3.601 \(\int \frac{(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=184 \[ -\frac{3 (c+d)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{16 \sqrt{2} a^{5/2} f \sqrt{c-d}}-\frac{(3 c+7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}} \]
[Out]
(-3*(c + d)^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]])])/(16*Sqrt[2]*a^(5/2)*Sqrt[c - d]*f) - ((c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*f*(a + a*Sin[
e + f*x])^(5/2)) - ((3*c + 7*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(16*a*f*(a + a*Sin[e + f*x])^(3/2))
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Rubi [A] time = 0.538081, antiderivative size = 184, 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 =
{2765, 2978, 12, 2782, 208} \[ -\frac{3 (c+d)^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{16 \sqrt{2} a^{5/2} f \sqrt{c-d}}-\frac{(3 c+7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a \sin (e+f x)+a)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x])^(5/2),x]
[Out]
(-3*(c + d)^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]])])/(16*Sqrt[2]*a^(5/2)*Sqrt[c - d]*f) - ((c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*f*(a + a*Sin[
e + f*x])^(5/2)) - ((3*c + 7*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(16*a*f*(a + a*Sin[e + f*x])^(3/2))
Rule 2765
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(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 - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*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] && GtQ[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{(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac{(c-d) \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 \left (3 c^2+6 c d-d^2\right )-a d (c+3 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{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac{\int \frac{3 a^2 (c-d) (c+d)^2}{4 \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{8 a^4 (c-d)}\\ &=-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac{\left (3 (c+d)^2\right ) \int \frac{1}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{32 a^2}\\ &=-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{\left (3 (c+d)^2\right ) \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 f}\\ &=-\frac{3 (c+d)^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 )}{16 \sqrt{2} a^{5/2} \sqrt{c-d} f}-\frac{(c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{(3 c+7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{16 a f (a+a \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [B] time = 7.1875, size = 396, normalized size = 2.15 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \left (\frac{3 (c+d)^2 \left (\log \left (\tan \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left ((d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d\right )\right )}{\frac{\sec ^2\left (\frac{1}{2} (e+f x)\right )}{2 \tan \left (\frac{1}{2} (e+f x)\right )+2}-\frac{\frac{\sqrt{c-d} \left (\frac{1}{\cos (e+f x)+1}\right )^{3/2} (c \sin (e+f x)+d \cos (e+f x)+d)}{\sqrt{c+d \sin (e+f x)}}-\frac{1}{2} (c-d) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d}}-\frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x)) ((3 c+7 d) \sin (e+f x)+7 c+3 d)}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}\right )}{32 f (a (\sin (e+f x)+1))^{5/2} \sqrt{c+d \sin (e+f x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c + d*Sin[e + f*x])^(3/2)/(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])*(c + d*Sin[e + f*x])*(7*c
+ 3*d + (3*c + 7*d)*Sin[e + f*x]))/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (3*(c + d)^2*(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]))))/(32*f*(a*(1 + Sin
[e + f*x]))^(5/2)*Sqrt[c + d*Sin[e + f*x]])
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Maple [B] time = 0.23, size = 3050, normalized size = 16.6 \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))^(3/2)/(a+a*sin(f*x+e))^(5/2),x)
[Out]
1/64/f/(c-d)*(-12*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c^2+12*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*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+6*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*d+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)))*d^2+28*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^2-28*cos(f*x+e)^3*((c+d*sin(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)))*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+12*sin(f*x+e)*cos(f*x+e)*((c+d*
sin(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+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)))*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^2-12*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)))*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+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*c
os(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*d^2+16*sin(f*x+e)*cos(f*x+e)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)
*c*d-24*(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^2-3*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)))*d^2-6*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+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)))*d^2+18*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-24*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+12*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)/sin(f*x+e)/(a*(1+sin(f*x+e)))^(5/2)/((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{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{{\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))^(3/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
integrate((d*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) + a)^(5/2), x)
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Fricas [B] time = 2.7812, size = 3136, normalized size = 17.04 \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))^(3/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
[1/128*(3*((c^2 + 2*c*d + d^2)*cos(f*x + e)^3 + 3*(c^2 + 2*c*d + d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 -
2*(c^2 + 2*c*d + d^2)*cos(f*x + e) + ((c^2 + 2*c*d + d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 - 2*(c^2 + 2
*c*d + 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)*s
qrt(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 - 7*d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 - 4*c*d - 3*d^
2)*cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 - (3*c^2 + 4*c*d - 7*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 - a^3*d)*f*cos(f*x + e)^3 + 3*(a^3*c - a^3*d)*f*cos(f*x + e)^2 - 2
*(a^3*c - a^3*d)*f*cos(f*x + e) - 4*(a^3*c - a^3*d)*f + ((a^3*c - a^3*d)*f*cos(f*x + e)^2 - 2*(a^3*c - a^3*d)*
f*cos(f*x + e) - 4*(a^3*c - a^3*d)*f)*sin(f*x + e)), -1/64*(3*((c^2 + 2*c*d + d^2)*cos(f*x + e)^3 + 3*(c^2 + 2
*c*d + d^2)*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 - 2*(c^2 + 2*c*d + d^2)*cos(f*x + e) + ((c^2 + 2*c*d + d^2)
*cos(f*x + e)^2 - 4*c^2 - 8*c*d - 4*d^2 - 2*(c^2 + 2*c*d + 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 - 7*
d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 - 4*c*d - 3*d^2)*cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 -
(3*c^2 + 4*c*d - 7*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
- a^3*d)*f*cos(f*x + e)^3 + 3*(a^3*c - a^3*d)*f*cos(f*x + e)^2 - 2*(a^3*c - a^3*d)*f*cos(f*x + e) - 4*(a^3*c
- a^3*d)*f + ((a^3*c - a^3*d)*f*cos(f*x + e)^2 - 2*(a^3*c - a^3*d)*f*cos(f*x + e) - 4*(a^3*c - a^3*d)*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))**(3/2)/(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{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{{\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))^(3/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")
[Out]
integrate((d*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) + a)^(5/2), x)