3.4.15 \(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx\) [315]

Optimal. Leaf size=203 \[ -\frac {(c-d) (A c+3 B c+7 A d-11 B d) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {d (3 A c-15 B c-9 A d+13 B d) \cos (e+f x)}{3 a f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) d^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{6 a^2 f}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}} \]

[Out]

-1/2*(A-B)*cos(f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*sin(f*x+e))^(3/2)-1/4*(c-d)*(A*c+7*A*d+3*B*c-11*B*d)*arctanh(1
/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(3/2)/f*2^(1/2)+1/3*d*(3*A*c-9*A*d-15*B*c+13*B*d)*cos(
f*x+e)/a/f/(a+a*sin(f*x+e))^(1/2)+1/6*(3*A-7*B)*d^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/a^2/f

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Rubi [A]
time = 0.39, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {3056, 3047, 3102, 2830, 2728, 212} \begin {gather*} -\frac {(c-d) (A c+7 A d+3 B c-11 B d) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {d^2 (3 A-7 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{6 a^2 f}+\frac {d (3 A c-9 A d-15 B c+13 B d) \cos (e+f x)}{3 a f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a \sin (e+f x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((c - d)*(A*c + 3*B*c + 7*A*d - 11*B*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])
])/(Sqrt[2]*a^(3/2)*f) + (d*(3*A*c - 15*B*c - 9*A*d + 13*B*d)*Cos[e + f*x])/(3*a*f*Sqrt[a + a*Sin[e + f*x]]) +
((3*A - 7*B)*d^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(6*a^2*f) - ((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x]
)^2)/(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 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*} \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {(c+d \sin (e+f x)) \left (\frac {1}{2} a (A c+3 B c+4 A d-4 B d)-\frac {1}{2} a (3 A-7 B) d \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {\frac {1}{2} a c (A c+3 B c+4 A d-4 B d)+\left (-\frac {1}{2} a (3 A-7 B) c d+\frac {1}{2} a d (A c+3 B c+4 A d-4 B d)\right ) \sin (e+f x)-\frac {1}{2} a (3 A-7 B) d^2 \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{2 a^2}\\ &=\frac {(3 A-7 B) d^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{6 a^2 f}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {-\frac {1}{4} a^2 \left ((3 A-7 B) d^2-3 c (A c+3 B c+4 A d-4 B d)\right )-\frac {1}{2} a^2 d (3 A c-15 B c-9 A d+13 B d) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{3 a^3}\\ &=\frac {d (3 A c-15 B c-9 A d+13 B d) \cos (e+f x)}{3 a f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) d^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{6 a^2 f}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}+\frac {((c-d) (A c+3 B c+7 A d-11 B d)) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{4 a}\\ &=\frac {d (3 A c-15 B c-9 A d+13 B d) \cos (e+f x)}{3 a f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) d^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{6 a^2 f}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {((c-d) (A c+3 B c+7 A d-11 B d)) \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 )}{2 a f}\\ &=-\frac {(c-d) (A c+3 B c+7 A d-11 B d) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {d (3 A c-15 B c-9 A d+13 B d) \cos (e+f x)}{3 a f \sqrt {a+a \sin (e+f x)}}+\frac {(3 A-7 B) d^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{6 a^2 f}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.51, size = 357, normalized size = 1.76 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 (A-B) (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 (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 )+(3+3 i) (-1)^{3/4} (c-d) (A c+3 B c+7 A d-11 B d) \tanh ^{-1}\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 )^2+6 d (-4 B c-2 A d+3 B d) \cos \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 )^2-2 B d^2 \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 )^2-6 d (-4 B c-2 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 )^2-2 B d^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {3}{2} (e+f x)\right )\right )}{6 f (a (1+\sin (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(6*(A - B)*(c - d)^2*Sin[(e + f*x)/2] - 3*(A - B)*(c - d)^2*(Cos[(e + f
*x)/2] + Sin[(e + f*x)/2]) + (3 + 3*I)*(-1)^(3/4)*(c - d)*(A*c + 3*B*c + 7*A*d - 11*B*d)*ArcTanh[(1/2 + I/2)*(
-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 6*d*(-4*B*c - 2*A*d + 3*B*d)*Cos[
(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 2*B*d^2*Cos[(3*(e + f*x))/2]*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^2 - 6*d*(-4*B*c - 2*A*d + 3*B*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 2*B*d^
2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Sin[(3*(e + f*x))/2]))/(6*f*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(693\) vs. \(2(180)=360\).
time = 7.08, size = 694, normalized size = 3.42

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right ) \left (3 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2}+18 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c d -21 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{2}+24 A \,d^{2} a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}+9 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2}-42 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c d +33 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{2}-8 B \,d^{2} \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}+48 B c d \,a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}-24 B \,d^{2} a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}\right )+3 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2}+18 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c d -21 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{2}+6 A \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c^{2}-12 A \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c d +30 A \,d^{2} a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}+9 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2}-42 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c d +33 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{2}-8 B \,d^{2} \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}-6 B \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c^{2}+60 B c d \,a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}-30 B \,d^{2} a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{12 a^{\frac {7}{2}} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(694\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/a^(7/2)*(sin(f*x+e)*(3*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+18*A*2^(1/2
)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d-21*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*
2^(1/2)/a^(1/2))*a^2*d^2+24*A*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2)+9*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2
)*2^(1/2)/a^(1/2))*a^2*c^2-42*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d+33*B*2^(1/
2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2-8*B*d^2*(a-a*sin(f*x+e))^(3/2)*a^(1/2)+48*B*c*d
*a^(3/2)*(a-a*sin(f*x+e))^(1/2)-24*B*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2))+3*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+
e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+18*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d-21
*A*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2+6*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2-
12*A*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c*d+30*A*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2)+9*B*2^(1/2)*arctanh(1/2*(a-a*s
in(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2-42*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2
*c*d+33*B*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2-8*B*d^2*(a-a*sin(f*x+e))^(3/2)*a
^(1/2)-6*B*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2+60*B*c*d*a^(3/2)*(a-a*sin(f*x+e))^(1/2)-30*B*d^2*a^(3/2)*(a-a*si
n(f*x+e))^(1/2))*(-a*(sin(f*x+e)-1))^(1/2)/cos(f*x+e)/(a+a*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))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 610 vs. \(2 (189) = 378\).
time = 0.46, size = 610, normalized size = 3.00 \begin {gather*} -\frac {3 \, \sqrt {2} {\left (2 \, {\left (A + 3 \, B\right )} c^{2} + 4 \, {\left (3 \, A - 7 \, B\right )} c d - 2 \, {\left (7 \, A - 11 \, B\right )} d^{2} - {\left ({\left (A + 3 \, B\right )} c^{2} + 2 \, {\left (3 \, A - 7 \, B\right )} c d - {\left (7 \, A - 11 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} + {\left ({\left (A + 3 \, B\right )} c^{2} + 2 \, {\left (3 \, A - 7 \, B\right )} c d - {\left (7 \, A - 11 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) + {\left (2 \, {\left (A + 3 \, B\right )} c^{2} + 4 \, {\left (3 \, A - 7 \, B\right )} c d - 2 \, {\left (7 \, A - 11 \, B\right )} d^{2} + {\left ({\left (A + 3 \, B\right )} c^{2} + 2 \, {\left (3 \, A - 7 \, B\right )} c d - {\left (7 \, A - 11 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\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 (4 \, B d^{2} \cos \left (f x + e\right )^{3} - 3 \, {\left (A - B\right )} c^{2} + 6 \, {\left (A - B\right )} c d - 3 \, {\left (A - B\right )} d^{2} - 4 \, {\left (6 \, B c d + {\left (3 \, A - 4 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left ({\left (A - B\right )} c^{2} - 2 \, {\left (A - 5 \, B\right )} c d + 5 \, {\left (A - B\right )} d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, B d^{2} \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} c^{2} + 6 \, {\left (A - B\right )} c d - 3 \, {\left (A - B\right )} d^{2} + 12 \, {\left (2 \, B c d + {\left (A - B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{24 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \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))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/24*(3*sqrt(2)*(2*(A + 3*B)*c^2 + 4*(3*A - 7*B)*c*d - 2*(7*A - 11*B)*d^2 - ((A + 3*B)*c^2 + 2*(3*A - 7*B)*c*
d - (7*A - 11*B)*d^2)*cos(f*x + e)^2 + ((A + 3*B)*c^2 + 2*(3*A - 7*B)*c*d - (7*A - 11*B)*d^2)*cos(f*x + e) + (
2*(A + 3*B)*c^2 + 4*(3*A - 7*B)*c*d - 2*(7*A - 11*B)*d^2 + ((A + 3*B)*c^2 + 2*(3*A - 7*B)*c*d - (7*A - 11*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)*(c
os(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*(4*B*d^2*cos(f*x + e)^3 - 3*(A - B)*c^2 + 6*(A -
B)*c*d - 3*(A - B)*d^2 - 4*(6*B*c*d + (3*A - 4*B)*d^2)*cos(f*x + e)^2 - 3*((A - B)*c^2 - 2*(A - 5*B)*c*d + 5*(
A - B)*d^2)*cos(f*x + e) - (4*B*d^2*cos(f*x + e)^2 - 3*(A - B)*c^2 + 6*(A - B)*c*d - 3*(A - B)*d^2 + 12*(2*B*c
*d + (A - B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x
+ e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{2}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sin(e + f*x))*(c + d*sin(e + f*x))**2/(a*(sin(e + f*x) + 1))**(3/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (189) = 378\).
time = 0.61, size = 478, normalized size = 2.35 \begin {gather*} \frac {\frac {3 \, \sqrt {2} {\left (A \sqrt {a} c^{2} + 3 \, B \sqrt {a} c^{2} + 6 \, A \sqrt {a} c d - 14 \, B \sqrt {a} c d - 7 \, A \sqrt {a} d^{2} + 11 \, B \sqrt {a} d^{2}\right )} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {3 \, \sqrt {2} {\left (A \sqrt {a} c^{2} + 3 \, B \sqrt {a} c^{2} + 6 \, A \sqrt {a} c d - 14 \, B \sqrt {a} c d - 7 \, A \sqrt {a} d^{2} + 11 \, B \sqrt {a} d^{2}\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {6 \, \sqrt {2} {\left (A \sqrt {a} c^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - B \sqrt {a} c^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, A \sqrt {a} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, B \sqrt {a} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A \sqrt {a} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - B \sqrt {a} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {16 \, \sqrt {2} {\left (2 \, B a^{\frac {9}{2}} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, B a^{\frac {9}{2}} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, A a^{\frac {9}{2}} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, B a^{\frac {9}{2}} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{24 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*sqrt(2)*(A*sqrt(a)*c^2 + 3*B*sqrt(a)*c^2 + 6*A*sqrt(a)*c*d - 14*B*sqrt(a)*c*d - 7*A*sqrt(a)*d^2 + 11*B
*sqrt(a)*d^2)*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 3*sqrt(2)*(A
*sqrt(a)*c^2 + 3*B*sqrt(a)*c^2 + 6*A*sqrt(a)*c*d - 14*B*sqrt(a)*c*d - 7*A*sqrt(a)*d^2 + 11*B*sqrt(a)*d^2)*log(
-sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 6*sqrt(2)*(A*sqrt(a)*c^2*sin(
-1/4*pi + 1/2*f*x + 1/2*e) - B*sqrt(a)*c^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 2*A*sqrt(a)*c*d*sin(-1/4*pi + 1/2*
f*x + 1/2*e) + 2*B*sqrt(a)*c*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + A*sqrt(a)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) -
B*sqrt(a)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)*a^2*sgn(cos(-1/4*pi + 1
/2*f*x + 1/2*e))) - 16*sqrt(2)*(2*B*a^(9/2)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 6*B*a^(9/2)*c*d*sin(-1/4*pi
+ 1/2*f*x + 1/2*e) - 3*A*a^(9/2)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 3*B*a^(9/2)*d^2*sin(-1/4*pi + 1/2*f*x +
1/2*e))/(a^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f

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

[Out]

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

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