\(\int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{11/2}} \, dx\) [587]

Optimal result

Integrand size = 29, antiderivative size = 300 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{11/2}} \, dx=\frac {2 (c-d) \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))^{9/2}}+\frac {6 (c-d) (3 c+19 d) \cos (e+f x)}{7 d^2 (c+d)^2 f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{7/2}}-\frac {18 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{35 d^2 (c+d)^3 f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac {24 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{35 d^2 (c+d)^4 f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {48 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{35 d^2 (c+d)^5 f \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]

[Out]

2/63*a^3*(c-d)*(3*c+19*d)*cos(f*x+e)/d^2/(c+d)^2/f/(c+d*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(1/2)-2/105*a^3*(c^
2+10*c*d+73*d^2)*cos(f*x+e)/d^2/(c+d)^3/f/(c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2)-8/315*a^3*(c^2+10*c*d+
73*d^2)*cos(f*x+e)/d^2/(c+d)^4/f/(c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2)+2/9*a^2*(c-d)*cos(f*x+e)*(a+a*s
in(f*x+e))^(1/2)/d/(c+d)/f/(c+d*sin(f*x+e))^(9/2)-16/315*a^3*(c^2+10*c*d+73*d^2)*cos(f*x+e)/d^2/(c+d)^5/f/(a+a
*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2841, 3059, 2851, 2850} \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{11/2}} \, dx=-\frac {16 a^3 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{315 d^2 f (c+d)^5 \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {8 a^3 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{315 d^2 f (c+d)^4 \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac {2 a^3 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{105 d^2 f (c+d)^3 \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}+\frac {2 a^3 (c-d) (3 c+19 d) \cos (e+f x)}{63 d^2 f (c+d)^2 \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{7/2}}+\frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{9 d f (c+d) (c+d \sin (e+f x))^{9/2}} \]

[In]

Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(11/2),x]

[Out]

(2*a^2*(c - d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(9*d*(c + d)*f*(c + d*Sin[e + f*x])^(9/2)) + (2*a^3*(c -
d)*(3*c + 19*d)*Cos[e + f*x])/(63*d^2*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(7/2)) - (2*a
^3*(c^2 + 10*c*d + 73*d^2)*Cos[e + f*x])/(105*d^2*(c + d)^3*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5
/2)) - (8*a^3*(c^2 + 10*c*d + 73*d^2)*Cos[e + f*x])/(315*d^2*(c + d)^4*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e
+ f*x])^(3/2)) - (16*a^3*(c^2 + 10*c*d + 73*d^2)*Cos[e + f*x])/(315*d^2*(c + d)^5*f*Sqrt[a + a*Sin[e + f*x]]*
Sqrt[c + d*Sin[e + f*x]])

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c
+ a*d))), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1
)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*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] && GtQ[m, 1] && LtQ[n, -1
] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2850

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*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 2851

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*c - a*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[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{9 d (c+d) f (c+d \sin (e+f x))^{9/2}}-\frac {(2 a) \int \frac {\sqrt {a+a \sin (e+f x)} \left (\frac {1}{2} a (c-19 d)-\frac {3}{2} a (c+5 d) \sin (e+f x)\right )}{(c+d \sin (e+f x))^{9/2}} \, dx}{9 d (c+d)} \\ & = \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{9 d (c+d) f (c+d \sin (e+f x))^{9/2}}+\frac {2 a^3 (c-d) (3 c+19 d) \cos (e+f x)}{63 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{7/2}}+\frac {\left (a^2 \left (c^2+10 c d+73 d^2\right )\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{7/2}} \, dx}{21 d^2 (c+d)^2} \\ & = \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{9 d (c+d) f (c+d \sin (e+f x))^{9/2}}+\frac {2 a^3 (c-d) (3 c+19 d) \cos (e+f x)}{63 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{7/2}}-\frac {2 a^3 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}+\frac {\left (4 a^2 \left (c^2+10 c d+73 d^2\right )\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx}{105 d^2 (c+d)^3} \\ & = \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{9 d (c+d) f (c+d \sin (e+f x))^{9/2}}+\frac {2 a^3 (c-d) (3 c+19 d) \cos (e+f x)}{63 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{7/2}}-\frac {2 a^3 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac {8 a^3 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{315 d^2 (c+d)^4 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac {\left (8 a^2 \left (c^2+10 c d+73 d^2\right )\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{3/2}} \, dx}{315 d^2 (c+d)^4} \\ & = \frac {2 a^2 (c-d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{9 d (c+d) f (c+d \sin (e+f x))^{9/2}}+\frac {2 a^3 (c-d) (3 c+19 d) \cos (e+f x)}{63 d^2 (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{7/2}}-\frac {2 a^3 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{105 d^2 (c+d)^3 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}-\frac {8 a^3 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{315 d^2 (c+d)^4 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {16 a^3 \left (c^2+10 c d+73 d^2\right ) \cos (e+f x)}{315 d^2 (c+d)^5 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.24 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.01 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{11/2}} \, dx=-\frac {\sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{5/2} \left (1869 c^4+2088 c^3 d+5776 c^2 d^2+1804 c d^3+727 d^4-\left (63 c^4+648 c^3 d+4790 c^2 d^2+1424 c d^3+803 d^4\right ) \cos (2 (e+f x))+2 d^2 \left (c^2+10 c d+73 d^2\right ) \cos (4 (e+f x))+588 c^4 \sin (e+f x)+7326 c^3 d \sin (e+f x)+4370 c^2 d^2 \sin (e+f x)+5498 c d^3 \sin (e+f x)+698 d^4 \sin (e+f x)-18 c^3 d \sin (3 (e+f x))-182 c^2 d^2 \sin (3 (e+f x))-1334 c d^3 \sin (3 (e+f x))-146 d^4 \sin (3 (e+f x))\right )}{35 (c+d)^5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (c+d \sin (e+f x))^{9/2}} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x])^(11/2),x]

[Out]

-1/35*(Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(5/2)*(1869*c^4 + 2088*c^3*d + 5776*c^
2*d^2 + 1804*c*d^3 + 727*d^4 - (63*c^4 + 648*c^3*d + 4790*c^2*d^2 + 1424*c*d^3 + 803*d^4)*Cos[2*(e + f*x)] + 2
*d^2*(c^2 + 10*c*d + 73*d^2)*Cos[4*(e + f*x)] + 588*c^4*Sin[e + f*x] + 7326*c^3*d*Sin[e + f*x] + 4370*c^2*d^2*
Sin[e + f*x] + 5498*c*d^3*Sin[e + f*x] + 698*d^4*Sin[e + f*x] - 18*c^3*d*Sin[3*(e + f*x)] - 182*c^2*d^2*Sin[3*
(e + f*x)] - 1334*c*d^3*Sin[3*(e + f*x)] - 146*d^4*Sin[3*(e + f*x)]))/((c + d)^5*f*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^5*(c + d*Sin[e + f*x])^(9/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1279\) vs. \(2(287)=574\).

Time = 3.02 (sec) , antiderivative size = 1280, normalized size of antiderivative = 4.27

method	result	size
default	\(\text {Expression too large to display}\)	\(1280\)

[In]

int((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE)

[Out]

2/315/f*sec(f*x+e)*(a*(sin(f*x+e)+1))^(1/2)*(c+d*sin(f*x+e))^(1/2)*(-672*c^9-1056*c^8*d+615*c^8*d*cos(f*x+e)^2
+44*cos(f*x+e)^4*sin(f*x+e)^5*c^2*d^7+332*cos(f*x+e)^4*sin(f*x+e)^5*c*d^8+2658*cos(f*x+e)^4*sin(f*x+e)^4*c^2*d
^7-376*cos(f*x+e)^4*sin(f*x+e)^4*c*d^8+69*cos(f*x+e)^4*sin(f*x+e)^3*c^4*d^5-926*cos(f*x+e)^4*sin(f*x+e)^3*c^3*
d^6+742*cos(f*x+e)^2*sin(f*x+e)^5*c^2*d^7+1792*c^7*d^2-1856*c^5*d^4+3072*c^6*d^3-895*cos(f*x+e)^2*sin(f*x+e)^5
*c*d^8-4508*cos(f*x+e)^2*sin(f*x+e)^4*c^2*d^7+409*cos(f*x+e)^2*sin(f*x+e)^4*c*d^8-3381*cos(f*x+e)^2*sin(f*x+e)
^3*c^4*d^5+3004*cos(f*x+e)^2*sin(f*x+e)^3*c^3*d^6+2884*cos(f*x+e)^2*sin(f*x+e)*c^6*d^3+2110*cos(f*x+e)^4*sin(f
*x+e)*c^6*d^3+5094*cos(f*x+e)^4*sin(f*x+e)*c^5*d^4-37*sin(f*x+e)^2*c^4*d^5*cos(f*x+e)^6+3516*sin(f*x+e)*c^7*d^
2*cos(f*x+e)^2+543*sin(f*x+e)*c^8*d*cos(f*x+e)^2+8*cos(f*x+e)^6*sin(f*x+e)^4*c^2*d^7+80*cos(f*x+e)^6*sin(f*x+e
)^4*c*d^8-4*cos(f*x+e)^6*sin(f*x+e)^3*c^3*d^6+8682*cos(f*x+e)^2*sin(f*x+e)^2*c^4*d^5-996*cos(f*x+e)^2*sin(f*x+
e)^2*c^3*d^6+458*cos(f*x+e)^4*sin(f*x+e)*c^7*d^2-6778*cos(f*x+e)^4*sin(f*x+e)^2*c^4*d^5-10*cos(f*x+e)^4*sin(f*
x+e)^2*c^3*d^6-35*cos(f*x+e)^6*sin(f*x+e)*c^5*d^4-368*sin(f*x+e)^2*c^3*d^6*cos(f*x+e)^6+381*sin(f*x+e)^7*d^9-4
16*sin(f*x+e)^6*d^9+35*sin(f*x+e)^5*d^9+672*sin(f*x+e)*c^9-231*c^9*cos(f*x+e)^2-394*c^7*d^2*cos(f*x+e)^4-63*si
n(f*x+e)*c^9*cos(f*x+e)^2-279*cos(f*x+e)^4*c^8*d+1024*sin(f*x+e)^2*c^3*d^6-1892*c^7*d^2*cos(f*x+e)^2-584*cos(f
*x+e)^4*sin(f*x+e)^6*d^9-292*cos(f*x+e)^2*sin(f*x+e)^7*d^9+1095*cos(f*x+e)^2*sin(f*x+e)^6*d^9-2142*sin(f*x+e)^
5*c^2*d^7+288*sin(f*x+e)^5*c*d^8+1792*sin(f*x+e)^4*c^2*d^7-288*sin(f*x+e)^4*c*d^8+3217*sin(f*x+e)^3*c^4*d^5+35
0*sin(f*x+e)^3*c^2*d^7+310*c^6*d^3*cos(f*x+e)^6-3142*cos(f*x+e)^4*c^5*d^4+1805*cos(f*x+e)^6*c^5*d^4+3358*cos(f
*x+e)^4*c^6*d^3+1056*sin(f*x+e)*c^8*d-1024*sin(f*x+e)^3*c^3*d^6-3392*sin(f*x+e)^2*c^4*d^5+1856*sin(f*x+e)*c^5*
d^4-6010*sin(f*x+e)*cos(f*x+e)^2*c^5*d^4-6940*cos(f*x+e)^2*c^6*d^3+3158*cos(f*x+e)^2*c^5*d^4-1792*sin(f*x+e)*c
^7*d^2-3072*sin(f*x+e)*c^6*d^3+175*sin(f*x+e)*c^4*d^5)*a^2/(cos(f*x+e)^2*d^2+c^2-d^2)^5/(c+d)^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1492 vs. \(2 (287) = 574\).

Time = 0.42 (sec) , antiderivative size = 1492, normalized size of antiderivative = 4.97 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{11/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

2/315*(672*a^2*c^4 - 2304*a^2*c^3*d + 3008*a^2*c^2*d^2 - 1792*a^2*c*d^3 + 416*a^2*d^4 + 8*(a^2*c^2*d^2 + 10*a^
2*c*d^3 + 73*a^2*d^4)*cos(f*x + e)^5 - 4*(9*a^2*c^3*d + 89*a^2*c^2*d^2 + 647*a^2*c*d^3 - 73*a^2*d^4)*cos(f*x +
e)^4 - (63*a^2*c^4 + 648*a^2*c^3*d + 4798*a^2*c^2*d^2 + 1504*a^2*c*d^3 + 1387*a^2*d^4)*cos(f*x + e)^3 + (231*
a^2*c^4 + 3060*a^2*c^3*d - 2158*a^2*c^2*d^2 + 4580*a^2*c*d^3 - 673*a^2*d^4)*cos(f*x + e)^2 + 2*(483*a^2*c^4 +
684*a^2*c^3*d + 2642*a^2*c^2*d^2 + 812*a^2*c*d^3 + 419*a^2*d^4)*cos(f*x + e) - (672*a^2*c^4 - 2304*a^2*c^3*d +
3008*a^2*c^2*d^2 - 1792*a^2*c*d^3 + 416*a^2*d^4 + 8*(a^2*c^2*d^2 + 10*a^2*c*d^3 + 73*a^2*d^4)*cos(f*x + e)^4
+ 4*(9*a^2*c^3*d + 91*a^2*c^2*d^2 + 667*a^2*c*d^3 + 73*a^2*d^4)*cos(f*x + e)^3 - 3*(21*a^2*c^4 + 204*a^2*c^3*d
+ 1478*a^2*c^2*d^2 - 388*a^2*c*d^3 + 365*a^2*d^4)*cos(f*x + e)^2 - 2*(147*a^2*c^4 + 1836*a^2*c^3*d + 1138*a^2
*c^2*d^2 + 1708*a^2*c*d^3 + 211*a^2*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x +
e) + c)/((c^5*d^5 + 5*c^4*d^6 + 10*c^3*d^7 + 10*c^2*d^8 + 5*c*d^9 + d^10)*f*cos(f*x + e)^6 - 5*(c^6*d^4 + 5*c
^5*d^5 + 10*c^4*d^6 + 10*c^3*d^7 + 5*c^2*d^8 + c*d^9)*f*cos(f*x + e)^5 - (10*c^7*d^3 + 55*c^6*d^4 + 128*c^5*d^
5 + 165*c^4*d^6 + 130*c^3*d^7 + 65*c^2*d^8 + 20*c*d^9 + 3*d^10)*f*cos(f*x + e)^4 + 10*(c^8*d^2 + 5*c^7*d^3 + 1
1*c^6*d^4 + 15*c^5*d^5 + 15*c^4*d^6 + 11*c^3*d^7 + 5*c^2*d^8 + c*d^9)*f*cos(f*x + e)^3 + (5*c^9*d + 35*c^8*d^2
+ 120*c^7*d^3 + 260*c^6*d^4 + 378*c^5*d^5 + 370*c^4*d^6 + 240*c^3*d^7 + 100*c^2*d^8 + 25*c*d^9 + 3*d^10)*f*co
s(f*x + e)^2 - (c^10 + 5*c^9*d + 20*c^8*d^2 + 60*c^7*d^3 + 110*c^6*d^4 + 126*c^5*d^5 + 100*c^4*d^6 + 60*c^3*d^
7 + 25*c^2*d^8 + 5*c*d^9)*f*cos(f*x + e) - (c^10 + 10*c^9*d + 45*c^8*d^2 + 120*c^7*d^3 + 210*c^6*d^4 + 252*c^5
*d^5 + 210*c^4*d^6 + 120*c^3*d^7 + 45*c^2*d^8 + 10*c*d^9 + d^10)*f - ((c^5*d^5 + 5*c^4*d^6 + 10*c^3*d^7 + 10*c
^2*d^8 + 5*c*d^9 + d^10)*f*cos(f*x + e)^5 + (5*c^6*d^4 + 26*c^5*d^5 + 55*c^4*d^6 + 60*c^3*d^7 + 35*c^2*d^8 + 1
0*c*d^9 + d^10)*f*cos(f*x + e)^4 - 2*(5*c^7*d^3 + 25*c^6*d^4 + 51*c^5*d^5 + 55*c^4*d^6 + 35*c^3*d^7 + 15*c^2*d
^8 + 5*c*d^9 + d^10)*f*cos(f*x + e)^3 - 2*(5*c^8*d^2 + 30*c^7*d^3 + 80*c^6*d^4 + 126*c^5*d^5 + 130*c^4*d^6 + 9
0*c^3*d^7 + 40*c^2*d^8 + 10*c*d^9 + d^10)*f*cos(f*x + e)^2 + (5*c^9*d + 25*c^8*d^2 + 60*c^7*d^3 + 100*c^6*d^4
+ 126*c^5*d^5 + 110*c^4*d^6 + 60*c^3*d^7 + 20*c^2*d^8 + 5*c*d^9 + d^10)*f*cos(f*x + e) + (c^10 + 10*c^9*d + 45
*c^8*d^2 + 120*c^7*d^3 + 210*c^6*d^4 + 252*c^5*d^5 + 210*c^4*d^6 + 120*c^3*d^7 + 45*c^2*d^8 + 10*c*d^9 + d^10)
*f)*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(5/2)/(c+d*sin(f*x+e))**(11/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 984 vs. \(2 (287) = 574\).

Time = 0.45 (sec) , antiderivative size = 984, normalized size of antiderivative = 3.28 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{11/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

-2/315*((903*c^5 + 720*c^4*d + 494*c^3*d^2 + 200*c^2*d^3 + 35*c*d^4)*a^(5/2) - (315*c^5 - 8358*c^4*d - 4770*c^
3*d^2 - 2284*c^2*d^3 - 625*c*d^4 - 70*d^5)*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + (4179*c^5 - 1710*c^4*d +
30878*c^3*d^2 + 11540*c^2*d^3 + 3383*c*d^4 + 450*d^5)*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 3*(805*c^5
- 9912*c^4*d + 2330*c^3*d^2 - 18504*c^2*d^3 - 3895*c*d^4 - 504*d^5)*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)
^3 + 6*(1239*c^5 - 3100*c^4*d + 12918*c^3*d^2 - 3560*c^2*d^3 + 8043*c*d^4 + 700*d^5)*a^(5/2)*sin(f*x + e)^4/(c
os(f*x + e) + 1)^4 - 42*(149*c^5 - 894*c^4*d + 1402*c^3*d^2 - 2052*c^2*d^3 + 745*c*d^4 - 390*d^5)*a^(5/2)*sin(
f*x + e)^5/(cos(f*x + e) + 1)^5 + 42*(149*c^5 - 894*c^4*d + 1402*c^3*d^2 - 2052*c^2*d^3 + 745*c*d^4 - 390*d^5)
*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 6*(1239*c^5 - 3100*c^4*d + 12918*c^3*d^2 - 3560*c^2*d^3 + 8043*
c*d^4 + 700*d^5)*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 3*(805*c^5 - 9912*c^4*d + 2330*c^3*d^2 - 18504*
c^2*d^3 - 3895*c*d^4 - 504*d^5)*a^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - (4179*c^5 - 1710*c^4*d + 30878*c
^3*d^2 + 11540*c^2*d^3 + 3383*c*d^4 + 450*d^5)*a^(5/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + (315*c^5 - 8358*c
^4*d - 4770*c^3*d^2 - 2284*c^2*d^3 - 625*c*d^4 - 70*d^5)*a^(5/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - (903*
c^5 + 720*c^4*d + 494*c^3*d^2 + 200*c^2*d^3 + 35*c*d^4)*a^(5/2)*sin(f*x + e)^11/(cos(f*x + e) + 1)^11)*(sin(f*
x + e)^2/(cos(f*x + e) + 1)^2 + 1)^3/((c^5 + 5*c^4*d + 10*c^3*d^2 + 10*c^2*d^3 + 5*c*d^4 + d^5 + 3*(c^5 + 5*c^
4*d + 10*c^3*d^2 + 10*c^2*d^3 + 5*c*d^4 + d^5)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*(c^5 + 5*c^4*d + 10*c^3
*d^2 + 10*c^2*d^3 + 5*c*d^4 + d^5)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + (c^5 + 5*c^4*d + 10*c^3*d^2 + 10*c^2*
d^3 + 5*c*d^4 + d^5)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*(c + 2*d*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x
+ e)^2/(cos(f*x + e) + 1)^2)^(11/2)*f)

Giac [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 28.87 (sec) , antiderivative size = 1155, normalized size of antiderivative = 3.85 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c+d \sin (e+f x))^{11/2}} \, dx=\text {Too large to display} \]

[In]

int((a + a*sin(e + f*x))^(5/2)/(c + d*sin(e + f*x))^(11/2),x)

[Out]

-((c + d*sin(e + f*x))^(1/2)*((32*a^2*exp(e*1i + f*x*1i)*(a + a*sin(e + f*x))^(1/2)*(c*d*10i + c^2*1i + d^2*73
i))/(315*d^3*f*(c + d)^5) - (32*a^2*exp(e*10i + f*x*10i)*(a + a*sin(e + f*x))^(1/2)*(10*c*d + c^2 + 73*d^2))/(
315*d^3*f*(c + d)^5) - (32*a^2*exp(e*6i + f*x*6i)*(a + a*sin(e + f*x))^(1/2)*(25*c^4 - 25*c^3*d - 15*c*d^3 + 6
*d^4 + 57*c^2*d^2))/(5*d^5*f*(c + d)^5) + (32*a^2*exp(e*5i + f*x*5i)*(a + a*sin(e + f*x))^(1/2)*(c^4*25i - c^3
*d*25i - c*d^3*15i + d^4*6i + c^2*d^2*57i))/(5*d^5*f*(c + d)^5) - (16*a^2*exp(e*4i + f*x*4i)*(a + a*sin(e + f*
x))^(1/2)*(194*c*d^3 + 318*c^3*d + 25*c^4 - 5*d^4 - 20*c^2*d^2))/(15*d^5*f*(c + d)^5) + (16*a^2*exp(e*7i + f*x
*7i)*(a + a*sin(e + f*x))^(1/2)*(c*d^3*194i + c^3*d*318i + c^4*25i - d^4*5i - c^2*d^2*20i))/(15*d^5*f*(c + d)^
5) + (16*a^2*exp(e*8i + f*x*8i)*(a + a*sin(e + f*x))^(1/2)*(10*c*d^3 + 70*c^3*d + 7*c^4 + 73*d^4 + 512*c^2*d^2
))/(35*d^5*f*(c + d)^5) - (16*a^2*exp(e*3i + f*x*3i)*(a + a*sin(e + f*x))^(1/2)*(c*d^3*10i + c^3*d*70i + c^4*7
i + d^4*73i + c^2*d^2*512i))/(35*d^5*f*(c + d)^5) + (32*a^2*c*exp(e*2i + f*x*2i)*(a + a*sin(e + f*x))^(1/2)*(1
0*c*d + c^2 + 73*d^2))/(35*d^4*f*(c + d)^5) - (32*a^2*c*exp(e*9i + f*x*9i)*(a + a*sin(e + f*x))^(1/2)*(c*d*10i
+ c^2*1i + d^2*73i))/(35*d^4*f*(c + d)^5)))/(exp(e*11i + f*x*11i) - (c*1i + d*1i)^5/(c + d)^5 + (10*exp(e*7i
+ f*x*7i)*(4*c*d^3 + 8*c^3*d + 8*c^4 + d^4 + 12*c^2*d^2))/d^4 + (5*exp(e*3i + f*x*3i)*(8*c*d^2 + 8*c^2*d + 16*
c^3 + d^3))/d^3 - (5*exp(e*9i + f*x*9i)*(2*c*d + 8*c^2 + d^2))/d^2 - (2*exp(e*5i + f*x*5i)*(30*c*d^4 + 40*c^4*
d + 16*c^5 + 5*d^5 + 60*c^2*d^3 + 80*c^3*d^2))/d^5 - (exp(e*1i + f*x*1i)*(10*c + d))/d - (5*exp(e*8i + f*x*8i)
*(c*1i + d*1i)^5*(8*c*d^2 + 8*c^2*d + 16*c^3 + d^3))/(d^3*(c + d)^5) + (5*exp(e*2i + f*x*2i)*(c*1i + d*1i)^5*(
2*c*d + 8*c^2 + d^2))/(d^2*(c + d)^5) + (2*exp(e*6i + f*x*6i)*(c*1i + d*1i)^5*(30*c*d^4 + 40*c^4*d + 16*c^5 +
5*d^5 + 60*c^2*d^3 + 80*c^3*d^2))/(d^5*(c + d)^5) + (exp(e*10i + f*x*10i)*(10*c + d)*(c*1i + d*1i)^5)/(d*(c +
d)^5) - (10*exp(e*4i + f*x*4i)*(c*1i + d*1i)^5*(4*c*d^3 + 8*c^3*d + 8*c^4 + d^4 + 12*c^2*d^2))/(d^4*(c + d)^5)
)