3.1.85 \(\int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\) [85]

Optimal. Leaf size=122 \[ \frac {2 \sqrt {2} a (A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c f} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3046, 2937, 2830, 2728, 212} \begin {gather*} -\frac {2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {2 \sqrt {2} a (A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}+\frac {2 a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*a*(A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a*(3
*A + 5*B)*Cos[e + f*x])/(3*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a*B*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(3*c*f)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2830

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

Rule 2937

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[d*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 2)/(b^2*f*(m + 3))), x] - Dist[1/(b^2*(m + 3)
), Int[(a + b*Sin[e + f*x])^(m + 1)*(b*d*(m + 2) - a*c*(m + 3) + (b*c*(m + 3) - a*d*(m + 4))*Sin[e + f*x]), x]
, x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GeQ[m, -3/2] && LtQ[m, 0]

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx &=(a c) \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac {2 a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c f}-\frac {(2 a) \int \frac {-\frac {3 A c}{2}-\frac {B c}{2}+\left (-\frac {3 A c}{2}-\frac {5 B c}{2}\right ) \sin (e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{3 c}\\ &=-\frac {2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c f}+(2 a (A+B)) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c f}-\frac {(4 a (A+B)) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{f}\\ &=\frac {2 \sqrt {2} a (A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 c f}\\ \end {align*}

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Mathematica [A]
time = 0.88, size = 166, normalized size = 1.36 \begin {gather*} -\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 \sqrt {2} (A+B) \tan ^{-1}\left (\frac {\sqrt {-c (1+\sin (e+f x))}}{\sqrt {2} \sqrt {c}}\right ) \sqrt {-c (1+\sin (e+f x))}+\sqrt {c} (6 A+9 B-B \cos (2 (e+f x))+2 (3 A+5 B) \sin (e+f x))\right )}{3 \sqrt {c} f \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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(6*Sqrt[2]*(A + B)*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt[2]
*Sqrt[c])]*Sqrt[-(c*(1 + Sin[e + f*x]))] + Sqrt[c]*(6*A + 9*B - B*Cos[2*(e + f*x)] + 2*(3*A + 5*B)*Sin[e + f*x
])))/(Sqrt[c]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]
time = 8.15, size = 159, normalized size = 1.30

method	result	size
default	\(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, a \left (3 c^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) A +3 c^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) B -B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}-3 A c \sqrt {c \left (1+\sin \left (f x +e \right )\right )}-3 B c \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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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+a*sin(f*x+e))*(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)/sqrt(-c*sin(f*x + e) + c), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (111) = 222\).
time = 0.38, size = 274, normalized size = 2.25 \begin {gather*} \frac {\frac {3 \, \sqrt {2} {\left ({\left (A + B\right )} a c \cos \left (f x + e\right ) - {\left (A + B\right )} a c \sin \left (f x + e\right ) + {\left (A + B\right )} a c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} + 2 \, {\left (B a \cos \left (f x + e\right )^{2} - {\left (3 \, A + 4 \, B\right )} a \cos \left (f x + e\right ) - {\left (3 \, A + 5 \, B\right )} a - {\left (B a \cos \left (f x + e\right ) + {\left (3 \, A + 5 \, B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*sqrt(2)*((A + B)*a*c*cos(f*x + e) - (A + B)*a*c*sin(f*x + e) + (A + B)*a*c)*log(-(cos(f*x + e)^2 + (cos
(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) +
3*cos(f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(c) + 2*(B*a*co
s(f*x + e)^2 - (3*A + 4*B)*a*cos(f*x + e) - (3*A + 5*B)*a - (B*a*cos(f*x + e) + (3*A + 5*B)*a)*sin(f*x + e))*s
qrt(-c*sin(f*x + e) + c))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int \frac {A}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {A \sin {\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {B \sin {\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {B \sin ^{2}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (111) = 222\).
time = 0.47, size = 314, normalized size = 2.57 \begin {gather*} \frac {\frac {3 \, \sqrt {2} {\left (A a \sqrt {c} + B a \sqrt {c}\right )} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {4 \, \sqrt {2} {\left (3 \, A a \sqrt {c} + 5 \, B a \sqrt {c} - \frac {6 \, A a \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {6 \, B a \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {3 \, A a \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {9 \, B a \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}}{c {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*sqrt(2)*(A*a*sqrt(c) + B*a*sqrt(c))*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x +
1/2*e) + 1))/(c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 4*sqrt(2)*(3*A*a*sqrt(c) + 5*B*a*sqrt(c) - 6*A*a*sqrt(c
)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 6*B*a*sqrt(c)*(cos(-1/4*pi + 1/2
*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 3*A*a*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2
/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 9*B*a*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi +
1/2*f*x + 1/2*e) + 1)^2)/(c*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)^3*
sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

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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 (a+a\,\sin \left (e+f\,x\right )\right )}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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