\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx\) [321]

Optimal result

Integrand size = 37, antiderivative size = 308 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {(c-d) \left (B \left (5 c^2+62 c d-163 d^2\right )+3 A \left (c^2+6 c d+25 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}+\frac {d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {d^2 (9 A c+15 B c+39 A d-95 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{48 a^3 f}-\frac {(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}} \]

[Out]

-1/16*(3*A*c+9*A*d+5*B*c-17*B*d)*cos(f*x+e)*(c+d*sin(f*x+e))^2/a/f/(a+a*sin(f*x+e))^(3/2)-1/4*(A-B)*cos(f*x+e)
*(c+d*sin(f*x+e))^3/f/(a+a*sin(f*x+e))^(5/2)-1/32*(c-d)*(B*(5*c^2+62*c*d-163*d^2)+3*A*(c^2+6*c*d+25*d^2))*arct
anh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(5/2)/f*2^(1/2)+1/24*d*(A*(9*c^2+36*c*d-93*d^2)+B
*(15*c^2-228*c*d+197*d^2))*cos(f*x+e)/a^2/f/(a+a*sin(f*x+e))^(1/2)+1/48*d^2*(9*A*c+39*A*d+15*B*c-95*B*d)*cos(f
*x+e)*(a+a*sin(f*x+e))^(1/2)/a^3/f

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 308, 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 = {3056, 3047, 3102, 2830, 2728, 212} \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {(c-d) \left (3 A \left (c^2+6 c d+25 d^2\right )+B \left (5 c^2+62 c d-163 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}+\frac {d^2 (9 A c+39 A d+15 B c-95 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{48 a^3 f}+\frac {d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac {(3 A c+9 A d+5 B c-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a \sin (e+f x)+a)^{3/2}} \]

[In]

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

[Out]

-1/16*((c - d)*(B*(5*c^2 + 62*c*d - 163*d^2) + 3*A*(c^2 + 6*c*d + 25*d^2))*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqr
t[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[2]*a^(5/2)*f) + (d*(A*(9*c^2 + 36*c*d - 93*d^2) + B*(15*c^2 - 228*c*d +
197*d^2))*Cos[e + f*x])/(24*a^2*f*Sqrt[a + a*Sin[e + f*x]]) + (d^2*(9*A*c + 15*B*c + 39*A*d - 95*B*d)*Cos[e +
f*x]*Sqrt[a + a*Sin[e + f*x]])/(48*a^3*f) - ((3*A*c + 5*B*c + 9*A*d - 17*B*d)*Cos[e + f*x]*(c + d*Sin[e + f*x
])^2)/(16*a*f*(a + a*Sin[e + f*x])^(3/2)) - ((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(4*f*(a + a*Sin[e +
f*x])^(5/2))

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 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 2830

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

Rule 3047

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

Rule 3056

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[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\int \frac {(c+d \sin (e+f x))^2 \left (\frac {1}{2} a (3 A c+5 B c+6 A d-6 B d)-\frac {1}{2} a (3 A-11 B) d \sin (e+f x)\right )}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\int \frac {(c+d \sin (e+f x)) \left (\frac {1}{4} a^2 \left (B \left (5 c^2+47 c d-68 d^2\right )+3 A \left (c^2+3 c d+12 d^2\right )\right )-\frac {1}{4} a^2 d (9 A c+15 B c+39 A d-95 B d) \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{8 a^4} \\ & = -\frac {(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\int \frac {\frac {1}{4} a^2 c \left (B \left (5 c^2+47 c d-68 d^2\right )+3 A \left (c^2+3 c d+12 d^2\right )\right )+\left (-\frac {1}{4} a^2 c d (9 A c+15 B c+39 A d-95 B d)+\frac {1}{4} a^2 d \left (B \left (5 c^2+47 c d-68 d^2\right )+3 A \left (c^2+3 c d+12 d^2\right )\right )\right ) \sin (e+f x)-\frac {1}{4} a^2 d^2 (9 A c+15 B c+39 A d-95 B d) \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{8 a^4} \\ & = \frac {d^2 (9 A c+15 B c+39 A d-95 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{48 a^3 f}-\frac {(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\int \frac {\frac {1}{8} a^3 \left (3 A \left (3 c^3+9 c^2 d+33 c d^2-13 d^3\right )+B \left (15 c^3+141 c^2 d-219 c d^2+95 d^3\right )\right )-\frac {1}{4} a^3 d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{12 a^5} \\ & = \frac {d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {d^2 (9 A c+15 B c+39 A d-95 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{48 a^3 f}-\frac {(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left ((c-d) \left (B \left (5 c^2+62 c d-163 d^2\right )+3 A \left (c^2+6 c d+25 d^2\right )\right )\right ) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2} \\ & = \frac {d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {d^2 (9 A c+15 B c+39 A d-95 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{48 a^3 f}-\frac {(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\left ((c-d) \left (B \left (5 c^2+62 c d-163 d^2\right )+3 A \left (c^2+6 c d+25 d^2\right )\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 f} \\ & = -\frac {(c-d) \left (B \left (5 c^2+62 c d-163 d^2\right )+3 A \left (c^2+6 c d+25 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} f}+\frac {d \left (A \left (9 c^2+36 c d-93 d^2\right )+B \left (15 c^2-228 c d+197 d^2\right )\right ) \cos (e+f x)}{24 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {d^2 (9 A c+15 B c+39 A d-95 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{48 a^3 f}-\frac {(3 A c+5 B c+9 A d-17 B d) \cos (e+f x) (c+d \sin (e+f x))^2}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{4 f (a+a \sin (e+f x))^{5/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.77 (sec) , antiderivative size = 523, normalized size of antiderivative = 1.70 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (24 (A-B) (c-d)^3 \sin \left (\frac {1}{2} (e+f x)\right )-12 (A-B) (c-d)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+6 (c-d)^2 (B (5 c-29 d)+3 A (c+7 d)) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-3 (c-d)^2 (B (5 c-29 d)+3 A (c+7 d)) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+(3+3 i) (-1)^{3/4} (c-d) \left (B \left (5 c^2+62 c d-163 d^2\right )+3 A \left (c^2+6 c d+25 d^2\right )\right ) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-16 B d^3 \cos \left (\frac {3}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+(24+24 i) d^2 (-6 B c-2 A d+5 B d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+(24+24 i) d^2 (6 B c+2 A d-5 B d) \left (i \cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-16 B d^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {3}{2} (e+f x)\right )\right )}{48 f (a (1+\sin (e+f x)))^{5/2}} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(24*(A - B)*(c - d)^3*Sin[(e + f*x)/2] - 12*(A - B)*(c - d)^3*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2]) + 6*(c - d)^2*(B*(5*c - 29*d) + 3*A*(c + 7*d))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])^2 - 3*(c - d)^2*(B*(5*c - 29*d) + 3*A*(c + 7*d))*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3
+ (3 + 3*I)*(-1)^(3/4)*(c - d)*(B*(5*c^2 + 62*c*d - 163*d^2) + 3*A*(c^2 + 6*c*d + 25*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 - 16*B*d^3*Cos[(3*(e + f*x))/2]*(
Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 + (24 + 24*I)*d^2*(-6*B*c - 2*A*d + 5*B*d)*(Cos[(e + f*x)/2] + I*Sin[(e
+ f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 + (24 + 24*I)*d^2*(6*B*c + 2*A*d - 5*B*d)*(I*Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 16*B*d^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2
])^4*Sin[(3*(e + f*x))/2]))/(48*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. \(1156\) vs. \(2(281)=562\).

Time = 3.99 (sec) , antiderivative size = 1157, normalized size of antiderivative = 3.76

method	result	size
parts	\(\text {Expression too large to display}\)	\(1157\)
default	\(\text {Expression too large to display}\)	\(1438\)

[In]

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

[Out]

-1/32*A*c^3/a^(9/2)*(-3*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*cos(f*x+e)^2+6*2^(1/2)
*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*sin(f*x+e)+6*(a-a*sin(f*x+e))^(1/2)*sin(f*x+e)*a^(3/2
)+6*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2+14*(a-a*sin(f*x+e))^(1/2)*a^(3/2))*(-a*(si
n(f*x+e)-1))^(1/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f-1/32*c^2*(3*A*d+B*c)*(-5*2^(1/2)*arctanh
(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3*cos(f*x+e)^2+10*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^
(1/2)/a^(1/2))*sin(f*x+e)*a^3+12*(a-a*sin(f*x+e))^(1/2)*a^(5/2)-10*(a-a*sin(f*x+e))^(3/2)*a^(3/2)+10*2^(1/2)*a
rctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^3)*(-a*(sin(f*x+e)-1))^(1/2)/a^(11/2)/(1+sin(f*x+e))/cos(
f*x+e)/(a+a*sin(f*x+e))^(1/2)/f+1/32*d^2*(A*d+3*B*c)/a^(9/2)*(-75*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2
^(1/2)/a^(1/2))*a^2*cos(f*x+e)^2+64*a^(3/2)*(a-a*sin(f*x+e))^(1/2)*cos(f*x+e)^2+150*2^(1/2)*arctanh(1/2*(a-a*s
in(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*sin(f*x+e)-128*(a-a*sin(f*x+e))^(1/2)*sin(f*x+e)*a^(3/2)+150*2^(1/2)*arc
tanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2-204*(a-a*sin(f*x+e))^(1/2)*a^(3/2)+42*(a-a*sin(f*x+e))^(3
/2)*a^(1/2))*(-a*(sin(f*x+e)-1))^(1/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f-1/96*d^3*B*(64*(a-a*
sin(f*x+e))^(3/2)*a^(1/2)*cos(f*x+e)^2+384*a^(3/2)*(a-a*sin(f*x+e))^(1/2)*cos(f*x+e)^2-489*2^(1/2)*arctanh(1/2
*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*cos(f*x+e)^2-128*(a-a*sin(f*x+e))^(3/2)*sin(f*x+e)*a^(1/2)-768*(a
-a*sin(f*x+e))^(1/2)*sin(f*x+e)*a^(3/2)+978*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*si
n(f*x+e)+46*(a-a*sin(f*x+e))^(3/2)*a^(1/2)-1092*(a-a*sin(f*x+e))^(1/2)*a^(3/2)+978*2^(1/2)*arctanh(1/2*(a-a*si
n(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2)*(-a*(sin(f*x+e)-1))^(1/2)/a^(9/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x
+e))^(1/2)/f-3/32*c*d*(A*d+B*c)/a^(9/2)*(-19*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c
os(f*x+e)^2+38*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*sin(f*x+e)+38*2^(1/2)*arctanh(1
/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2-44*(a-a*sin(f*x+e))^(1/2)*a^(3/2)+26*(a-a*sin(f*x+e))^(3/2)*a^(
1/2))*(-a*(sin(f*x+e)-1))^(1/2)/(1+sin(f*x+e))/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. 980 vs. \(2 (282) = 564\).

Time = 0.30 (sec) , antiderivative size = 980, normalized size of antiderivative = 3.18 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/192*(3*sqrt(2)*(4*(3*A + 5*B)*c^3 + 12*(5*A + 19*B)*c^2*d + 12*(19*A - 75*B)*c*d^2 - 4*(75*A - 163*B)*d^3 -
((3*A + 5*B)*c^3 + 3*(5*A + 19*B)*c^2*d + 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e)^3 - 3*((3*
A + 5*B)*c^3 + 3*(5*A + 19*B)*c^2*d + 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e)^2 + 2*((3*A + 5
*B)*c^3 + 3*(5*A + 19*B)*c^2*d + 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e) + (4*(3*A + 5*B)*c^3
+ 12*(5*A + 19*B)*c^2*d + 12*(19*A - 75*B)*c*d^2 - 4*(75*A - 163*B)*d^3 - ((3*A + 5*B)*c^3 + 3*(5*A + 19*B)*c
^2*d + 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e)^2 + 2*((3*A + 5*B)*c^3 + 3*(5*A + 19*B)*c^2*d
+ 3*(19*A - 75*B)*c*d^2 - (75*A - 163*B)*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*s
qrt(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)) + 4*(32*B*d
^3*cos(f*x + e)^4 - 12*(A - B)*c^3 + 36*(A - B)*c^2*d - 36*(A - B)*c*d^2 + 12*(A - B)*d^3 + 32*(9*B*c*d^2 + (3
*A - 5*B)*d^3)*cos(f*x + e)^3 - 3*((3*A + 5*B)*c^3 + 3*(5*A - 13*B)*c^2*d - 3*(13*A - 53*B)*c*d^2 + (53*A - 93
*B)*d^3)*cos(f*x + e)^2 - 3*((7*A + B)*c^3 + 3*(A - 9*B)*c^2*d - 27*(A - 9*B)*c*d^2 + 9*(9*A - 17*B)*d^3)*cos(
f*x + e) + (32*B*d^3*cos(f*x + e)^3 + 12*(A - B)*c^3 - 36*(A - B)*c^2*d + 36*(A - B)*c*d^2 - 12*(A - B)*d^3 -
96*(3*B*c*d^2 + (A - 2*B)*d^3)*cos(f*x + e)^2 - 3*((3*A + 5*B)*c^3 + 3*(5*A - 13*B)*c^2*d - 3*(13*A - 85*B)*c*
d^2 + (85*A - 157*B)*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^3 + 3*a^3*
f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*si
n(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{3}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 751 vs. \(2 (282) = 564\).

Time = 0.55 (sec) , antiderivative size = 751, normalized size of antiderivative = 2.44 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/192*(3*sqrt(2)*(3*A*sqrt(a)*c^3 + 5*B*sqrt(a)*c^3 + 15*A*sqrt(a)*c^2*d + 57*B*sqrt(a)*c^2*d + 57*A*sqrt(a)*c
*d^2 - 225*B*sqrt(a)*c*d^2 - 75*A*sqrt(a)*d^3 + 163*B*sqrt(a)*d^3)*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^
3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 3*sqrt(2)*(3*A*sqrt(a)*c^3 + 5*B*sqrt(a)*c^3 + 15*A*sqrt(a)*c^2*d + 5
7*B*sqrt(a)*c^2*d + 57*A*sqrt(a)*c*d^2 - 225*B*sqrt(a)*c*d^2 - 75*A*sqrt(a)*d^3 + 163*B*sqrt(a)*d^3)*log(-sin(
-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 6*sqrt(2)*(3*A*sqrt(a)*c^3*sin(-1/
4*pi + 1/2*f*x + 1/2*e)^3 + 5*B*sqrt(a)*c^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 15*A*sqrt(a)*c^2*d*sin(-1/4*pi
+ 1/2*f*x + 1/2*e)^3 - 39*B*sqrt(a)*c^2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 39*A*sqrt(a)*c*d^2*sin(-1/4*pi +
1/2*f*x + 1/2*e)^3 + 63*B*sqrt(a)*c*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 21*A*sqrt(a)*d^3*sin(-1/4*pi + 1/2*
f*x + 1/2*e)^3 - 29*B*sqrt(a)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 5*A*sqrt(a)*c^3*sin(-1/4*pi + 1/2*f*x + 1
/2*e) - 3*B*sqrt(a)*c^3*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 9*A*sqrt(a)*c^2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 33
*B*sqrt(a)*c^2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 33*A*sqrt(a)*c*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 57*B*sqr
t(a)*c*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 19*A*sqrt(a)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 27*B*sqrt(a)*d^3
*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2
*e))) - 128*sqrt(2)*(2*B*a^(13/2)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 9*B*a^(13/2)*c*d^2*sin(-1/4*pi + 1/2*
f*x + 1/2*e) - 3*A*a^(13/2)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 6*B*a^(13/2)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*
e))/(a^9*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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