3.2.80 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\) [180]

Optimal. Leaf size=271 \[ -\frac {4 (3 A-5 B) c^4 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 (3 A-5 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}} \]

[Out]

-1/2*(A-B)*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/f/(a+a*sin(f*x+e))^(3/2)-1/2*(3*A-5*B)*c^2*cos(f*x+e)*(c-c*sin(f*
x+e))^(3/2)/a/f/(a+a*sin(f*x+e))^(1/2)-1/6*(3*A-5*B)*c*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a/f/(a+a*sin(f*x+e))^
(1/2)-4*(3*A-5*B)*c^4*cos(f*x+e)*ln(1+sin(f*x+e))/a/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)-2*(3*A-5*B
)*c^3*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.39, antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3051, 2819, 2816, 2746, 31} \begin {gather*} -\frac {4 c^4 (3 A-5 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 c^3 (3 A-5 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a \sin (e+f x)+a}}-\frac {c^2 (3 A-5 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a \sin (e+f x)+a}}-\frac {c (3 A-5 B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{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 - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(-4*(3*A - 5*B)*c^4*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]
) - (2*(3*A - 5*B)*c^3*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]) - ((3*A - 5*B)*c^
2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((3*A - 5*B)*c*Cos[e + f*x]*(c -
c*Sin[e + f*x])^(5/2))/(6*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(
2*f*(a + a*Sin[e + f*x])^(3/2))

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 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))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {(3 A-5 B) \int \frac {(c-c \sin (e+f x))^{7/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{2 a}\\ &=-\frac {(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {((3 A-5 B) c) \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac {(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (2 (3 A-5 B) c^2\right ) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac {2 (3 A-5 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (4 (3 A-5 B) c^3\right ) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac {2 (3 A-5 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (4 (3 A-5 B) c^4 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {2 (3 A-5 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (4 (3 A-5 B) c^4 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {4 (3 A-5 B) c^4 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 (3 A-5 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(3 A-5 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 2.32, size = 271, normalized size = 1.00 \begin {gather*} -\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (132 A-45 B+2 (27 A-59 B) \cos (2 (e+f x))+B \cos (4 (e+f x))+576 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-960 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-117 A \sin (e+f x)+279 B \sin (e+f x)+576 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x)-960 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x)-3 A \sin (3 (e+f x))+13 B \sin (3 (e+f x))\right )}{24 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-1/24*(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(132*A - 45*B + 2*(27*A - 59*B)*Cos[
2*(e + f*x)] + B*Cos[4*(e + f*x)] + 576*A*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 960*B*Log[Cos[(e + f*x)/2
] + Sin[(e + f*x)/2]] - 117*A*Sin[e + f*x] + 279*B*Sin[e + f*x] + 576*A*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2
]]*Sin[e + f*x] - 960*B*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sin[e + f*x] - 3*A*Sin[3*(e + f*x)] + 13*B*Si
n[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(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. \(947\) vs. \(2(245)=490\).
time = 0.32, size = 948, normalized size = 3.50

method	result	size
default	\(\text {Expression too large to display}\)	\(948\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/6/f*(288*A*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-480*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x
+e))/sin(f*x+e))-480*B*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+288*A*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x
+e))+24*A*cos(f*x+e)^2*sin(f*x+e)-48*B*cos(f*x+e)^2*sin(f*x+e)-102*A+166*B-120*B*ln(2/(1+cos(f*x+e)))*sin(f*x+
e)*cos(f*x+e)+27*A*cos(f*x+e)+166*B*sin(f*x+e)-102*A*sin(f*x+e)-27*A*cos(f*x+e)^3+61*B*cos(f*x+e)^3-59*B*cos(f
*x+e)-13*B*cos(f*x+e)^3*sin(f*x+e)-11*B*cos(f*x+e)^4+99*A*cos(f*x+e)^2-155*B*cos(f*x+e)^2-144*A*cos(f*x+e)*ln(
-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+240*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)^3*sin(f*x+e)+72*A*cos(f*x+e)^2*ln(2/(1+cos(f*x+e)))+72*A*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-2*B*cos(f*x+e)^5
+240*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)-107*B*sin(f*x+e)*cos(f*x+e)-144*A*ln(2/(1+cos(f*x+e)))*sin(f*x+e)-120*B
*cos(f*x+e)*ln(2/(1+cos(f*x+e)))+3*A*cos(f*x+e)^4+72*A*cos(f*x+e)*sin(f*x+e)*ln(2/(1+cos(f*x+e)))-144*A*cos(f*
x+e)*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+240*B*cos(f*x+e)*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(
f*x+e))/sin(f*x+e))-144*A*ln(2/(1+cos(f*x+e)))+240*B*ln(2/(1+cos(f*x+e)))-120*B*ln(2/(1+cos(f*x+e)))*cos(f*x+e
)^2+75*A*sin(f*x+e)*cos(f*x+e)-144*A*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+240*B*ln(-(-1+cos
(f*x+e)-sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+2*B*sin(f*x+e)*cos(f*x+e)^4)*(-c*(sin(f*x+e)-1))^(7/2)/(cos(f*x+e
)^4-cos(f*x+e)^3*sin(f*x+e)+3*cos(f*x+e)^3+4*cos(f*x+e)^2*sin(f*x+e)-8*cos(f*x+e)^2+4*cos(f*x+e)*sin(f*x+e)-4*
cos(f*x+e)-8*sin(f*x+e)+8)/(a*(1+sin(f*x+e)))^(3/2)

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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-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(-c*sin(f*x + e) + c)^(7/2)/(a*sin(f*x + e) + a)^(3/2), x)

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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))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral((B*c^3*cos(f*x + e)^4 + (3*A - 5*B)*c^3*cos(f*x + e)^2 - 4*(A - B)*c^3 - ((A - 3*B)*c^3*cos(f*x + e)^
2 - 4*(A - B)*c^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^2*cos(f*x + e)^2 - 2*a^
2*sin(f*x + e) - 2*a^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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))**(7/2)/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 0.60, size = 413, normalized size = 1.52 \begin {gather*} \frac {\sqrt {2} \sqrt {c} {\left (\frac {6 \, {\left (3 \, \sqrt {2} A \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, \sqrt {2} B \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \log \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 {6 \, {\left (\sqrt {2} A \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \sqrt {2} B \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\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 {\sqrt {2} {\left (4 \, B a^{\frac {9}{2}} c^{3} \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 )^{6} - 3 \, A a^{\frac {9}{2}} c^{3} \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 )^{4} + 9 \, B a^{\frac {9}{2}} c^{3} \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 )^{4} - 12 \, A a^{\frac {9}{2}} c^{3} \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} + 24 \, B a^{\frac {9}{2}} c^{3} \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}\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{3 \, 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))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

1/3*sqrt(2)*sqrt(c)*(6*(3*sqrt(2)*A*sqrt(a)*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*sqrt(2)*B*sqrt(a)*c^3*
sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*log(-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))) - 6*(sqrt(2)*A*sqrt(a)*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - sqrt(2)*B*sqrt(a)*c^3*sgn(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))) - s
qrt(2)*(4*B*a^(9/2)*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^6 - 3*A*a^(9/2)*c^3
*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4 + 9*B*a^(9/2)*c^3*sgn(sin(-1/4*pi + 1/2*
f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4 - 12*A*a^(9/2)*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4
*pi + 1/2*f*x + 1/2*e)^2 + 24*B*a^(9/2)*c^3*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^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-c\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\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 - c*sin(e + f*x))^(7/2))/(a + a*sin(e + f*x))^(3/2),x)

[Out]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(7/2))/(a + a*sin(e + f*x))^(3/2), x)

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