3.592 \(\int \frac{1}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=131 \[ \frac{2 d \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f (c-d)^{3/2}} \]
[Out]
-((Sqrt[2]*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
]])])/(Sqrt[a]*(c - d)^(3/2)*f)) + (2*d*Cos[e + f*x])/((c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]])
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Rubi [A] time = 0.242868, antiderivative size = 131, 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 =
{2779, 12, 2782, 208} \[ \frac{2 d \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f (c-d)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)),x]
[Out]
-((Sqrt[2]*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
]])])/(Sqrt[a]*(c - d)^(3/2)*f)) + (2*d*Cos[e + f*x])/((c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]])
Rule 2779
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/
(2*b*(n + 1)*(c^2 - d^2)), Int[((c + d*Sin[e + f*x])^(n + 1)*Simp[a*d - 2*b*c*(n + 1) + b*d*(2*n + 3)*Sin[e +
f*x], x])/Sqrt[a + b*Sin[e + f*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[n, -1] && IntegerQ[2*n]
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{1}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx &=\frac{2 d \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\int \frac{a (c+d)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{a \left (c^2-d^2\right )}\\ &=\frac{2 d \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\int \frac{1}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{c-d}\\ &=\frac{2 d \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{(2 a) \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 )}{(c-d) f}\\ &=-\frac{\sqrt{2} \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 )}{\sqrt{a} (c-d)^{3/2} f}+\frac{2 d \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [B] time = 6.35406, size = 306, normalized size = 2.34 \[ \frac{\frac{2 d \cos (e+f x)}{c+d}+\frac{\log \left (\tan \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left ((d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d\right )}{\frac{\sec ^2\left (\frac{1}{2} (e+f x)\right )}{2 \tan \left (\frac{1}{2} (e+f x)\right )+2}-\frac{\frac{\sqrt{c-d} \left (\frac{1}{\cos (e+f x)+1}\right )^{3/2} (c \sin (e+f x)+d \cos (e+f x)+d)}{\sqrt{c+d \sin (e+f x)}}-\frac{1}{2} (c-d) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d}}}{f (c-d) \sqrt{a (\sin (e+f x)+1)} \sqrt{c+d \sin (e+f x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)),x]
[Out]
((2*d*Cos[e + f*x])/(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])))/((c - d)*f*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c + d*Sin[e + f*x]])
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Maple [B] time = 0.224, size = 879, normalized size = 6.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x)
[Out]
1/f/(c+d)/(2*c-2*d)^(1/2)/(c-d)*(-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)*((c+d*sin(f*x
+e))/(cos(f*x+e)+1))^(1/2)*c*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)*((c+d*s
in(f*x+e))/(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)-cos(f*x+e)*2^(1/2)*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)))*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c-cos(f*x+e)*2^(1/2)*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)))*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*d-2^(1/2)*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)+si
n(f*x+e)))*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c-2^(1/2)*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)))*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*d+2*d*(2*c-2*d)^(1/2)*cos(f*x+e))/(c+d*sin(f*x+e))^(1/2)/(a*(1+s
in(f*x+e)))^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sin \left (f x + e\right ) + a}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate(1/(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(3/2)), x)
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Fricas [B] time = 3.16088, size = 2430, normalized size = 18.55 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
[-1/4*(8*(d*cos(f*x + e) - d*sin(f*x + e) + d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c) - sqrt(2)*(a*
c^2 + 2*a*c*d + a*d^2 - (a*c*d + a*d^2)*cos(f*x + e)^2 + (a*c^2 + a*c*d)*cos(f*x + e) + (a*c^2 + 2*a*c*d + a*d
^2 + (a*c*d + a*d^2)*cos(f*x + e))*sin(f*x + e))*log(((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^3 - (13*c^2 - 22*c*
d - 3*d^2)*cos(f*x + e)^2 + 4*sqrt(2)*((c^2 - 4*c*d + 3*d^2)*cos(f*x + e)^2 - 4*c^2 + 8*c*d - 4*d^2 - (3*c^2 -
4*c*d + d^2)*cos(f*x + e) + (4*c^2 - 8*c*d + 4*d^2 + (c^2 - 4*c*d + 3*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a
*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/sqrt(a*c - a*d) - 4*c^2 - 8*c*d - 4*d^2 - 2*(9*c^2 - 14*c*d + 9*d^
2)*cos(f*x + e) + ((c^2 - 14*c*d + 17*d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 + 2*(7*c^2 - 18*c*d + 7*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))/sqrt(a*c - a*d))/((a*c^2*d - a*d^3)*f*cos(f*x + e)^2 - (a*c^3 - a*c*d^2)*f*cos(f
*x + e) - (a*c^3 + a*c^2*d - a*c*d^2 - a*d^3)*f - ((a*c^2*d - a*d^3)*f*cos(f*x + e) + (a*c^3 + a*c^2*d - a*c*d
^2 - a*d^3)*f)*sin(f*x + e)), -1/2*(sqrt(2)*(a*c^2 + 2*a*c*d + a*d^2 - (a*c*d + a*d^2)*cos(f*x + e)^2 + (a*c^2
+ a*c*d)*cos(f*x + e) + (a*c^2 + 2*a*c*d + a*d^2 + (a*c*d + a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-1/(a*c -
a*d))*arctan(-1/4*sqrt(2)*sqrt(a*sin(f*x + e) + a)*((c - 3*d)*sin(f*x + e) - 3*c + d)*sqrt(d*sin(f*x + e) + c
)*sqrt(-1/(a*c - a*d))/(d*cos(f*x + e)*sin(f*x + e) + c*cos(f*x + e))) + 4*(d*cos(f*x + e) - d*sin(f*x + e) +
d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((a*c^2*d - a*d^3)*f*cos(f*x + e)^2 - (a*c^3 - a*c*d^2)*
f*cos(f*x + e) - (a*c^3 + a*c^2*d - a*c*d^2 - a*d^3)*f - ((a*c^2*d - a*d^3)*f*cos(f*x + e) + (a*c^3 + a*c^2*d
- a*c*d^2 - a*d^3)*f)*sin(f*x + e))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )} \left (c + d \sin{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**(3/2),x)
[Out]
Integral(1/(sqrt(a*(sin(e + f*x) + 1))*(c + d*sin(e + f*x))**(3/2)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sin \left (f x + e\right ) + a}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate(1/(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(3/2)), x)