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

Optimal. Leaf size=174 \[ \frac{(A+B+C) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{(A-B-3 C) \cos (e+f x) \log (1-\sin (e+f x))}{4 c f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(A-B+C) \cos (e+f x) \log (\sin (e+f x)+1)}{4 c f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.686202, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 50, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3035, 2969, 2737, 2667, 31} \[ \frac{(A+B+C) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{(A-B-3 C) \cos (e+f x) \log (1-\sin (e+f x))}{4 c f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{(A-B+C) \cos (e+f x) \log (\sin (e+f x)+1)}{4 c f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3035

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + a*C)*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(2*b*c*f*(2*m + 1)), x] - Dist[1/(2*b*c*d*(2*m + 1)), Int[
(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(c^2*(m + 1) + d^2*(2*m + n + 2)) - B*c*d*(m - n -
1) - C*(c^2*m - d^2*(n + 1)) + d*((A*c + B*d)*(m + n + 2) - c*C*(3*m - n))*Sin[e + f*x], x], x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (LtQ[m, -2^(-1)] || (EqQ[m +
n + 2, 0] && NeQ[2*m + 1, 0]))

Rule 2969

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[
(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b + a*B)/(2*a*b), Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e +
f*x]], x], x] + Dist[(B*c + A*d)/(2*c*d), Int[Sqrt[c + d*Sin[e + f*x]]/Sqrt[a + b*Sin[e + f*x]], x], x] /; Fre
eQ[{a, b, c, d, e, f, A, B}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 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 \sin ^2(e+f x)}{\sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx &=\frac{(A+B+C) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{\int \frac{-2 a^2 (A-B-C)+4 a^2 C \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}} \, dx}{4 a^2 c}\\ &=\frac{(A+B+C) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac{(A-B-3 C) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c-c \sin (e+f x)}} \, dx}{4 a c}+\frac{(A-B+C) \int \frac{\sqrt{c-c \sin (e+f x)}}{\sqrt{a+a \sin (e+f x)}} \, dx}{4 c^2}\\ &=\frac{(A+B+C) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac{((A-B-3 C) \cos (e+f x)) \int \frac{\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{4 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}+\frac{(a (A-B+C) \cos (e+f x)) \int \frac{\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{4 c \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B+C) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{((A-B-3 C) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{4 c f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}+\frac{((A-B+C) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,a \sin (e+f x)\right )}{4 c f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B+C) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac{(A-B-3 C) \cos (e+f x) \log (1-\sin (e+f x))}{4 c f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}+\frac{(A-B+C) \cos (e+f x) \log (1+\sin (e+f x))}{4 c f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.696643, size = 196, normalized size = 1.13 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((-A+B+3 C) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(A-B+C) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+A+B+C\right )}{2 f \sqrt{a (\sin (e+f x)+1)} (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.404, size = 479, normalized size = 2.8 \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*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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