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

Optimal. Leaf size=145 \[ -\frac{2 a^2 (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a (A+B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{f \sqrt{c-c \sin (e+f x)}}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.381569, antiderivative size = 145, 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 = {2973, 2740, 2737, 2667, 31} \[ -\frac{2 a^2 (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a (A+B) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{f \sqrt{c-c \sin (e+f x)}}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2973

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*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)] && NeQ[m + n + 1, 0]

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(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] && E
qQ[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 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+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{\sqrt{c-c \sin (e+f x)}} \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}+(A+B) \int \frac{(a+a \sin (e+f x))^{3/2}}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=-\frac{a (A+B) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{f \sqrt{c-c \sin (e+f x)}}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}+(2 a (A+B)) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=-\frac{a (A+B) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{f \sqrt{c-c \sin (e+f x)}}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}+\frac{\left (2 a^2 (A+B) c \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 (A+B) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{f \sqrt{c-c \sin (e+f x)}}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}-\frac{\left (2 a^2 (A+B) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{2 a^2 (A+B) \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 (A+B) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{f \sqrt{c-c \sin (e+f x)}}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.709697, size = 136, normalized size = 0.94 \[ -\frac{(a (\sin (e+f x)+1))^{3/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (4 (A+2 B) \sin (e+f x)+16 (A+B) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-B \cos (2 (e+f x))\right )}{4 f \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2)*(-(B*Cos[2*(e + f*x)]) + 16*(A + B)*Log[C
os[(e + f*x)/2] - Sin[(e + f*x)/2]] + 4*(A + 2*B)*Sin[e + f*x]))/(4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*
Sqrt[c - c*Sin[e + f*x]])

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Maple [B]  time = 0.342, size = 495, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

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))^(1/2),x)

[Out]

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

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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 (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}{\sqrt{-c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

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))^(1/2),x, algorithm="maxima")

[Out]

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

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

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))^(1/2),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*sin(f*x + e) - c), 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+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(1/2),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 (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}{\sqrt{-c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

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))^(1/2),x, algorithm="giac")

[Out]

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