3.596 \(\int \frac{\sqrt{c+d \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=126 \[ -\frac{(c+d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f \sqrt{c-d}}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}} \]
[Out]
-((c + d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]
])])/(2*Sqrt[2]*a^(3/2)*Sqrt[c - d]*f) - (Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(2*f*(a + a*Sin[e + f*x])^(3/
2))
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Rubi [A] time = 0.220565, antiderivative size = 126, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used =
{2764, 12, 2782, 208} \[ -\frac{(c+d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f \sqrt{c-d}}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[c + d*Sin[e + f*x]]/(a + a*Sin[e + f*x])^(3/2),x]
[Out]
-((c + d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]
])])/(2*Sqrt[2]*a^(3/2)*Sqrt[c - d]*f) - (Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(2*f*(a + a*Sin[e + f*x])^(3/
2))
Rule 2764
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a*d*n - b*c*(m + 1) - b*d*(m + n + 1)*Sin[
e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && LtQ[m, -1] && LtQ[0, n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 2782
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{\sqrt{c+d \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{a (c+d)}{2 \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{2 a^2}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 f (a+a \sin (e+f x))^{3/2}}+\frac{(c+d) \int \frac{1}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{4 a}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 f (a+a \sin (e+f x))^{3/2}}-\frac{(c+d) \operatorname{Subst}\left (\int \frac{1}{2 a^2-(a c-a d) x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{2 f}\\ &=-\frac{(c+d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} \sqrt{c-d} f}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [B] time = 5.24486, size = 372, normalized size = 2.95 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\frac{(c+d) \left (\log \left (\tan \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left ((d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d\right )\right )}{\frac{\sec ^2\left (\frac{1}{2} (e+f x)\right )}{2 \tan \left (\frac{1}{2} (e+f x)\right )+2}-\frac{\frac{\sqrt{c-d} \left (\frac{1}{\cos (e+f x)+1}\right )^{3/2} (c \sin (e+f x)+d \cos (e+f x)+d)}{\sqrt{c+d \sin (e+f x)}}-\frac{1}{2} (c-d) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d}}-\frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}\right )}{4 f (a (\sin (e+f x)+1))^{3/2} \sqrt{c+d \sin (e+f x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sqrt[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*((-2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(c + d*Sin[e + f*x]))/(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2]) + ((c + d)*(Log[1 + Tan[(e + f*x)/2]] - Log[c - d + 2*Sqrt[c - d]*Sqrt[(1 +
Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]]))/(Sec[(e + f*x)/2]^2/(2 + 2*Tan[(e
+ f*x)/2]) - (-((c - d)*Sec[(e + f*x)/2]^2)/2 + (Sqrt[c - d]*((1 + Cos[e + f*x])^(-1))^(3/2)*(d + d*Cos[e + f*
x] + c*Sin[e + f*x]))/Sqrt[c + d*Sin[e + f*x]])/(c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c +
d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]))))/(4*f*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c + d*Sin[e + f*x]])
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Maple [B] time = 0.229, size = 1373, normalized size = 10.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(3/2),x)
[Out]
1/8/f/(c-d)*(sin(f*x+e)*cos(f*x+e)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f
*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*2^(1/2)*(2*c-2*d)^(1
/2)*c+sin(f*x+e)*cos(f*x+e)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c
*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*2^(1/2)*(2*c-2*d)^(1/2)*d+c
os(f*x+e)^2*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*si
n(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*2^(1/2)*(2*c-2*d)^(1/2)*c+cos(f*x+e)^2*ln(2
*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f
*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*2^(1/2)*(2*c-2*d)^(1/2)*d-2*sin(f*x+e)*ln(2*((2*c-2*d)^(1/2
)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+
e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*2^(1/2)*(2*c-2*d)^(1/2)*c-2*sin(f*x+e)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*
sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f
*x+e)+sin(f*x+e)))*2^(1/2)*(2*c-2*d)^(1/2)*d+cos(f*x+e)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f
*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))
*2^(1/2)*(2*c-2*d)^(1/2)*c+cos(f*x+e)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*si
n(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*2^(1/2)*(2*c-2*d)
^(1/2)*d-4*sin(f*x+e)*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c+4*sin(f*x+e)*cos(f*x+e)*((c+d*sin(f
*x+e))/(cos(f*x+e)+1))^(1/2)*d-2*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x
+e)+c*sin(f*x+e)-d*sin(f*x+e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*2^(1/2)*(2*c-2*d)^(1/2
)*c-2*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+c*sin(f*x+e)-d*sin(f*x+
e)+c*cos(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*2^(1/2)*(2*c-2*d)^(1/2)*d)*(c+d*sin(f*x+e))^(1/2)
/sin(f*x+e)/(a*(1+sin(f*x+e)))^(3/2)/((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{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))^(1/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate(sqrt(d*sin(f*x + e) + c)/(a*sin(f*x + e) + a)^(3/2), x)
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Fricas [B] time = 2.44585, size = 2199, normalized size = 17.45 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
[1/16*(((c + d)*cos(f*x + e)^2 - (c + d)*cos(f*x + e) - ((c + d)*cos(f*x + e) + 2*c + 2*d)*sin(f*x + e) - 2*c
- 2*d)*sqrt(2*a*c - 2*a*d)*log(((a*c^2 - 14*a*c*d + 17*a*d^2)*cos(f*x + e)^3 - 4*a*c^2 - 8*a*c*d - 4*a*d^2 - (
13*a*c^2 - 22*a*c*d - 3*a*d^2)*cos(f*x + e)^2 - 4*((c - 3*d)*cos(f*x + e)^2 - (3*c - d)*cos(f*x + e) + ((c - 3
*d)*cos(f*x + e) + 4*c - 4*d)*sin(f*x + e) - 4*c + 4*d)*sqrt(2*a*c - 2*a*d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*si
n(f*x + e) + c) - 2*(9*a*c^2 - 14*a*c*d + 9*a*d^2)*cos(f*x + e) - (4*a*c^2 + 8*a*c*d + 4*a*d^2 - (a*c^2 - 14*a
*c*d + 17*a*d^2)*cos(f*x + e)^2 - 2*(7*a*c^2 - 18*a*c*d + 7*a*d^2)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e)^3
+ 3*cos(f*x + e)^2 + (cos(f*x + e)^2 - 2*cos(f*x + e) - 4)*sin(f*x + e) - 2*cos(f*x + e) - 4)) + 8*((c - d)*c
os(f*x + e) - (c - d)*sin(f*x + e) + c - d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((a^2*c - a^2*d
)*f*cos(f*x + e)^2 - (a^2*c - a^2*d)*f*cos(f*x + e) - 2*(a^2*c - a^2*d)*f - ((a^2*c - a^2*d)*f*cos(f*x + e) +
2*(a^2*c - a^2*d)*f)*sin(f*x + e)), -1/8*(((c + d)*cos(f*x + e)^2 - (c + d)*cos(f*x + e) - ((c + d)*cos(f*x +
e) + 2*c + 2*d)*sin(f*x + e) - 2*c - 2*d)*sqrt(-2*a*c + 2*a*d)*arctan(1/4*sqrt(-2*a*c + 2*a*d)*sqrt(a*sin(f*x
+ e) + a)*((c - 3*d)*sin(f*x + e) - 3*c + d)*sqrt(d*sin(f*x + e) + c)/((a*c*d - a*d^2)*cos(f*x + e)*sin(f*x +
e) + (a*c^2 - a*c*d)*cos(f*x + e))) - 4*((c - d)*cos(f*x + e) - (c - d)*sin(f*x + e) + c - d)*sqrt(a*sin(f*x +
e) + a)*sqrt(d*sin(f*x + e) + c))/((a^2*c - a^2*d)*f*cos(f*x + e)^2 - (a^2*c - a^2*d)*f*cos(f*x + e) - 2*(a^2
*c - a^2*d)*f - ((a^2*c - a^2*d)*f*cos(f*x + e) + 2*(a^2*c - a^2*d)*f)*sin(f*x + e))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{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))**(1/2)/(a+a*sin(f*x+e))**(3/2),x)
[Out]
Integral(sqrt(c + d*sin(e + f*x))/(a*(sin(e + f*x) + 1))**(3/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{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))^(1/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate(sqrt(d*sin(f*x + e) + c)/(a*sin(f*x + e) + a)^(3/2), x)