3.2.44 \(\int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\) [144]

Optimal. Leaf size=158 \[ \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 (A+3 B) \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {a (A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {a^2 (A+3 B) \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a (A+3 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 c f \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.60, size = 210, normalized size = 1.33 \begin {gather*} -\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (4 A+3 B+B \cos (2 (e+f x))+4 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+12 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-2 \left (-B+2 (A+3 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)\right )}{2 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(4*A + 3*B + B*Cos[2*(e + f*x)] + 4*A
*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 12*B*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - 2*(-B + 2*(A + 3*B
)*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x]))/(c*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Si
n[e + f*x])*Sqrt[c - c*Sin[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(747\) vs. \(2(142)=284\).
time = 0.35, size = 748, normalized size = 4.73

method	result	size
default	\(\frac {\left (-2 A -4 B +6 B \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-3 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 )+A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-2 A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+4 B \sin \left (f x +e \right )+2 A \sin \left (f x +e \right )+B \cos \left (f x +e \right )-B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+A \cos \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 )-6 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 )+3 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+6 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-3 B \sin \left (f x +e \right ) \cos \left (f x +e \right )-6 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 )-12 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 )+2 A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-2 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 )-4 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 )+3 B \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+2 A \left (\cos ^{2}\left (f x +e \right )\right )+2 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 )-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 )-2 A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+4 A \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 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+12 B \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+4 B \left (\cos ^{2}\left (f x +e \right )\right )-2 A \sin \left (f x +e \right ) \cos \left (f x +e \right )\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}{f \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )+\cos ^{2}\left (f x +e \right )-2 \sin \left (f x +e \right )+\cos \left (f x +e \right )-2\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}}\)	\(748\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-B*cos(f*x+e)^2*sin(f*x+e)-2*A-4*B+6*B*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*sin(f*x+e)*cos(f*x+e)-3
*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)*cos(f*x+e)+4*B*sin(f*x+e)+2*A*sin(f*x+e)-B*cos(f*x+e)^3+B*cos(f*x+e)+2*A*co
s(f*x+e)^2+4*B*cos(f*x+e)^2+A*cos(f*x+e)^2*ln(2/(1+cos(f*x+e)))+A*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-6*B*ln(-(-1+
cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+6*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)-3*B*sin(f*x+e)*cos(f*x+e)-
6*B*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-12*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f
*x+e))+2*A*ln(2/(1+cos(f*x+e)))*sin(f*x+e)-2*A*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-4*A*sin(f
*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+3*B*cos(f*x+e)*ln(2/(1+cos(f*x+e)))+2*A*cos(f*x+e)*sin(f*x+e)
*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-A*cos(f*x+e)*sin(f*x+e)*ln(2/(1+cos(f*x+e)))-2*A*ln(2/(1+cos(f*x+e
)))+4*A*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-6*B*ln(2/(1+cos(f*x+e)))+12*B*ln(-(-1+cos(f*x+e)+sin(f*x+e)
)/sin(f*x+e))-2*A*cos(f*x+e)^2*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+3*B*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^
2-2*A*sin(f*x+e)*cos(f*x+e))*(a*(1+sin(f*x+e)))^(3/2)/(cos(f*x+e)*sin(f*x+e)+cos(f*x+e)^2-2*sin(f*x+e)+cos(f*x
+e)-2)/(-c*(sin(f*x+e)-1))^(3/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (152) = 304\).
time = 0.53, size = 394, normalized size = 2.49 \begin {gather*} -\frac {B {\left (\frac {6 \, a^{\frac {3}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {3}{2}}} - \frac {3 \, a^{\frac {3}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {3}{2}}} + \frac {2 \, {\left (\frac {3 \, a^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, a^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{c^{\frac {3}{2}} - \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}\right )} + A {\left (\frac {2 \, a^{\frac {3}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {3}{2}}} - \frac {a^{\frac {3}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {3}{2}}} + \frac {4 \, a^{\frac {3}{2}} \sqrt {c} \sin \left (f x + e\right )}{{\left (c^{2} - \frac {2 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*(6*a^(3/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(3/2) - 3*a^(3/2)*log(sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + 1)/c^(3/2) + 2*(3*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1
)^2 + 3*a^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(c^(3/2) - 2*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 2*
c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + c^(3/2)*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4)) + A*(2*a^(3/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(3/2) - a^(3/2)*log(si
n(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/c^(3/2) + 4*a^(3/2)*sqrt(c)*sin(f*x + e)/((c^2 - 2*c^2*sin(f*x + e)/(co
s(f*x + e) + 1) + c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))))/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+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]
time = 0.52, size = 236, normalized size = 1.49 \begin {gather*} -\frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} B a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{c^{\frac {3}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \log \left (-8 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 8\right )}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (A a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \sqrt {a}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*(2*sqrt(2)*B*a*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))/(c^(3/2)*sgn(
sin(-1/4*pi + 1/2*f*x + 1/2*e))) + sqrt(2)*(A*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*a*sqrt(c)*sg
n(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*log(-8*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 8)/(c^2*sgn(sin(-1/4*pi + 1/2*f*x
+ 1/2*e))) - sqrt(2)*(A*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + B*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e)))/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))*sqrt(a)/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c-c\,\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))*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x))^(3/2),x)

[Out]

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

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