3.597 \(\int \frac{1}{(a+a \sin (e+f x))^{3/2} \sqrt{c+d \sin (e+f x)}} \, dx\)
Optimal. Leaf size=135 \[ -\frac{(c-3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f (c-d)^{3/2}}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}} \]
[Out]
-((c - 3*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)*(c - d)^(3/2)*f) - (Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(2*(c - d)*f*(a + a*Sin[e
+ f*x])^(3/2))
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Rubi [A] time = 0.236837, antiderivative size = 135, 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 =
{2766, 12, 2782, 208} \[ -\frac{(c-3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f (c-d)^{3/2}}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/((a + a*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]),x]
[Out]
-((c - 3*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)*(c - d)^(3/2)*f) - (Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(2*(c - d)*f*(a + a*Sin[e
+ f*x])^(3/2))
Rule 2766
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (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{1}{(a+a \sin (e+f x))^{3/2} \sqrt{c+d \sin (e+f x)}} \, dx &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}-\frac{\int -\frac{a (c-3 d)}{2 \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{2 a^2 (c-d)}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}+\frac{(c-3 d) \int \frac{1}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{4 a (c-d)}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}-\frac{(c-3 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 (c-d) f}\\ &=-\frac{(c-3 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} (c-d)^{3/2} f}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 (c-d) f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [B] time = 5.53513, size = 381, normalized size = 2.82 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\frac{(c-3 d) \left (\log \left (\tan \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left ((d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d\right )\right )}{\frac{\sec ^2\left (\frac{1}{2} (e+f x)\right )}{2 \tan \left (\frac{1}{2} (e+f x)\right )+2}-\frac{\frac{\sqrt{c-d} \left (\frac{1}{\cos (e+f x)+1}\right )^{3/2} (c \sin (e+f x)+d \cos (e+f x)+d)}{\sqrt{c+d \sin (e+f x)}}-\frac{1}{2} (c-d) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d}}-\frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}\right )}{4 f (c-d) (a (\sin (e+f x)+1))^{3/2} \sqrt{c+d \sin (e+f x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/((a + a*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]),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 - 3*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*(c - d)*f*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c + d*Sin[e +
f*x]])
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Maple [B] time = 0.224, size = 1268, normalized size = 9.4 \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))^(3/2)/(c+d*sin(f*x+e))^(1/2),x)
[Out]
-1/4/f/(2*c-2*d)^(1/2)/(c-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+3*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)))*d-cos(f*x+e)^2*
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+3*cos(f*x+e)^2*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)))*d+2*sin(f*x+e)*cos(f*x+e)*(2*c-2*d)^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(
1/2)+2*sin(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*s
in(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-6*sin(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)))*d-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+3*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)*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)))*d+2*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-6*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)))*d)*(c+d*sin(f*x+e))^(1/2)/(a*(1+sin(f*x+e)))^(3/2)/sin(f*x+e)/((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{1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \sqrt{d \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate(1/((a*sin(f*x + e) + a)^(3/2)*sqrt(d*sin(f*x + e) + c)), x)
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Fricas [B] time = 2.81277, size = 2431, normalized size = 18.01 \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))^(3/2)/(c+d*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
[1/16*(((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(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)*sqr
t(d*sin(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)*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^2
- 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e)^2 - (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e) - 2*(a^2*c^2 - 2*a^2
*c*d + a^2*d^2)*f - ((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e) + 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f)*sin
(f*x + e)), -1/8*(((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)*si
n(f*x + e) - 2*c + 6*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*si
n(f*x + e) + c))/((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e)^2 - (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x
+ e) - 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f - ((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e) + 2*(a^2*c^2 - 2
*a^2*c*d + a^2*d^2)*f)*sin(f*x + e))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}} \sqrt{c + d \sin{\left (e + f x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))**(3/2)/(c+d*sin(f*x+e))**(1/2),x)
[Out]
Integral(1/((a*(sin(e + f*x) + 1))**(3/2)*sqrt(c + d*sin(e + f*x))), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \sqrt{d \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate(1/((a*sin(f*x + e) + a)^(3/2)*sqrt(d*sin(f*x + e) + c)), x)