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

Optimal. Leaf size=201 \[ -\frac {\left (3 c^2-10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f (c-d)^{5/2}}-\frac {3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{16 a f (c-d)^2 (a \sin (e+f x)+a)^{3/2}}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}} \]

[Out]

-1/32*(3*c^2-10*c*d+19*d^2)*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)^(5/2)/f*2^(1/2)-1/4*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/(c-d)/f/(a+a*sin(f*x+e))^(
5/2)-3/16*(c-3*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a/(c-d)^2/f/(a+a*sin(f*x+e))^(3/2)

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Rubi [A]  time = 0.49, antiderivative size = 201, 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 = {2766, 2978, 12, 2782, 208} \[ -\frac {\left (3 c^2-10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f (c-d)^{5/2}}-\frac {3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{16 a f (c-d)^2 (a \sin (e+f x)+a)^{3/2}}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*c^2 - 10*c*d + 19*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)*(c - d)^(5/2)*f) - (Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*(c -
d)*f*(a + a*Sin[e + f*x])^(5/2)) - (3*(c - 3*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(16*a*(c - d)^2*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 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]

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 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 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])

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sin (e+f x))^{5/2} \sqrt {c+d \sin (e+f x)}} \, dx &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {\int \frac {-\frac {1}{2} a (3 c-7 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 (c-d)}\\ &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {a^2 \left (3 c^2-10 c d+19 d^2\right )}{4 \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx}{8 a^4 (c-d)^2}\\ &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}+\frac {\left (3 c^2-10 c d+19 d^2\right ) \int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \, dx}{32 a^2 (c-d)^2}\\ &=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}-\frac {\left (3 c^2-10 c d+19 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 (c-d)^2 f}\\ &=-\frac {\left (3 c^2-10 c d+19 d^2\right ) \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)^{5/2} f}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {3 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [B]  time = 7.13, size = 411, normalized size = 2.04 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\frac {\left (3 c^2-10 c d+19 d^2\right ) \left (\log \left (\tan \left (\frac {1}{2} (e+f x)\right )+1\right )-\log \left ((d-c) \tan \left (\frac {1}{2} (e+f x)\right )+2 \sqrt {c-d} \sqrt {\frac {1}{\cos (e+f x)+1}} \sqrt {c+d \sin (e+f x)}+c-d\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2 \tan \left (\frac {1}{2} (e+f x)\right )+2}-\frac {\frac {\sqrt {c-d} \left (\frac {1}{\cos (e+f x)+1}\right )^{3/2} (c \sin (e+f x)+d \cos (e+f x)+d)}{\sqrt {c+d \sin (e+f x)}}-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{(d-c) \tan \left (\frac {1}{2} (e+f x)\right )+2 \sqrt {c-d} \sqrt {\frac {1}{\cos (e+f x)+1}} \sqrt {c+d \sin (e+f x)}+c-d}}-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (3 (c-3 d) \sin (e+f x)+7 c-13 d) (c+d \sin (e+f x))}{\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3}\right )}{32 f (c-d)^2 (a (\sin (e+f x)+1))^{5/2} \sqrt {c+d \sin (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + a*Sin[e + f*x])^(5/2)*Sqrt[c + d*Sin[e + f*x]]),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 - 13*d + 3*(c - 3*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 - 10*c*d + 19*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]) - (-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)^2*f*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c + d*Sin[e + f*x]])

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fricas [B]  time = 1.14, size = 1644, normalized size = 8.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/128*(((3*c^2 - 10*c*d + 19*d^2)*cos(f*x + e)^3 + 3*(3*c^2 - 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*c^2 + 40*c
*d - 76*d^2 - 2*(3*c^2 - 10*c*d + 19*d^2)*cos(f*x + e) + ((3*c^2 - 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*c^2 +
40*c*d - 76*d^2 - 2*(3*c^2 - 10*c*d + 19*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*c
os(f*x + e) - 4)*sin(f*x + e) - 2*cos(f*x + e) - 4)) + 8*(3*(c^2 - 4*c*d + 3*d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c
*d + 4*d^2 + (7*c^2 - 20*c*d + 13*d^2)*cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 - 3*(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))/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2
- a^3*d^3)*f*cos(f*x + e)^3 + 3*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3
- 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f +
((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a
^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f)*sin(f*x + e)), -1/64*(((3*c^2 -
10*c*d + 19*d^2)*cos(f*x + e)^3 + 3*(3*c^2 - 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*c^2 + 40*c*d - 76*d^2 - 2*(3
*c^2 - 10*c*d + 19*d^2)*cos(f*x + e) + ((3*c^2 - 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*c^2 + 40*c*d - 76*d^2 -
2*(3*c^2 - 10*c*d + 19*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)*s
qrt(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 + 3*d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d
+ 4*d^2 + (7*c^2 - 20*c*d + 13*d^2)*cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 - 3*(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))/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a
^3*d^3)*f*cos(f*x + e)^3 + 3*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3
*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f + ((a
^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*
d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f)*sin(f*x + e))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {d \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.36, size = 2805, normalized size = 13.96 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/f*(-12*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^2-76*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)*d^2+6*cos(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^2+38*cos(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)*d^
2+40*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+6*cos(f*x+e)^3*(2*c-2*d)^(1/2)*((c+d*s
in(f*x+e))/(cos(f*x+e)+1))^(1/2)*c-18*cos(f*x+e)^3*(2*c-2*d)^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*d-1
0*sin(f*x+e)*cos(f*x+e)^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)))*2^(1/2)*c*d-20*sin(f*x+e)*cos
(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-6*cos(f*x+e)*(2*c-2*d)^(1/2)*((c+d*
sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c+18*cos(f*x+e)*(2*c-2*d)^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*d-3*
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*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)))*2^(1/2)*c^2-19*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*co
s(f*x+e)-c+d)/(1-cos(f*x+e)+sin(f*x+e)))*2^(1/2)*d^2+9*cos(f*x+e)^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+57*cos(f*x+e)^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)*d^2-12
*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^2-76*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)*d^2-20*cos(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+3
*sin(f*x+e)*cos(f*x+e)^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*si
n(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+19*sin(f*x+e)*cos(
f*x+e)^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)*d^2+10*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)))*2^(1/2)*c*d+6*sin(f*x+e)*cos(f*x+e)*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)))*2^(1/2)*c^2+38*sin(f*x+e)*cos(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)*d^2-30*cos(f*x+e)^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-14*sin(f*x+e)*c
os(f*x+e)*(2*c-2*d)^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*c+26*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)*d+40*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)*(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
)/(2*c-2*d)^(1/2)/(c-d)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {d \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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