3.602 \(\int \frac{\sqrt{c+d \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=191 \[ -\frac{(3 c-5 d) (c+d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{16 \sqrt{2} a^{5/2} f (c-d)^{3/2}}-\frac{(3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a f (c-d) (a \sin (e+f x)+a)^{3/2}}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}} \]
[Out]
-((3*c - 5*d)*(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]])])/(16*Sqrt[2]*a^(5/2)*(c - d)^(3/2)*f) - (Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*f*(a + a*Si
n[e + f*x])^(5/2)) - ((3*c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(16*a*(c - d)*f*(a + a*Sin[e + f*x])^(3
/2))
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Rubi [A] time = 0.488392, 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 =
{2764, 2978, 12, 2782, 208} \[ -\frac{(3 c-5 d) (c+d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{16 \sqrt{2} a^{5/2} f (c-d)^{3/2}}-\frac{(3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a f (c-d) (a \sin (e+f x)+a)^{3/2}}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[c + d*Sin[e + f*x]]/(a + a*Sin[e + f*x])^(5/2),x]
[Out]
-((3*c - 5*d)*(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]])])/(16*Sqrt[2]*a^(5/2)*(c - d)^(3/2)*f) - (Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*f*(a + a*Si
n[e + f*x])^(5/2)) - ((3*c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(16*a*(c - d)*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 2978
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*(A*b - a*B)*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] + Dist[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*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || 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))^{5/2}} \, dx &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\int \frac{\frac{1}{2} a (3 c+d)+a d \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt{c+d \sin (e+f x)}} \, dx}{4 a^2}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a (c-d) f (a+a \sin (e+f x))^{3/2}}-\frac{\int -\frac{a^2 (3 c-5 d) (c+d)}{4 \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{8 a^4 (c-d)}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a (c-d) f (a+a \sin (e+f x))^{3/2}}+\frac{((3 c-5 d) (c+d)) \int \frac{1}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{32 a^2 (c-d)}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a (c-d) f (a+a \sin (e+f x))^{3/2}}-\frac{((3 c-5 d) (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 )}{16 a (c-d) f}\\ &=-\frac{(3 c-5 d) (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 )}{16 \sqrt{2} a^{5/2} (c-d)^{3/2} f}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a (c-d) f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [B] time = 6.80194, size = 412, normalized size = 2.16 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \left (\frac{\left (3 c^2-2 c d-5 d^2\right ) \left (\log \left (\tan \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left ((d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d\right )\right )}{\frac{\sec ^2\left (\frac{1}{2} (e+f x)\right )}{2 \tan \left (\frac{1}{2} (e+f x)\right )+2}-\frac{\frac{\sqrt{c-d} \left (\frac{1}{\cos (e+f x)+1}\right )^{3/2} (c \sin (e+f x)+d \cos (e+f x)+d)}{\sqrt{c+d \sin (e+f x)}}-\frac{1}{2} (c-d) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d}}-\frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) ((3 c-d) \sin (e+f x)+7 c-5 d) (c+d \sin (e+f x))}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}\right )}{32 f (c-d) (a (\sin (e+f x)+1))^{5/2} \sqrt{c+d \sin (e+f x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sqrt[c + d*Sin[e + f*x]]/(a + a*Sin[e + f*x])^(5/2),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*((-2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(7*c - 5*d + (3*c - d)*Sin
[e + f*x])*(c + d*Sin[e + f*x]))/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + ((3*c^2 - 2*c*d - 5*d^2)*(Log[1 + T
an[(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]))))/(32*(c
- d)*f*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c + d*Sin[e + f*x]])
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Maple [B] time = 0.234, size = 3050, normalized size = 16. \begin{align*} \text{output 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))^(5/2),x)
[Out]
1/64/f/(c-d)^2*(-12*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c^2-4*sin(f*x+e)*cos(f*x+e)*(2*c-2*d)^(
1/2)*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*s
in(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*sin(f*x+e)*cos(f*x+e)^2*(2*c-2*d)^(1
/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*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)))*c*d-5*sin(f*x+e)*cos(f*x+e)^2*(2*c-2*d)^(1/
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^2-4*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+
e)+1))^(1/2)*d^2+12*cos(f*x+e)^3*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c^2+4*cos(f*x+e)^3*((c+d*sin(f*x+e))/
(cos(f*x+e)+1))^(1/2)*d^2+20*(2*c-2*d)^(1/2)*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+3*sin(f*x+e)*cos(f*x+e)^2*(2*c-2*d)^(1/2)*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^2
+16*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c*d-16*cos(f*x+e)^3*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(
1/2)*c*d-28*sin(f*x+e)*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c^2-20*sin(f*x+e)*cos(f*x+e)*((c+d*s
in(f*x+e))/(cos(f*x+e)+1))^(1/2)*d^2-12*(2*c-2*d)^(1/2)*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^2+9*cos(f*x+e)^2*(2*c-2*d)^(1/2)*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^2
-15*cos(f*x+e)^2*(2*c-2*d)^(1/2)*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^2-12*sin(f*x
+e)*(2*c-2*d)^(1/2)*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^2+20*sin(f*x+e)*(2*c-2*d)
^(1/2)*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+6*cos(f*x+e)*(2*c-2*d)^(1/2)*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^2-10*cos(f*x+e)*(2*c-2*d)^(1/2)*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+48*sin(f*x+e)*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2
)*c*d+8*(2*c-2*d)^(1/2)*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*d-3*cos(f*x+e)^3*(2*c
-2*d)^(1/2)*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^2+5*cos(f*x+e)^3*(2*c-2*d)^(1/2)*
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+2*cos(f*x+e)^3*(2*c-2*d)^(1/2)*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*d+6*sin(f*x+e)*cos(f*x+e)*(2*c-2*d)^(1/2)*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^2-10*sin(f*x+e)*cos(f*x+e)*(2*c-2*d)^(1/2)*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^2-6*cos(f*x+e)^2*(2*c-2*d)^(1/2)*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*d+8*sin(f*x+e)*(2*c-2*d)^(1/2)*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*d-4*cos(f*x+e)*(2*c-2*d)^(1/2)*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*d)*(c+d*sin(f*x+e))^(1/2)/(a*(1+sin(f*x+e)))^(5/2)/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{\sqrt{d \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{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))^(5/2),x, algorithm="maxima")
[Out]
integrate(sqrt(d*sin(f*x + e) + c)/(a*sin(f*x + e) + a)^(5/2), x)
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Fricas [B] time = 3.20349, size = 3490, normalized size = 18.27 \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))^(5/2),x, algorithm="fricas")
[Out]
[1/128*(((3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e)^3 + 3*(3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e)^2 - 12*c^2 + 8*c*d +
20*d^2 - 2*(3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e) + ((3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e)^2 - 12*c^2 + 8*c*d + 2
0*d^2 - 2*(3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e))*sin(f*x + e))*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*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*((3*c^2 - 4*c*d + d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (
7*c^2 - 12*c*d + 5*d^2)*cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 - (3*c^2 - 4*c*d + d^2)*cos(f*x + e))*sin(f*x +
e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e)^3 + 3*(
a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e)^2 - 2*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e) - 4*(a^3*c^
2 - 2*a^3*c*d + a^3*d^2)*f + ((a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e)^2 - 2*(a^3*c^2 - 2*a^3*c*d + a^3*
d^2)*f*cos(f*x + e) - 4*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f)*sin(f*x + e)), -1/64*(((3*c^2 - 2*c*d - 5*d^2)*cos(
f*x + e)^3 + 3*(3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e)^2 - 12*c^2 + 8*c*d + 20*d^2 - 2*(3*c^2 - 2*c*d - 5*d^2)*co
s(f*x + e) + ((3*c^2 - 2*c*d - 5*d^2)*cos(f*x + e)^2 - 12*c^2 + 8*c*d + 20*d^2 - 2*(3*c^2 - 2*c*d - 5*d^2)*cos
(f*x + e))*sin(f*x + e))*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*((3*c^2 - 4*c*d + d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 - 12*c*d + 5*d^2
)*cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 - (3*c^2 - 4*c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e
) + a)*sqrt(d*sin(f*x + e) + c))/((a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e)^3 + 3*(a^3*c^2 - 2*a^3*c*d +
a^3*d^2)*f*cos(f*x + e)^2 - 2*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e) - 4*(a^3*c^2 - 2*a^3*c*d + a^3*d^
2)*f + ((a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e)^2 - 2*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*f*cos(f*x + e) -
4*(a^3*c^2 - 2*a^3*c*d + a^3*d^2)*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((c+d*sin(f*x+e))**(1/2)/(a+a*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{\sqrt{d \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{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))^(5/2),x, algorithm="giac")
[Out]
integrate(sqrt(d*sin(f*x + e) + c)/(a*sin(f*x + e) + a)^(5/2), x)