3.2.75 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx\) [175]

Optimal. Leaf size=146 \[ \frac {2 (A-B) c^2 \cos (e+f x) \log (1+\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {(A-B) c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 146, 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 = {3052, 2819, 2816, 2746, 31} \begin {gather*} \frac {2 c^2 (A-B) \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c (A-B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A - B)*c^2*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + ((A
- B)*c*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]) - (B*Cos[e + f*x]*(c - c*Sin[e + f
*x])^(3/2))/(2*f*Sqrt[a + a*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 3052

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Si
n[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)] && NeQ[m + n + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.47, size = 146, normalized size = 1.00 \begin {gather*} -\frac {c \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)} \left (B \cos (2 (e+f x))-4 \left (4 (-A+B) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+(A-2 B) \sin (e+f x)\right )\right )}{4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {a (1+\sin (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(506\) vs. \(2(132)=264\).
time = 0.33, size = 507, normalized size = 3.47

method	result	size
default	\(\frac {\left (-B \left (\cos ^{3}\left (f x +e \right )\right )+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 )+2 A \sin \left (f x +e \right ) \cos \left (f x +e \right )+4 A \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-8 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 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )+8 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 \left (\cos ^{2}\left (f x +e \right )\right )-4 B \sin \left (f x +e \right ) \cos \left (f x +e \right )-4 B \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+8 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 )+4 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-8 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 \sin \left (f x +e \right )-4 A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+8 A \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+B \cos \left (f x +e \right )+3 B \sin \left (f x +e \right )+4 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-8 B \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 +3 B \right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}}{2 f \left (\cos ^{2}\left (f x +e \right )-\cos \left (f x +e \right ) \sin \left (f x +e \right )+\cos \left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}\)	\(507\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2/f*(-B*cos(f*x+e)^3+B*cos(f*x+e)^2*sin(f*x+e)+2*A*cos(f*x+e)^2+2*A*sin(f*x+e)*cos(f*x+e)+4*A*cos(f*x+e)*ln(
2/(1+cos(f*x+e)))-8*A*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-4*A*ln(2/(1+cos(f*x+e)))*sin(f*x+e
)+8*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)^2-4*B*sin(f*x+e)*cos(f*x+e)-4*B*cos
(f*x+e)*ln(2/(1+cos(f*x+e)))+8*B*cos(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+4*B*ln(2/(1+cos(f*x+e))
)*sin(f*x+e)-8*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-2*A*sin(f*x+e)-4*A*ln(2/(1+cos(f*x+e)))
+8*A*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+B*cos(f*x+e)+3*B*sin(f*x+e)+4*B*ln(2/(1+cos(f*x+e)))-8*B*ln(-(
-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-2*A+3*B)*(-c*(sin(f*x+e)-1))^(3/2)/(cos(f*x+e)^2-cos(f*x+e)*sin(f*x+e)+c
os(f*x+e)+2*sin(f*x+e)-2)/(a*(1+sin(f*x+e)))^(1/2)

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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))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(-c*sin(f*x + e) + c)^(3/2)/sqrt(a*sin(f*x + e) + a), 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))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [A]
time = 0.55, size = 260, normalized size = 1.78 \begin {gather*} -\frac {\sqrt {2} \sqrt {c} {\left (\frac {\sqrt {2} {\left (A \sqrt {a} c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B \sqrt {a} c \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 \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} B a^{\frac {3}{2}} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \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} - \sqrt {2} A a^{\frac {3}{2}} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \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} + \sqrt {2} B a^{\frac {3}{2}} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \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^{2}}\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))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-sqrt(2)*sqrt(c)*(sqrt(2)*(A*sqrt(a)*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B*sqrt(a)*c*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*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - (sqrt(2)*
B*a^(3/2)*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/
2*e)^4 - sqrt(2)*A*a^(3/2)*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*
pi + 1/2*f*x + 1/2*e)^2 + sqrt(2)*B*a^(3/2)*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*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^2)/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 (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,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))^(3/2))/(a + a*sin(e + f*x))^(1/2),x)

[Out]

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

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