3.7.2 \(\int \frac {\sqrt {c+d \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [602]

Optimal. Leaf size=191 \[ -\frac {(3 c-5 d) (c+d) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} (c-d)^{3/2} f}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {(3 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{16 a (c-d) f (a+a \sin (e+f x))^{3/2}} \]

[Out]

-1/32*(3*c-5*d)*(c+d)*arctanh(1/2*cos(f*x+e)*a^(1/2)*(c-d)^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e
))^(1/2))/a^(5/2)/(c-d)^(3/2)/f*2^(1/2)-1/4*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/(a+a*sin(f*x+e))^(5/2)-1/16*(3
*c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a/(c-d)/f/(a+a*sin(f*x+e))^(3/2)

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Rubi [A]
time = 0.32, antiderivative size = 191, normalized size of antiderivative = 1.00, 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 = {2843, 3057, 12, 2861, 214} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((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]])])/(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*
Sin[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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2843

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 2861

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 3057

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])

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)) \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(412\) vs. \(2(191)=382\).
time = 6.68, size = 412, normalized size = 2.16 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (7 c-5 d+(3 c-d) \sin (e+f x)) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+\frac {\left (3 c^2-2 c d-5 d^2\right ) \left (\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{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 {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}\right )}{32 (c-d) f (a (1+\sin (e+f x)))^{5/2} \sqrt {c+d \sin (e+f x)}} \end {gather*}

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]) - (-1/2*((c - d)*Sec[(e + f*x)/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] Leaf count of result is larger than twice the leaf count of optimal. \(3049\) vs. \(2(162)=324\).
time = 11.13, size = 3050, normalized size = 15.97

method	result	size
default	\(\text {Expression too large to display}\)	\(3050\)

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,method=_RETURNVERBOSE)

[Out]

-1/64/f*(-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)+cos(f*x+e
)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*sin(f*x+e)*cos(f*x+
e)*c^2+12*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*c^2+4*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*cos
(f*x+e)*d^2-12*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^3*c^2-4*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1
/2)*cos(f*x+e)^3*d^2-16*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*c*d+16*((c+d*sin(f*x+e))/(cos(f*x+e
)+1))^(1/2)*cos(f*x+e)^3*c*d+12*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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)
*c^2-20*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)+cos(f*x+e)*c-
d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*d^2+28*((c+d*sin(f*x+e)
)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*cos(f*x+e)*c^2+20*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*cos(f*
x+e)*d^2+3*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)+cos(f*x+e)
*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*cos(f*x+e)^3*c^2-5*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)+cos(f*x+e)*c-d*cos(f*x
+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*cos(f*x+e)^3*d^2-9*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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*
x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*cos(f*x+e)^2*c^2+15*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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(
f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*cos(f*x+e)^2*d^2+12*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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)
/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*sin(f*x+e)*c^2-20*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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e
)+sin(f*x+e)))*(2*c-2*d)^(1/2)*sin(f*x+e)*d^2-6*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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))
*(2*c-2*d)^(1/2)*cos(f*x+e)*c^2+10*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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1
/2)*cos(f*x+e)*d^2-8*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)+
cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*c*d-48*((c
+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*cos(f*x+e)*c*d+10*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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f
*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*sin(f*x+e)*cos(f*x+e)*d^2+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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+
e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*cos(f*x+e)^2*c*d-8*2^(1/2)*ln(2*((2*c-2*d)^(1/2)*2^(1/2)*((c+d*sin(f*x+e))/(co
s(f*x+e)+1))^(1/2)*sin(f*x+e)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e
)))*(2*c-2*d)^(1/2)*sin(f*x+e)*c*d+4*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)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^
(1/2)*cos(f*x+e)*c*d-3*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
)+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*sin(f*x+
e)*cos(f*x+e)^2*c^2+5*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)
+cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*sin(f*x+e
)*cos(f*x+e)^2*d^2-2*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)+
cos(f*x+e)*c-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*cos(f*x+e)
^3*c*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)+cos(f*x+e)*c
-d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*sin(f*x+e)*cos(f*x+e)^
2*c*d+4*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)+cos(f*x+e)*c-
d*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*(2*c-2*d)^(1/2)*sin(f*x+e)*cos(f*x+e)*c
*d)*(c+d*sin(f*x+e))^(1/2)/((c+d*sin(f*x+e))/(c...

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 644 vs. \(2 (171) = 342\).
time = 0.63, size = 1534, normalized size = 8.03 \begin {gather*} \left [\frac {{\left ({\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} + 8 \, c d + 20 \, d^{2} - 2 \, {\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left ({\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} + 8 \, c d + 20 \, d^{2} - 2 \, {\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {2 \, a c - 2 \, a d} \log \left (\frac {{\left (a c^{2} - 14 \, a c d + 17 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} - 4 \, a c^{2} - 8 \, a c d - 4 \, a d^{2} - {\left (13 \, a c^{2} - 22 \, a c d - 3 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (c - 3 \, d\right )} \cos \left (f x + e\right )^{2} - {\left (3 \, c - d\right )} \cos \left (f x + e\right ) + {\left ({\left (c - 3 \, d\right )} \cos \left (f x + e\right ) + 4 \, c - 4 \, d\right )} \sin \left (f x + e\right ) - 4 \, c + 4 \, d\right )} \sqrt {2 \, a c - 2 \, a d} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c} - 2 \, {\left (9 \, a c^{2} - 14 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, a c^{2} + 8 \, a c d + 4 \, a d^{2} - {\left (a c^{2} - 14 \, a c d + 17 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (7 \, a c^{2} - 18 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) - 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 4}\right ) + 8 \, {\left ({\left (3 \, c^{2} - 4 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, c^{2} - 8 \, c d + 4 \, d^{2} + {\left (7 \, c^{2} - 12 \, c d + 5 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, c^{2} - 8 \, c d + 4 \, d^{2} - {\left (3 \, c^{2} - 4 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c}}{128 \, {\left ({\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right ) - 4 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f + {\left ({\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right ) - 4 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f\right )} \sin \left (f x + e\right )\right )}}, -\frac {{\left ({\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} + 8 \, c d + 20 \, d^{2} - 2 \, {\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left ({\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} + 8 \, c d + 20 \, d^{2} - 2 \, {\left (3 \, c^{2} - 2 \, c d - 5 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {-2 \, a c + 2 \, a d} \arctan \left (\frac {\sqrt {-2 \, a c + 2 \, a d} \sqrt {a \sin \left (f x + e\right ) + a} {\left ({\left (c - 3 \, d\right )} \sin \left (f x + e\right ) - 3 \, c + d\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{4 \, {\left ({\left (a c d - a d^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (a c^{2} - a c d\right )} \cos \left (f x + e\right )\right )}}\right ) - 4 \, {\left ({\left (3 \, c^{2} - 4 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, c^{2} - 8 \, c d + 4 \, d^{2} + {\left (7 \, c^{2} - 12 \, c d + 5 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, c^{2} - 8 \, c d + 4 \, d^{2} - {\left (3 \, c^{2} - 4 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c}}{64 \, {\left ({\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right ) - 4 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f + {\left ({\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f \cos \left (f x + e\right ) - 4 \, {\left (a^{3} c^{2} - 2 \, a^{3} c d + a^{3} d^{2}\right )} f\right )} \sin \left (f x + e\right )\right )}}\right ] \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + d \sin {\left (e + f x \right )}}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

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]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c+d\,\sin \left (e+f\,x\right )}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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