3.593 \(\int \frac{1}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=191 \[ \frac{2 d (5 c+d) \cos (e+f x)}{3 f \left (c^2-d^2\right )^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}+\frac{2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f (c-d)^{5/2}} \]

[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)^(5/2)*f)) + (2*d*Cos[e + f*x])/(3*(c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e +
f*x])^(3/2)) + (2*d*(5*c + d)*Cos[e + f*x])/(3*(c^2 - d^2)^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]])

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Rubi [A]  time = 0.475902, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2779, 2984, 12, 2782, 208} \[ \frac{2 d (5 c+d) \cos (e+f x)}{3 f \left (c^2-d^2\right )^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}+\frac{2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{\sqrt{a} f (c-d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)),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)^(5/2)*f)) + (2*d*Cos[e + f*x])/(3*(c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e +
f*x])^(3/2)) + (2*d*(5*c + d)*Cos[e + f*x])/(3*(c^2 - d^2)^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 2984

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 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}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx &=\frac{2 d \cos (e+f x)}{3 \left (c^2-d^2\right ) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{\int \frac{a (3 c+d)-2 a d \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx}{3 a \left (c^2-d^2\right )}\\ &=\frac{2 d \cos (e+f x)}{3 \left (c^2-d^2\right ) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{2 d (5 c+d) \cos (e+f x)}{3 \left (c^2-d^2\right )^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{2 \int -\frac{3 a^2 (c+d)^2}{2 \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{3 a^2 \left (c^2-d^2\right )^2}\\ &=\frac{2 d \cos (e+f x)}{3 \left (c^2-d^2\right ) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{2 d (5 c+d) \cos (e+f x)}{3 \left (c^2-d^2\right )^2 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)^2}\\ &=\frac{2 d \cos (e+f x)}{3 \left (c^2-d^2\right ) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{2 d (5 c+d) \cos (e+f x)}{3 \left (c^2-d^2\right )^2 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)^2 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)^{5/2} f}+\frac{2 d \cos (e+f x)}{3 \left (c^2-d^2\right ) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{2 d (5 c+d) \cos (e+f x)}{3 \left (c^2-d^2\right )^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 6.19057, size = 387, normalized size = 2.03 \[ \frac{\frac{2 d \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (6 c^2+d (5 c+d) \sin (e+f x)+c d-d^2\right )}{(c+d)^2 (c+d \sin (e+f x))}+\frac{3 \left (\log \left (\tan \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left ((d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d\right )\right )}{\frac{\sec ^2\left (\frac{1}{2} (e+f x)\right )}{2 \tan \left (\frac{1}{2} (e+f x)\right )+2}-\frac{\frac{\sqrt{c-d} \left (\frac{1}{\cos (e+f x)+1}\right )^{3/2} (c \sin (e+f x)+d \cos (e+f x)+d)}{\sqrt{c+d \sin (e+f x)}}-\frac{1}{2} (c-d) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d}}}{3 f (c-d)^2 \sqrt{a (\sin (e+f x)+1)} \sqrt{c+d \sin (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)),x]

[Out]

((2*d*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(6*c^2 + c*d - d^2 + d*(5*c
+ d)*Sin[e + f*x]))/((c + d)^2*(c + d*Sin[e + f*x])) + (3*(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])))/(3*(c - d)^2*f*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c +
d*Sin[e + f*x]])

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Maple [B]  time = 0.278, size = 2575, normalized size = 13.5 \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))^(5/2),x)

[Out]

-1/3/f/(c+d)^2/(2*c-2*d)^(1/2)/(c-d)^2*(3*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(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)))*cos(f*x+e)^2*2^(1/2)*c^2*d+6*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(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*co
s(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*cos(f*x+e)^2*2^(1/2)*c*d^2+3*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(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*c
os(f*x+e)-d*cos(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*cos(f*x+e)^2*2^(1/2)*d^3-3*cos(f*x+e)*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*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c^2*d-6*cos(f
*x+e)*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*sin(f*x+e))/(cos(f*x+e)+
1))^(1/2)*c*d^2-3*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(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)
))*cos(f*x+e)*sin(f*x+e)*2^(1/2)*d^3-3*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(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)))*cos(f*x+e)*2^(1/2)*c^3-6*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^2*d*
2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)-3*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^2*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)-3*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*s
in(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*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-9*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^2*d*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-9*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^2*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-3*((c
+d*sin(f*x+e))/(cos(f*x+e)+1))^(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)))*sin(f*x+e)*2^(1/2)
*d^3-3*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(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)))*2^(1/2)*
c^3-9*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(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)))*2^(1/2)*c
^2*d-9*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(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)))*2^(1/2)*
c*d^2-3*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^3*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(
1/2)+10*cos(f*x+e)*sin(f*x+e)*(2*c-2*d)^(1/2)*c*d^2+2*cos(f*x+e)*sin(f*x+e)*(2*c-2*d)^(1/2)*d^3+12*cos(f*x+e)*
(2*c-2*d)^(1/2)*c^2*d+2*cos(f*x+e)*(2*c-2*d)^(1/2)*c*d^2-2*cos(f*x+e)*(2*c-2*d)^(1/2)*d^3)*(c+d*sin(f*x+e))^(1
/2)/(cos(f*x+e)^2*d^2-2*sin(f*x+e)*c*d-c^2-d^2)/(a*(1+sin(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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(5/2)), x)

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Fricas [B]  time = 5.0642, size = 4224, normalized size = 22.12 \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))^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(8*(6*c^2*d - 4*c*d^2 - 2*d^3 + (5*c*d^2 + d^3)*cos(f*x + e)^2 + (6*c^2*d + c*d^2 - d^3)*cos(f*x + e) -
(6*c^2*d - 4*c*d^2 - 2*d^3 - (5*c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(
f*x + e) + c) + 3*sqrt(2)*(a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^
4)*cos(f*x + e)^3 - (2*a*c^3*d + 5*a*c^2*d^2 + 4*a*c*d^3 + a*d^4)*cos(f*x + e)^2 + (a*c^4 + 2*a*c^3*d + 2*a*c^
2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e) + (a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - (a*c^2*d^2 +
2*a*c*d^3 + a*d^4)*cos(f*x + e)^2 + 2*(a*c^3*d + 2*a*c^2*d^2 + a*c*d^3)*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^4*d^2
- 2*a*c^2*d^4 + a*d^6)*f*cos(f*x + e)^3 + (2*a*c^5*d + a*c^4*d^2 - 4*a*c^3*d^3 - 2*a*c^2*d^4 + 2*a*c*d^5 + a*d
^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^4*d^2 - a*c^2*d^4 + a*d^6)*f*cos(f*x + e) - (a*c^6 + 2*a*c^5*d - a*c^4*d^2
- 4*a*c^3*d^3 - a*c^2*d^4 + 2*a*c*d^5 + a*d^6)*f + ((a*c^4*d^2 - 2*a*c^2*d^4 + a*d^6)*f*cos(f*x + e)^2 - 2*(a
*c^5*d - 2*a*c^3*d^3 + a*c*d^5)*f*cos(f*x + e) - (a*c^6 + 2*a*c^5*d - a*c^4*d^2 - 4*a*c^3*d^3 - a*c^2*d^4 + 2*
a*c*d^5 + a*d^6)*f)*sin(f*x + e)), -1/6*(3*sqrt(2)*(a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - (a*c
^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^3 - (2*a*c^3*d + 5*a*c^2*d^2 + 4*a*c*d^3 + a*d^4)*cos(f*x + e)^2 + (a
*c^4 + 2*a*c^3*d + 2*a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e) + (a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^
3 + a*d^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^2 + 2*(a*c^3*d + 2*a*c^2*d^2 + a*c*d^3)*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*(6*c^2
*d - 4*c*d^2 - 2*d^3 + (5*c*d^2 + d^3)*cos(f*x + e)^2 + (6*c^2*d + c*d^2 - d^3)*cos(f*x + e) - (6*c^2*d - 4*c*
d^2 - 2*d^3 - (5*c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/(
(a*c^4*d^2 - 2*a*c^2*d^4 + a*d^6)*f*cos(f*x + e)^3 + (2*a*c^5*d + a*c^4*d^2 - 4*a*c^3*d^3 - 2*a*c^2*d^4 + 2*a*
c*d^5 + a*d^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^4*d^2 - a*c^2*d^4 + a*d^6)*f*cos(f*x + e) - (a*c^6 + 2*a*c^5*d
- a*c^4*d^2 - 4*a*c^3*d^3 - a*c^2*d^4 + 2*a*c*d^5 + a*d^6)*f + ((a*c^4*d^2 - 2*a*c^2*d^4 + a*d^6)*f*cos(f*x +
e)^2 - 2*(a*c^5*d - 2*a*c^3*d^3 + a*c*d^5)*f*cos(f*x + e) - (a*c^6 + 2*a*c^5*d - a*c^4*d^2 - 4*a*c^3*d^3 - a*c
^2*d^4 + 2*a*c*d^5 + a*d^6)*f)*sin(f*x + e))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(5/2),x)

[Out]

Timed out

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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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(5/2)), x)