3.2.89 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [189]

Optimal. Leaf size=211 \[ \frac {(A-5 B) c^3 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {(A-5 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{2 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a+a \sin (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.33, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3051, 2818, 2819, 2816, 2746, 31} \begin {gather*} \frac {c^3 (A-5 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c^2 (A-5 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{2 a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {c (A-5 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a f (a \sin (e+f x)+a)^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{4 f (a \sin (e+f x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2816

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[a
*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), Int[Cos[e + f*x]/(c + d*Sin[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 2818

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

Rule 2819

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 - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Dist[a*((2*m - 1)/(
m + n)), 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, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m
]) &&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

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]

Rubi steps

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

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Mathematica [A]
time = 0.77, size = 199, normalized size = 0.94 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c-c \sin (e+f x))^{5/2} \left (-2 A+2 B+4 (A-2 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+2 (A-5 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 )^4+B \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (a (1+\sin (e+f x)))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1119\) vs. \(2(189)=378\).
time = 0.30, size = 1120, normalized size = 5.31

method	result	size
default	\(\text {Expression too large to display}\)	\(1120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-8*A*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+40*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))
/sin(f*x+e))+40*B*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-8*A*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-5*
B*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^3-2*A*cos(f*x+e)^2*sin(f*x+e)+9*B*cos(f*x+e)^2*sin(f*x+e)+2*A-14*B+10*B*ln(2
/(1+cos(f*x+e)))*sin(f*x+e)*cos(f*x+e)+A*cos(f*x+e)^3*ln(2/(1+cos(f*x+e)))-2*A*cos(f*x+e)-14*B*sin(f*x+e)+2*A*
sin(f*x+e)+2*A*cos(f*x+e)^3-8*B*cos(f*x+e)^3+8*B*cos(f*x+e)-B*cos(f*x+e)^3*sin(f*x+e)+5*B*ln(2/(1+cos(f*x+e)))
*cos(f*x+e)^2*sin(f*x+e)-B*cos(f*x+e)^4-2*A*cos(f*x+e)^2+15*B*cos(f*x+e)^2+4*A*cos(f*x+e)*ln(-(-1+cos(f*x+e)-s
in(f*x+e))/sin(f*x+e))-20*B*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-3*A*cos(f*x+e)^2*ln(2/(1+cos
(f*x+e)))-2*A*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-20*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)+6*B*sin(f*x+e)*cos(f*x+e)-A
*cos(f*x+e)^2*sin(f*x+e)*ln(2/(1+cos(f*x+e)))+4*A*ln(2/(1+cos(f*x+e)))*sin(f*x+e)+10*B*cos(f*x+e)*ln(2/(1+cos(
f*x+e)))-2*A*cos(f*x+e)*sin(f*x+e)*ln(2/(1+cos(f*x+e)))+2*A*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))*cos(f*x
+e)^2*sin(f*x+e)-10*B*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2*sin(f*x+e)+4*A*cos(f*x+e)*sin(f*
x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-20*B*cos(f*x+e)*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(
f*x+e))-2*A*cos(f*x+e)^3*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+10*B*cos(f*x+e)^3*ln(-(-1+cos(f*x+e)-sin(f
*x+e))/sin(f*x+e))+4*A*ln(2/(1+cos(f*x+e)))-20*B*ln(2/(1+cos(f*x+e)))+15*B*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2+6
*A*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2-30*B*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))*cos
(f*x+e)^2)*(-c*(sin(f*x+e)-1))^(5/2)/(cos(f*x+e)^3+cos(f*x+e)^2*sin(f*x+e)-3*cos(f*x+e)^2+2*cos(f*x+e)*sin(f*x
+e)-2*cos(f*x+e)-4*sin(f*x+e)+4)/(a*(1+sin(f*x+e)))^(5/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 544 vs. \(2 (202) = 404\).
time = 0.53, size = 544, normalized size = 2.58 \begin {gather*} \frac {{\left (\frac {8 \, \sqrt {a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (a^{3} + \frac {4 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {4 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{\frac {5}{2}}} + \frac {c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{a^{\frac {5}{2}}}\right )} A + B {\left (\frac {10 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{\frac {5}{2}}} - \frac {5 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{a^{\frac {5}{2}}} - \frac {2 \, {\left (\frac {5 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {16 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {14 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {16 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {5 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{\frac {5}{2}} + \frac {4 \, a^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {7 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {8 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {7 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {4 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {a^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((8*sqrt(a)*c^(5/2)*sin(f*x + e)^2/((a^3 + 4*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 6*a^3*sin(f*x + e)^2/(cos(f
*x + e) + 1)^2 + 4*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*(cos(f*x
+ e) + 1)^2) - 2*c^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^(5/2) + c^(5/2)*log(sin(f*x + e)^2/(cos(f
*x + e) + 1)^2 + 1)/a^(5/2))*A + B*(10*c^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^(5/2) - 5*c^(5/2)*lo
g(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/a^(5/2) - 2*(5*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 16*c^(5/2)
*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 14*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 16*c^(5/2)*sin(f*x + e
)^4/(cos(f*x + e) + 1)^4 + 5*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^(5/2) + 4*a^(5/2)*sin(f*x + e)/(c
os(f*x + e) + 1) + 7*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)
^3 + 7*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4*a^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + a^(5/2)*s
in(f*x + e)^6/(cos(f*x + e) + 1)^6)))/f

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((A - 2*B)*c^2*cos(f*x + e)^2 - 2*(A - B)*c^2 + (B*c^2*cos(f*x + e)^2 + 2*(A - B)*c^2)*sin(f*x + e))*
sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3
)*sin(f*x + e)), x)

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [A]
time = 0.53, size = 312, normalized size = 1.48 \begin {gather*} \frac {\sqrt {2} {\left (\frac {4 \, \sqrt {2} B c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, \sqrt {2} {\left (A \sqrt {a} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, B \sqrt {a} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \log \left (-32 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 32\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (3 \, A \sqrt {a} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 7 \, B \sqrt {a} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 4 \, {\left (A \sqrt {a} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, B \sqrt {a} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \sqrt {c}}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*(4*sqrt(2)*B*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2/(a^(5/2)*sgn
(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 2*sqrt(2)*(A*sqrt(a)*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*B*sqrt(a)
*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*log(-32*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 32)/(a^3*sgn(cos(-1/4*pi
+ 1/2*f*x + 1/2*e))) - sqrt(2)*(3*A*sqrt(a)*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 7*B*sqrt(a)*c^2*sgn(sin(
-1/4*pi + 1/2*f*x + 1/2*e)) - 4*(A*sqrt(a)*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*B*sqrt(a)*c^2*sgn(sin(-
1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2)/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*a^3*sg
n(cos(-1/4*pi + 1/2*f*x + 1/2*e))))*sqrt(c)/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-c\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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