3.552 \(\int \frac{c+d \sin (e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=87 \[ -\frac{(c+3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}-\frac{(c-d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0727975, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2750, 2649, 206} \[ -\frac{(c+3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f}-\frac{(c-d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

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

Rule 2750

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, 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)}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac{(c-d) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{(c+3 d) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{4 a}\\ &=-\frac{(c-d) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{(c+3 d) \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 )}{2 a f}\\ &=-\frac{(c+3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f}-\frac{(c-d) \cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.197537, size = 150, normalized size = 1.72 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (2 (c-d) \sin \left (\frac{1}{2} (e+f x)\right )+(d-c) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+(1+i) (-1)^{3/4} (c+3 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{2 f (a (\sin (e+f x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Sin[e + f*x])/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

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

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Maple [B]  time = 0.54, size = 176, normalized size = 2. \begin{align*} -{\frac{1}{4\,f\cos \left ( fx+e \right ) } \left ( \sin \left ( fx+e \right ) \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) a \left ( c+3\,d \right ) +\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) ac+3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) ad+2\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}c-2\,\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{a}d \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{a}^{-{\frac{5}{2}}}{\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))/(a+a*sin(f*x+e))^(3/2),x)

[Out]

-1/4/a^(5/2)*(sin(f*x+e)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*(c+3*d)+2^(1/2)*arctanh
(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c+3*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))
*a*d+2*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*c-2*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*d)*(-a*(-1+sin(f*x+e)))^(1/2)/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{d \sin \left (f x + e\right ) + c}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)/(a*sin(f*x + e) + a)^(3/2), x)

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Fricas [B]  time = 1.69843, size = 755, normalized size = 8.68 \begin{align*} \frac{\sqrt{2}{\left ({\left (c + 3 \, d\right )} \cos \left (f x + e\right )^{2} -{\left (c + 3 \, d\right )} \cos \left (f x + e\right ) -{\left ({\left (c + 3 \, d\right )} \cos \left (f x + e\right ) + 2 \, c + 6 \, d\right )} \sin \left (f x + e\right ) - 2 \, c - 6 \, d\right )} \sqrt{a} \log \left (-\frac{a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\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 ) + 4 \,{\left ({\left (c - d\right )} \cos \left (f x + e\right ) -{\left (c - d\right )} \sin \left (f x + e\right ) + c - d\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{8 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/8*(sqrt(2)*((c + 3*d)*cos(f*x + e)^2 - (c + 3*d)*cos(f*x + e) - ((c + 3*d)*cos(f*x + e) + 2*c + 6*d)*sin(f*x
+ e) - 2*c - 6*d)*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) -
sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x
+ e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*((c - d)*cos(f*x + e) - (c - d)*sin(f*x + e) + c - d)*sqrt(a*s
in(f*x + e) + a))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*
x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c + d \sin{\left (e + f x \right )}}{\left (a \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((c+d*sin(f*x+e))/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Integral((c + d*sin(e + f*x))/(a*(sin(e + f*x) + 1))**(3/2), x)

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Giac [B]  time = 2.93497, size = 599, normalized size = 6.89 \begin{align*} \frac{\frac{\sqrt{2}{\left (c + 3 \, d\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} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )} + \frac{2 \,{\left (3 \,{\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}\right )}^{3} c - 3 \,{\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}\right )}^{3} d +{\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}\right )}^{2} \sqrt{a} c -{\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}\right )}^{2} \sqrt{a} d -{\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}\right )} a c +{\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}\right )} a d + a^{\frac{3}{2}} c - a^{\frac{3}{2}} d\right )}}{{\left ({\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}\right )}^{2} + 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}\right )} \sqrt{a} - a\right )}^{2} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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