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]

-1/2*(c-d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(3/2)-1/4*(c+3*d)*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x
+e))^(1/2))/a^(3/2)/f*2^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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])

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 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)]

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*}

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Mathematica [C]  time = 0.20, 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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fricas [B]  time = 0.44, size = 293, normalized size = 3.37 \[ \frac {\sqrt {2} {\left ({\left (c + 3 \, d\right )} \cos \left (f x + e\right )^{2} - {\left (c + 3 \, d\right )} \cos \left (f x + e\right ) - {\left ({\left (c + 3 \, d\right )} \cos \left (f x + e\right ) + 2 \, c + 6 \, d\right )} \sin \left (f x + e\right ) - 2 \, c - 6 \, d\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left ({\left (c - d\right )} \cos \left (f x + e\right ) - {\left (c - d\right )} \sin \left (f x + e\right ) + c - d\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{8 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*2*(1/4*(-3*c*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x
+exp(1))/2)^2+a))^3+3*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3+a*c*(-sqrt(a)*tan((f*
x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-a*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a
))+sqrt(a)*c*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-sqrt(a)*d*(-sqrt(a)*tan((f*x+exp
(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+sqrt(a)*a*c-sqrt(a)*a*d)/(-(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan
((f*x+exp(1))/2)^2+a))^2+2*sqrt(a)*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+a)^2/a/sign(
tan((f*x+exp(1))/2)+1)+1/4*(c+3*d)*atan((-sqrt(a)*tan((f*x+exp(1))/2)-sqrt(a)+sqrt(a*tan((f*x+exp(1))/2)^2+a))
/sqrt(2)/sqrt(-a))/sqrt(2)/sqrt(-a)/a/sign(tan((f*x+exp(1))/2)+1))

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maple [B]  time = 0.86, size = 176, normalized size = 2.02 \[ -\frac {\left (\sin \left (f x +e \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \left (c +3 d \right )+\sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a c +3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a d +2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, c -2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, d \right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{4 a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

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*(sin(f*x+e)-1))^(1/2)/cos(f*x+
e)/(a+a*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d \sin \left (f x + e\right ) + c}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {c+d\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {c + d \sin {\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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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