*(A - B)*c*d + 4*(A - B)*d^2 - ((3*
A + 5*B)*c^2 - 2*(7*A + B)*c*d + (11*A - 3*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a^3
*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^3 + 3*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^
3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^
2*d + 3*a^3*c*d^2 - a^3*d^3)*f + ((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^
3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f)
*sin(f*x + e))]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 760 vs. \(2 (228) = 456\).

Time = 0.43 (sec) , antiderivative size = 760, normalized size of antiderivative = 2.91 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/32*(64*sqrt(2)*(B*c*d^2 - A*d^3)*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sqrt(-c*d - d^2))/((sqrt(2
)*a^(5/2)*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*sqrt(2)*a^(5/2)*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)
) + 3*sqrt(2)*a^(5/2)*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - sqrt(2)*a^(5/2)*d^3*sgn(cos(-1/4*pi + 1/2*f*
x + 1/2*e)))*sqrt(-c*d - d^2)) - (3*A*sqrt(a)*c^2 + 5*B*sqrt(a)*c^2 - 14*A*sqrt(a)*c*d - 34*B*sqrt(a)*c*d + 43
*A*sqrt(a)*d^2 - 3*B*sqrt(a)*d^2)*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(sqrt(2)*a^3*c^3*sgn(cos(-1/4*pi + 1
/2*f*x + 1/2*e)) - 3*sqrt(2)*a^3*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*sqrt(2)*a^3*c*d^2*sgn(cos(-1/4*
pi + 1/2*f*x + 1/2*e)) - sqrt(2)*a^3*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + (3*A*sqrt(a)*c^2 + 5*B*sqrt(a)
*c^2 - 14*A*sqrt(a)*c*d - 34*B*sqrt(a)*c*d + 43*A*sqrt(a)*d^2 - 3*B*sqrt(a)*d^2)*log(-sin(-1/4*pi + 1/2*f*x +
1/2*e) + 1)/(sqrt(2)*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*sqrt(2)*a^3*c^2*d*sgn(cos(-1/4*pi + 1/2*f
*x + 1/2*e)) + 3*sqrt(2)*a^3*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - sqrt(2)*a^3*d^3*sgn(cos(-1/4*pi + 1/2
*f*x + 1/2*e))) + 2*(3*A*sqrt(a)*c*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 5*B*sqrt(a)*c*sin(-1/4*pi + 1/2*f*x + 1/
2*e)^3 - 11*A*sqrt(a)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 3*B*sqrt(a)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 5*
A*sqrt(a)*c*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 3*B*sqrt(a)*c*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 13*A*sqrt(a)*d*sin
(-1/4*pi + 1/2*f*x + 1/2*e) - 5*B*sqrt(a)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((sqrt(2)*a^3*c^2*sgn(cos(-1/4*pi
+ 1/2*f*x + 1/2*e)) - 2*sqrt(2)*a^3*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + sqrt(2)*a^3*d^2*sgn(cos(-1/4*pi
+ 1/2*f*x + 1/2*e)))*(sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2))/f

Mupad [F(-1)]

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

[In]

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

[Out]

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