\(\int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx\) [593]

Optimal result

Integrand size = 29, antiderivative size = 185 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{(c-d)^{5/2} f}+\frac {2 d \cos (e+f x)}{3 \left (c^2-d^2\right ) f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac {2 d (5 c+d) \cos (e+f x)}{3 \left (c^2-d^2\right )^2 f \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]

[Out]

-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))*2^(1/2)/(c-
d)^(5/2)/f/a^(1/2)+2/3*d*cos(f*x+e)/(c^2-d^2)/f/(c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2)+2/3*d*(5*c+d)*co
s(f*x+e)/(c^2-d^2)^2/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2858, 3063, 12, 2861, 214} \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{\sqrt {a} f (c-d)^{5/2}}+\frac {2 d (5 c+d) \cos (e+f x)}{3 f \left (c^2-d^2\right )^2 \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}+\frac {2 d \cos (e+f x)}{3 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}} \]

[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 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 2858

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-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 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 3063

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])

Rubi steps \begin{align*} \text {integral}& = \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) \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 )}{(c-d)^2 f} \\ & = -\frac {\sqrt {2} \text {arctanh}\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] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(390\) vs. \(2(185)=370\).

Time = 3.10 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.11 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\frac {\frac {2 d \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 c^2+c d-d^2+d (5 c+d) \sin (e+f x)\right )}{(c+d)^2 (c+d \sin (e+f x))}+\frac {3 \left (\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{2+2 \tan \left (\frac {1}{2} (e+f x)\right )}-\frac {-\frac {1}{2} (c-d) \sec ^2\left (\frac {1}{2} (e+f x)\right )+\frac {\sqrt {c-d} \left (\frac {1}{1+\cos (e+f x)}\right )^{3/2} (d+d \cos (e+f x)+c \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}}{c-d+2 \sqrt {c-d} \sqrt {\frac {1}{1+\cos (e+f x)}} \sqrt {c+d \sin (e+f x)}+(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}}}{3 \sqrt {3} (c-d)^2 f \sqrt {1+\sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]

[In]

Integrate[1/(Sqrt[3 + 3*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]) - (-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])))/(3*Sqrt[3]*(c - d)^2*f*Sqrt[1 + Sin[e + f*x]]*Sq
rt[c + d*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2406\) vs. \(2(164)=328\).

Time = 5.53 (sec) , antiderivative size = 2407, normalized size of antiderivative = 13.01

method	result	size
default	\(\text {Expression too large to display}\)	\(2407\)

[In]

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

[Out]

1/3/f*(3*2^(1/2)*((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)/(cos(f*x+e)-1-sin(f*x+e)
))*c^2*cos(f*x+e)^2*d+6*2^(1/2)*((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*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)/(cos(f*x+
e)-1-sin(f*x+e)))*c*cos(f*x+e)^2*d^2-3*cos(f*x+e)*sin(f*x+e)*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*l
n(-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)/(cos(f*x+e)-1-sin(f*x+e)))*c^2*d-6*cos(f*x+e)*sin(f*x+e)*2^(1/2)*((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)/(cos(f*x+e)-1-sin(f*x+e)))*c*d^2-3*2^(1/2)*cos(f*x+e)*sin(
f*x+e)*((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)/(cos(f*x+e)-1-sin(f*x+e)))*d^3-3*l
n(-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)/(cos(f*x+e)-1-sin(f*x+e)))*d^3*2^(1/2)*((c+d*sin(f*x+e))/(cos(f*x+e)+1))^(1/2)*sin
(f*x+e)^2-3*2^(1/2)*((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)/(cos(f*x+e)-1-sin(f*x
+e)))*c^3*cos(f*x+e)-6*2^(1/2)*((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*si
n(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)/(cos(f*x+e
)-1-sin(f*x+e)))*c^2*cos(f*x+e)*d-3*2^(1/2)*((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)/(cos(f*x+e)-1-sin(f*x+e)))*c*cos(f*x+e)*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)/(cos(f*x+e)-1-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)/(cos(f*x+e)-1-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)/(cos(f*x
+e)-1-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*2^(1/2)*sin(f*x+e)*((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)/(cos(f*x+e)-1-sin(f*x+e)))*d^3-3*2^(1/2)*((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)/(cos(f*x+e)-1-sin(f*x+e)))*c^3-9*2^(1/2)*((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)/(cos(f*x+e)-1-sin(f*x+e)))*c^2*d-9*2^(1/2)*((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)*si
n(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)/(cos(f*x+e)-1-sin(f*x+e)))*c*d^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))^(3/2)/(a*(sin(f*x+
e)+1))^(1/2)/(c+d)^2/(2*c-2*d)^(1/2)/(c-d)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 798 vs. \(2 (164) = 328\).

Time = 0.58 (sec) , antiderivative size = 1855, normalized size of antiderivative = 10.03 \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[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))]

Sympy [F]

\[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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