3.1.16 \(\int \frac {A+B \sin (e+f x)+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx\) [16]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.41, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3114, 3048, 2816, 2746, 31} \begin {gather*} \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)}} \end {gather*}

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 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 3048

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 3114

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]))

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)) \text {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)) \text {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 = 0.51, size = 196, normalized size = 1.13 \begin {gather*} \frac {\left (A+B+C+(-A+B+3 C) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+(A-B+C) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 f \sqrt {a (1+\sin (e+f x))} (c-c \sin (e+f x))^{3/2}} \end {gather*}

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))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(475\) vs. \(2(156)=312\).
time = 25.38, size = 476, normalized size = 2.74

method	result	size
default	\(-\frac {\left (A \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sin \left (f x +e \right )-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 )-B \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sin \left (f x +e \right )+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 )+C \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sin \left (f x +e \right )+3 C \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sin \left (f x +e \right )-2 C \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-A \sin \left (f x +e \right )-A \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+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 \sin \left (f x +e \right )+B \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 {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\sin \left (f x +e \right ) C -C \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-3 C \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 C \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )\right ) \cos \left (f x +e \right )}{2 f \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}}}\)	\(476\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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*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)

________________________________________________________________________________________

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*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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

Giac [A]
time = 0.56, size = 252, normalized size = 1.45 \begin {gather*} -\frac {\frac {{\left (A \sqrt {a} \sqrt {c} - B \sqrt {a} \sqrt {c} + C \sqrt {a} \sqrt {c}\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a c^{2} \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 )} - \frac {2 \, {\left (A \sqrt {a} \sqrt {c} - B \sqrt {a} \sqrt {c} - 3 \, C \sqrt {a} \sqrt {c}\right )} \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{a c^{2} \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 )} + \frac {A \sqrt {a} \sqrt {c} + B \sqrt {a} \sqrt {c} + C \sqrt {a} \sqrt {c}}{a c^{2} \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}}}{4 \, f} \end {gather*}

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]

-1/4*((A*sqrt(a)*sqrt(c) - B*sqrt(a)*sqrt(c) + C*sqrt(a)*sqrt(c))*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(
a*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 2*(A*sqrt(a)*sqrt(c) - B*sqrt
(a)*sqrt(c) - 3*C*sqrt(a)*sqrt(c))*log(abs(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/(a*c^2*sgn(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)*sqrt(c) + B*sqrt(a)*sqrt(c) + C*sqrt(a)*sqrt(c))/(a
*c^2*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)
)/f

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________