\(\int \frac {(e \cos (c+d x))^{5/2}}{a+b \sin (c+d x)} \, dx\) [577]

Optimal result

Integrand size = 25, antiderivative size = 384 \[ \int \frac {(e \cos (c+d x))^{5/2}}{a+b \sin (c+d x)} \, dx=\frac {\left (-a^2+b^2\right )^{3/4} e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{5/2} d}-\frac {\left (-a^2+b^2\right )^{3/4} e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{5/2} d}+\frac {2 e (e \cos (c+d x))^{3/2}}{3 b d}+\frac {2 a e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^3 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^3 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}} \]

[Out]

(-a^2+b^2)^(3/4)*e^(5/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(5/2)/d-(-a^2+b^2)^(3
/4)*e^(5/2)*arctanh(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(5/2)/d+2/3*e*(e*cos(d*x+c))^(3/2
)/b/d-a*(a^2-b^2)*e^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b-(-a
^2+b^2)^(1/2)),2^(1/2))*cos(d*x+c)^(1/2)/b^3/d/(b-(-a^2+b^2)^(1/2))/(e*cos(d*x+c))^(1/2)-a*(a^2-b^2)*e^3*(cos(
1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos
(d*x+c)^(1/2)/b^3/d/(b+(-a^2+b^2)^(1/2))/(e*cos(d*x+c))^(1/2)+2*a*e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x
+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/b^2/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2774, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {(e \cos (c+d x))^{5/2}}{a+b \sin (c+d x)} \, dx=\frac {e^{5/2} \left (b^2-a^2\right )^{3/4} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} d}-\frac {e^{5/2} \left (b^2-a^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{5/2} d}-\frac {a e^3 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b^3 d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}-\frac {a e^3 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (c+d x),2\right )}{b^3 d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}}+\frac {2 a e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}+\frac {2 e (e \cos (c+d x))^{3/2}}{3 b d} \]

[In]

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

[Out]

((-a^2 + b^2)^(3/4)*e^(5/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(5/2)*d) -
((-a^2 + b^2)^(3/4)*e^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(5/2)*d)
+ (2*e*(e*Cos[c + d*x])^(3/2))/(3*b*d) + (2*a*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b^2*d*Sqrt
[Cos[c + d*x]]) - (a*(a^2 - b^2)*e^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c + d*x)/2,
2])/(b^3*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) - (a*(a^2 - b^2)*e^3*Sqrt[Cos[c + d*x]]*EllipticPi[(2*
b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(b^3*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]])

Rule 211

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

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2774

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(b*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2780

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> With[{q = Rt[-a^2
+ b^2, 2]}, Dist[a*(g/(2*b)), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (-Dist[a*(g/(2*b)),
Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x] + Dist[b*(g/f), Subst[Int[Sqrt[x]/(g^2*(a^2 - b^2)
+ b^2*x^2), x], x, g*Cos[e + f*x]], x])] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2886

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/
(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2946

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]))/((a_) + (b_.)*sin[(e_.) + (
f_.)*(x_)]), x_Symbol] :> Dist[d/b, Int[(g*Cos[e + f*x])^p, x], x] + Dist[(b*c - a*d)/b, Int[(g*Cos[e + f*x])^
p/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 e (e \cos (c+d x))^{3/2}}{3 b d}+\frac {e^2 \int \frac {\sqrt {e \cos (c+d x)} (b+a \sin (c+d x))}{a+b \sin (c+d x)} \, dx}{b} \\ & = \frac {2 e (e \cos (c+d x))^{3/2}}{3 b d}+\frac {\left (a e^2\right ) \int \sqrt {e \cos (c+d x)} \, dx}{b^2}+\frac {\left (\left (-a^2+b^2\right ) e^2\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx}{b^2} \\ & = \frac {2 e (e \cos (c+d x))^{3/2}}{3 b d}+\frac {\left (a \left (a^2-b^2\right ) e^3\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{2 b^3}-\frac {\left (a \left (a^2-b^2\right ) e^3\right ) \int \frac {1}{\sqrt {e \cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{2 b^3}-\frac {\left (\left (a^2-b^2\right ) e^3\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \cos (c+d x)\right )}{b d}+\frac {\left (a e^2 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{b^2 \sqrt {\cos (c+d x)}} \\ & = \frac {2 e (e \cos (c+d x))^{3/2}}{3 b d}+\frac {2 a e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}-\frac {\left (2 \left (a^2-b^2\right ) e^3\right ) \text {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{b d}+\frac {\left (a \left (a^2-b^2\right ) e^3 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{2 b^3 \sqrt {e \cos (c+d x)}}-\frac {\left (a \left (a^2-b^2\right ) e^3 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \left (\sqrt {-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{2 b^3 \sqrt {e \cos (c+d x)}} \\ & = \frac {2 e (e \cos (c+d x))^{3/2}}{3 b d}+\frac {2 a e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^3 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^3 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {\left (\left (a^2-b^2\right ) e^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{b^2 d}-\frac {\left (\left (a^2-b^2\right ) e^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{b^2 d} \\ & = \frac {\left (-a^2+b^2\right )^{3/4} e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{5/2} d}-\frac {\left (-a^2+b^2\right )^{3/4} e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{5/2} d}+\frac {2 e (e \cos (c+d x))^{3/2}}{3 b d}+\frac {2 a e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {\cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^3 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^3 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {a \left (a^2-b^2\right ) e^3 \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (c+d x),2\right )}{b^3 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 15.13 (sec) , antiderivative size = 709, normalized size of antiderivative = 1.85 \[ \int \frac {(e \cos (c+d x))^{5/2}}{a+b \sin (c+d x)} \, dx=\frac {(e \cos (c+d x))^{5/2} \left (2 \cos ^{\frac {3}{2}}(c+d x)-\frac {a \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(c+d x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+b \cos (c+d x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(c+d x)}\right )}{4 b^{3/2} \left (-a^2+b^2\right ) (a+b \sin (c+d x))}-\frac {6 b \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (c+d x)}+i b \cos (c+d x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (c+d x) \left (a+b \sqrt {\sin ^2(c+d x)}\right )}{\sqrt {\sin ^2(c+d x)} (a+b \sin (c+d x))}\right )}{3 b d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[(e*Cos[c + d*x])^(5/2)/(a + b*Sin[c + d*x]),x]

[Out]

((e*Cos[c + d*x])^(5/2)*(2*Cos[c + d*x]^(3/2) - (a*(8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Cos[c + d*x]^2, (b^2
*Cos[c + d*x]^2)/(-a^2 + b^2)]*Cos[c + d*x]^(3/2) + 3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[
b]*Sqrt[Cos[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[c + d*x]])/(a^2 - b^2)^(1/4
)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c + d*x]] + Log[Sqrt[a
^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[c + d*x]] + b*Cos[c + d*x]]))*(a + b*Sqrt[Sin[c + d*x]^
2]))/(4*b^(3/2)*(-a^2 + b^2)*(a + b*Sin[c + d*x])) - (6*b*((a*AppellF1[3/4, 1/2, 1, 7/4, Cos[c + d*x]^2, (b^2*
Cos[c + d*x]^2)/(-a^2 + b^2)]*Cos[c + d*x]^(3/2))/(3*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[b
]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/
4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]] + I*b*Cos[c + d*x]] + Log[S
qrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[c + d*x]] + I*b*Cos[c + d*x]]))/(Sqrt[b]*(-a^2 +
b^2)^(1/4)))*Sin[c + d*x]*(a + b*Sqrt[Sin[c + d*x]^2]))/(Sqrt[Sin[c + d*x]^2]*(a + b*Sin[c + d*x]))))/(3*b*d*
Cos[c + d*x]^(5/2))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 5.28 (sec) , antiderivative size = 1025, normalized size of antiderivative = 2.67

method	result	size
default	\(\text {Expression too large to display}\)	\(1025\)

[In]

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

[Out]

(4*e^3*b*(1/6/b^2*(-2*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*sin(1/2*d*x+1/2*c)^2+(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(
1/2))/e-1/16/b^4*(a^2-b^2)/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*e*cos(1/2*d*x+1/2*c)^2-e-(e^2*(a^2-b^2)/b^
2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(2*e*cos(1/2*d*x+1/2*c)^2-e+(e^
2*(a^2-b^2)/b^2)^(1/4)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)*2^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2
)*(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))-2*arctan((-2^(1/2)*
(2*e*cos(1/2*d*x+1/2*c)^2-e)^(1/2)+(e^2*(a^2-b^2)/b^2)^(1/4))/(e^2*(a^2-b^2)/b^2)^(1/4))))+1/8*(e*(2*cos(1/2*d
*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*e^3/a*(16*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^
(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2*a^2+sum((a^2-b^2)/_alpha*(8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*c
os(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*(e*(2*_alpha^2*b^2
+a^2-2*b^2)/b^2)^(1/2)*_alpha^3*b^2-8*b^2*_alpha*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2
)*EllipticPi(cos(1/2*d*x+1/2*c),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+a^2*
2^(1/2)*arctanh(1/2*e*(4*_alpha^2-3)/(4*a^2-3*b^2)*(4*a^2*cos(1/2*d*x+1/2*c)^2-3*cos(1/2*d*x+1/2*c)^2*b^2+b^2*
_alpha^2-3*a^2+2*b^2)*2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x
+1/2*c)^2))^(1/2))*(-e*sin(1/2*d*x+1/2*c)^2*(2*sin(1/2*d*x+1/2*c)^2-1))^(1/2))/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b
^2)^(1/2)/(-e*sin(1/2*d*x+1/2*c)^2*(2*sin(1/2*d*x+1/2*c)^2-1))^(1/2),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2))
*(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2))/b^4/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^
2))^(1/2)/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2))/d

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

integrate((e*cos(d*x+c))**(5/2)/(a+b*sin(d*x+c)),x)

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

integrate((e*cos(d*x + c))^(5/2)/(b*sin(d*x + c) + a), x)

Giac [F]

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

[In]

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

[Out]

integrate((e*cos(d*x + c))^(5/2)/(b*sin(d*x + c) + a), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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