3.2.90 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [190]

Optimal. Leaf size=149 \[ -\frac {B c^2 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {B c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 f (a+a \sin (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 149, 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, 2818, 2816, 2746, 31} \begin {gather*} -\frac {B c^2 \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 {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac {B c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a 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])^(3/2))/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

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

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Mathematica [A]
time = 0.66, size = 179, normalized size = 1.20 \begin {gather*} \frac {c \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 (B \cos (2 (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-B \left (2+3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+\left (A-3 B-4 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 )}{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)))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2))/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(B*Cos[2*(e + f*x)]*Log[Cos[(e + f*x)/2] + S
in[(e + f*x)/2]] - B*(2 + 3*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + (A - 3*B - 4*B*Log[Cos[(e + f*x)/2] +
Sin[(e + f*x)/2]])*Sin[e + f*x]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(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. \(610\) vs. \(2(135)=270\).
time = 0.31, size = 611, normalized size = 4.10

method	result	size
default	\(-\frac {\left (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 )+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 )-A -3 B +2 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 )-3 B \sin \left (f x +e \right )-A \sin \left (f x +e \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 )+2 B \cos \left (f x +e \right )-4 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 )+B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+2 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+2 B \left (\cos ^{3}\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 )-2 B \left (\cos ^{3}\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 )-4 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )+B \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 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 ) \sin \left (f x +e \right )+2 B \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+A \left (\cos ^{2}\left (f x +e \right )\right )-4 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 )-4 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+3 B \left (\cos ^{2}\left (f x +e \right )\right )-B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \left (\cos ^{3}\left (f x +e \right )\right )+A \sin \left (f x +e \right ) \cos \left (f x +e \right )\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{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 ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}\)	\(611\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/f*(8*B*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))+8*B*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))
-B*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^3+2*B*cos(f*x+e)^2*sin(f*x+e)-A-3*B+2*B*ln(2/(1+cos(f*x+e)))*sin(f*x+e)*cos
(f*x+e)-3*B*sin(f*x+e)-A*sin(f*x+e)-2*B*cos(f*x+e)^3+2*B*cos(f*x+e)+B*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2*sin(f*
x+e)+A*cos(f*x+e)^2+3*B*cos(f*x+e)^2-4*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)+B*sin(f*x+e)*cos(f*x+e)+2*B*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-2*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*B*cos(f*x+e)*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))
+2*B*cos(f*x+e)^3*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-4*B*ln(2/(1+cos(f*x+e)))+3*B*ln(2/(1+cos(f*x+e)))
*cos(f*x+e)^2+A*sin(f*x+e)*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)*(-c*(sin(f*
x+e)-1))^(3/2)/(cos(f*x+e)^2-cos(f*x+e)*sin(f*x+e)+cos(f*x+e)+2*sin(f*x+e)-2)/(a*(1+sin(f*x+e)))^(5/2)

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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))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(-c*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) + a)^(5/2), 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))^(3/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(-(B*c*cos(f*x + e)^2 - (A - B)*c*sin(f*x + e) + (A - B)*c)*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]
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 )}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, 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))**(5/2),x)

[Out]

Integral((-c*(sin(e + f*x) - 1))**(3/2)*(A + B*sin(e + f*x))/(a*(sin(e + f*x) + 1))**(5/2), x)

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Giac [A]
time = 0.52, size = 181, normalized size = 1.21 \begin {gather*} \frac {{\left (8 \, B \sqrt {a} c \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {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 ) - 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 ) - 3 \, 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 )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}\right )} \sqrt {c}}{4 \, a^{3} 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))^(3/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/4*(8*B*sqrt(a)*c*log(abs(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + (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)) - 2*(A*sqrt(a)*c*sgn(si
n(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*B*sqrt(a)*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-1/4*pi + 1/2*f*x + 1/2
*e)^2)/cos(-1/4*pi + 1/2*f*x + 1/2*e)^4)*sqrt(c)/(a^3*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 = -0.01 \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}}{{\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))^(3/2))/(a + a*sin(e + f*x))^(5/2),x)

[Out]

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

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