3.2.81 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\) [181]

Optimal. Leaf size=210 \[ -\frac {4 (A-2 B) c^3 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 (A-2 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-2 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{3/2}} \]

[Out]

-1/2*(A-B)*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(3/2)-1/2*(A-2*B)*c*cos(f*x+e)*(c-c*sin(f*x+e)
)^(3/2)/a/f/(a+a*sin(f*x+e))^(1/2)-4*(A-2*B)*c^3*cos(f*x+e)*ln(1+sin(f*x+e))/a/f/(a+a*sin(f*x+e))^(1/2)/(c-c*s
in(f*x+e))^(1/2)-2*(A-2*B)*c^2*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 = 0.33, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 6, 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^3 (A-2 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^2 (A-2 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a \sin (e+f x)+a}}-\frac {c (A-2 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a \sin (e+f x)+a}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{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])^(5/2))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(-4*(A - 2*B)*c^3*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*(A - 2*B)*c^2*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]) - ((A - 2*B)*c*Cos[e
+ f*x]*(c - c*Sin[e + f*x])^(3/2))/(2*a*f*Sqrt[a + a*Sin[e + f*x]]) - ((A - B)*Cos[e + f*x]*(c - c*Sin[e + f*x
])^(5/2))/(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))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {(A-2 B) \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac {(A-2 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {(2 (A-2 B) c) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac {2 (A-2 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-2 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (4 (A-2 B) c^2\right ) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a}\\ &=-\frac {2 (A-2 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-2 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (4 (A-2 B) c^3 \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 (A-2 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-2 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (4 (A-2 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 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {4 (A-2 B) c^3 \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 (A-2 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-2 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 1.07, size = 212, normalized size = 1.01 \begin {gather*} -\frac {c^2 \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 (28 A-16 B+2 (2 A-7 B) \cos (2 (e+f x))+64 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-128 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (-8 A+31 B+64 (A-2 B) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)+B \sin (3 (e+f x))\right )}{8 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])^(5/2))/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-1/8*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(28*A - 16*B + 2*(2*A - 7*B)*Cos[2*(e
+ f*x)] + 64*A*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 128*B*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (-
8*A + 31*B + 64*(A - 2*B)*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sin[e + f*x] + B*Sin[3*(e + f*x)]))/(f*(Co
s[(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. \(864\) vs. \(2(190)=380\).
time = 0.32, size = 865, normalized size = 4.12

method	result	size
default	\(\frac {\left (32 A \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-64 B \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-64 B \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+32 A \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-12 A +22 B -16 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+8 A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+2 A \cos \left (f x +e \right )+22 B \sin \left (f x +e \right )-12 A \sin \left (f x +e \right )-16 A \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+32 B \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-7 B \cos \left (f x +e \right )-16 A \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+32 B \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-6 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+2 A \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+8 A \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+7 B \left (\cos ^{3}\left (f x +e \right )\right )-16 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+32 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-15 B \sin \left (f x +e \right ) \cos \left (f x +e \right )-16 A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-16 B \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+12 A \left (\cos ^{2}\left (f x +e \right )\right )+8 A \cos \left (f x +e \right ) \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-B \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-16 A \cos \left (f x +e \right ) \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+32 B \cos \left (f x +e \right ) \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-16 A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+32 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-21 B \left (\cos ^{2}\left (f x +e \right )\right )-B \left (\cos ^{4}\left (f x +e \right )\right )+10 A \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 A \left (\cos ^{3}\left (f x +e \right )\right )\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}}}{2 f \left (\cos ^{3}\left (f x +e \right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right ) \sin \left (f x +e \right )-2 \cos \left (f x +e \right )-4 \sin \left (f x +e \right )+4\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}\)	\(865\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2/f*(32*A*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-64*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e
))/sin(f*x+e))-64*B*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+32*A*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))
+2*A*cos(f*x+e)^2*sin(f*x+e)-6*B*cos(f*x+e)^2*sin(f*x+e)-12*A+22*B-16*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)*cos(f*
x+e)+2*A*cos(f*x+e)+22*B*sin(f*x+e)-12*A*sin(f*x+e)-2*A*cos(f*x+e)^3+7*B*cos(f*x+e)^3-7*B*cos(f*x+e)-B*cos(f*x
+e)^3*sin(f*x+e)-B*cos(f*x+e)^4+12*A*cos(f*x+e)^2-21*B*cos(f*x+e)^2-16*A*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x
+e))/sin(f*x+e))+32*B*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+8*A*cos(f*x+e)^2*ln(2/(1+cos(f*x+e
)))+8*A*cos(f*x+e)*ln(2/(1+cos(f*x+e)))+32*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)-15*B*sin(f*x+e)*cos(f*x+e)-16*A*l
n(2/(1+cos(f*x+e)))*sin(f*x+e)-16*B*cos(f*x+e)*ln(2/(1+cos(f*x+e)))+8*A*cos(f*x+e)*sin(f*x+e)*ln(2/(1+cos(f*x+
e)))-16*A*cos(f*x+e)*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+32*B*cos(f*x+e)*sin(f*x+e)*ln(-(-1+
cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-16*A*ln(2/(1+cos(f*x+e)))+32*B*ln(2/(1+cos(f*x+e)))-16*B*ln(2/(1+cos(f*x+e)
))*cos(f*x+e)^2+10*A*sin(f*x+e)*cos(f*x+e)-16*A*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+32*B*l
n(-(-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*s
in(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)))^(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))^(5/2)/(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)^(5/2)/(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))^(5/2)/(a+a*sin(f*x+e))^(3/2),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)/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), 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))**(3/2),x)

[Out]

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

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Giac [A]
time = 0.50, size = 266, normalized size = 1.27 \begin {gather*} -\frac {2 \, {\left (B \sqrt {a} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + A \sqrt {a} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{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} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{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 )} \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) - \frac {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 ) - 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 )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}\right )} \sqrt {c}}{a^{2} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \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))^(3/2),x, algorithm="giac")

[Out]

-2*(B*sqrt(a)*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + A*sqrt(a)*c^2*cos(-1/
4*pi + 1/2*f*x + 1/2*e)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*B*sqrt(a)*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)
^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)))*log(abs(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - (A*sqrt(a)*c^2*sgn(sin(-1/4*
pi + 1/2*f*x + 1/2*e)) - B*sqrt(a)*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/cos(-1/4*pi + 1/2*f*x + 1/2*e)^2)*
sqrt(c)/(a^2*f*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))

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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 )}^{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))^(5/2))/(a + a*sin(e + f*x))^(3/2),x)

[Out]

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

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