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

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

[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))

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Rubi [A]  time = 0.39236, antiderivative size = 149, normalized size of antiderivative = 1., 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 = {2972, 2739, 2737, 2667, 31} \[ -\frac{B c^2 \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac{B c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{a f (a \sin (e+f x)+a)^{3/2}} \]

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 2972

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]

Rule 2739

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 2737

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 2667

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 ) \operatorname{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*}

Mathematica [A]  time = 0.975258, size = 179, normalized size = 1.2 \[ \frac{c \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin (e+f x) \left (A-4 B \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-3 B\right )+B \cos (2 (e+f x)) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-B \left (3 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+2\right )\right )}{f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

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]  time = 0.28, size = 604, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

-1/f*(-A-3*B-A*sin(f*x+e)+8*B*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+A*cos(f*x+e)^2+3*B*cos(f*x+e)^2*ln(2/(c
os(f*x+e)+1))+2*B*cos(f*x+e)^2*sin(f*x+e)-2*B*cos(f*x+e)^3+2*B*cos(f*x+e)+2*B*ln(2/(cos(f*x+e)+1))*sin(f*x+e)*
cos(f*x+e)+B*sin(f*x+e)*cos(f*x+e)+2*B*cos(f*x+e)*ln(2/(cos(f*x+e)+1))-4*B*sin(f*x+e)*ln(2/(cos(f*x+e)+1))+A*s
in(f*x+e)*cos(f*x+e)+2*B*cos(f*x+e)^3*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))-6*B*ln((1-cos(f*x+e)+sin(f*x+e)
)/sin(f*x+e))*cos(f*x+e)^2-2*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)
*ln((1-cos(f*x+e)+sin(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))-4*B*cos(f*x+
e)*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*ln(2/(cos(f*x+e)+1))+B*cos(f*x+e)^
2*sin(f*x+e)*ln(2/(cos(f*x+e)+1))-B*cos(f*x+e)^3*ln(2/(cos(f*x+e)+1))-3*B*sin(f*x+e))*(-c*(-1+sin(f*x+e)))^(3/
2)/(cos(f*x+e)^2-sin(f*x+e)*cos(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (B c \cos \left (f x + e\right )^{2} -{\left (A - B\right )} c \sin \left (f x + e\right ) +{\left (A - B\right )} c\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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]

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