3.544 \(\int \frac{(c+d \sin (e+f x))^2}{\sqrt{a+a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=123 \[ -\frac{4 d (3 c-d) \cos (e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}-\frac{\sqrt{2} (c-d)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{2 d^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 a f} \]

[Out]

-((Sqrt[2]*(c - d)^2*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*f)) - (4*(3*
c - d)*d*Cos[e + f*x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) - (2*d^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*a*f)

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Rubi [A]  time = 0.200624, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2761, 2751, 2649, 206} \[ -\frac{4 d (3 c-d) \cos (e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}-\frac{\sqrt{2} (c-d)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{2 d^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 a f} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])^2/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

-((Sqrt[2]*(c - d)^2*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*f)) - (4*(3*
c - d)*d*Cos[e + f*x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) - (2*d^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*a*f)

Rule 2761

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[(
d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x]
)^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], 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, -1]

Rule 2751

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*Sin
[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 2649

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 206

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

Rubi steps

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

Mathematica [C]  time = 0.384894, size = 125, normalized size = 1.02 \[ -\frac{2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (d \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (6 c+d \sin (e+f x)-d)-(3+3 i) (-1)^{3/4} (c-d)^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{3 f \sqrt{a (\sin (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Sin[e + f*x])^2/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

(-2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((-3 - 3*I)*(-1)^(3/4)*(c - d)^2*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1
+ Tan[(e + f*x)/4])] + d*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(6*c - d + d*Sin[e + f*x])))/(3*f*Sqrt[a*(1 + S
in[e + f*x])])

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Maple [A]  time = 0.839, size = 185, normalized size = 1.5 \begin{align*} -{\frac{1+\sin \left ( fx+e \right ) }{3\,{a}^{2}\cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( 3\,{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){c}^{2}-6\,{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) cd+3\,{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){d}^{2}-2\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}{d}^{2}+12\,acd\sqrt{a-a\sin \left ( fx+e \right ) } \right ){\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(3*a^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a
^(1/2))*c^2-6*a^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c*d+3*a^(3/2)*2^(1/2)*arctan
h(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*d^2-2*(a-a*sin(f*x+e))^(3/2)*d^2+12*a*c*d*(a-a*sin(f*x+e))^(1/2)
)/a^2/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^2/sqrt(a*sin(f*x + e) + a), x)

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Fricas [B]  time = 1.69025, size = 741, normalized size = 6.02 \begin{align*} \frac{\frac{3 \, \sqrt{2}{\left (a c^{2} - 2 \, a c d + a d^{2} +{\left (a c^{2} - 2 \, a c d + a d^{2}\right )} \cos \left (f x + e\right ) +{\left (a c^{2} - 2 \, a c d + a d^{2}\right )} \sin \left (f x + e\right )\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{a \sin \left (f x + e\right ) + a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt{a}} + 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{a}} - 4 \,{\left (d^{2} \cos \left (f x + e\right )^{2} + 6 \, c d - 2 \, d^{2} +{\left (6 \, c d - d^{2}\right )} \cos \left (f x + e\right ) +{\left (d^{2} \cos \left (f x + e\right ) - 6 \, c d + 2 \, d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{6 \,{\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*sqrt(2)*(a*c^2 - 2*a*c*d + a*d^2 + (a*c^2 - 2*a*c*d + a*d^2)*cos(f*x + e) + (a*c^2 - 2*a*c*d + a*d^2)*s
in(f*x + e))*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(
f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 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(a) - 4*(d^2*cos(f*x + e)^2 + 6*c*d - 2*d^2 + (6*c*d - d^2)*cos(f*x + e) + (d^2*cos(f*
x + e) - 6*c*d + 2*d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a*f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral((c + d*sin(e + f*x))**2/sqrt(a*(sin(e + f*x) + 1)), x)

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Giac [B]  time = 2.33619, size = 583, normalized size = 4.74 \begin{align*} \frac{\frac{6 \, \sqrt{2}{\left (c^{2} - 2 \, c d + d^{2}\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a} + \sqrt{a}\right )}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )} + \frac{{\left ({\left (\frac{{\left (6 \, a c d \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right ) - a d^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{6}} - \frac{3 \,{\left (2 \, a c d \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right ) - a d^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )\right )}}{a^{6}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{3 \,{\left (2 \, a c d \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right ) - a d^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )\right )}}{a^{6}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \frac{6 \, a c d \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right ) - a d^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}{a^{6}}}{{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (3 \, \sqrt{2} a^{7} c^{2} \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) - 6 \, \sqrt{2} a^{7} c d \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) + 3 \, \sqrt{2} a^{7} d^{2} \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) - 3 \, \sqrt{2} \sqrt{-a} \sqrt{a} c d + \sqrt{2} \sqrt{-a} \sqrt{a} d^{2}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}{\sqrt{-a} a^{7}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(6*sqrt(2)*(c^2 - 2*c*d + d^2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/
2*e)^2 + a) + sqrt(a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*f*x + 1/2*e) + 1)) + ((((6*a*c*d*sgn(tan(1/2*f*x + 1/2*
e) + 1) - a*d^2*sgn(tan(1/2*f*x + 1/2*e) + 1))*tan(1/2*f*x + 1/2*e)/a^6 - 3*(2*a*c*d*sgn(tan(1/2*f*x + 1/2*e)
+ 1) - a*d^2*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^6)*tan(1/2*f*x + 1/2*e) + 3*(2*a*c*d*sgn(tan(1/2*f*x + 1/2*e) +
1) - a*d^2*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^6)*tan(1/2*f*x + 1/2*e) - (6*a*c*d*sgn(tan(1/2*f*x + 1/2*e) + 1) -
a*d^2*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^6)/(a*tan(1/2*f*x + 1/2*e)^2 + a)^(3/2) - 2*(3*sqrt(2)*a^7*c^2*arctan(
sqrt(a)/sqrt(-a)) - 6*sqrt(2)*a^7*c*d*arctan(sqrt(a)/sqrt(-a)) + 3*sqrt(2)*a^7*d^2*arctan(sqrt(a)/sqrt(-a)) -
3*sqrt(2)*sqrt(-a)*sqrt(a)*c*d + sqrt(2)*sqrt(-a)*sqrt(a)*d^2)*sgn(tan(1/2*f*x + 1/2*e) + 1)/(sqrt(-a)*a^7))/f