3.6.75 \(\int \frac {(e \cos (c+d x))^{9/2}}{a+b \sin (c+d x)} \, dx\) [575]

Optimal. Leaf size=446 \[ \frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}-\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 b d}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt {\cos (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {2 e^3 (e \cos (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \sin (c+d x)\right )}{15 b^3 d} \]

[Out]

(-a^2+b^2)^(7/4)*e^(9/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(9/2)/d-(-a^2+b^2)^(7
/4)*e^(9/2)*arctanh(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(9/2)/d+2/7*e*(e*cos(d*x+c))^(7/2
)/b/d-2/15*e^3*(e*cos(d*x+c))^(3/2)*(5*a^2-5*b^2-3*a*b*sin(d*x+c))/b^3/d+a*(a^2-b^2)^2*e^5*(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^5/d/(b-(-a^2+b^2)^(1/2))/(e*cos(d*x+c))^(1/2)+a*(a^2-b^2)^2*e^5*(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^5/d/(b+(-a^2+b^2)^(1/2)
)/(e*cos(d*x+c))^(1/2)-2/5*a*(5*a^2-8*b^2)*e^4*(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^4/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.83, antiderivative size = 446, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2774, 2944, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \begin {gather*} \frac {e^{9/2} \left (b^2-a^2\right )^{7/4} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}-\frac {e^{9/2} \left (b^2-a^2\right )^{7/4} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}+\frac {a e^5 \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \cos (c+d x)}}+\frac {a e^5 \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \cos (c+d x)}}-\frac {2 a e^4 \left (5 a^2-8 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 b^4 d \sqrt {\cos (c+d x)}}-\frac {2 e^3 (e \cos (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \sin (c+d x)\right )}{15 b^3 d}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(9/2)/(a + b*Sin[c + d*x]),x]

[Out]

((-a^2 + b^2)^(7/4)*e^(9/2)*ArcTan[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(9/2)*d) -
((-a^2 + b^2)^(7/4)*e^(9/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Cos[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(b^(9/2)*d)
+ (2*e*(e*Cos[c + d*x])^(7/2))/(7*b*d) - (2*a*(5*a^2 - 8*b^2)*e^4*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
2])/(5*b^4*d*Sqrt[Cos[c + d*x]]) + (a*(a^2 - b^2)^2*e^5*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 +
b^2]), (c + d*x)/2, 2])/(b^5*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c + d*x]]) + (a*(a^2 - b^2)^2*e^5*Sqrt[Cos[c
+ d*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c + d*x)/2, 2])/(b^5*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Cos[c +
d*x]]) - (2*e^3*(e*Cos[c + d*x])^(3/2)*(5*(a^2 - b^2) - 3*a*b*Sin[c + d*x]))/(15*b^3*d)

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 2944

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

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*} \int \frac {(e \cos (c+d x))^{9/2}}{a+b \sin (c+d x)} \, dx &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 b d}+\frac {e^2 \int \frac {(e \cos (c+d x))^{5/2} (b+a \sin (c+d x))}{a+b \sin (c+d x)} \, dx}{b}\\ &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 b d}-\frac {2 e^3 (e \cos (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \sin (c+d x)\right )}{15 b^3 d}+\frac {\left (2 e^4\right ) \int \frac {\sqrt {e \cos (c+d x)} \left (-\frac {1}{2} b \left (2 a^2-5 b^2\right )-\frac {1}{2} a \left (5 a^2-8 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{5 b^3}\\ &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 b d}-\frac {2 e^3 (e \cos (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \sin (c+d x)\right )}{15 b^3 d}-\frac {\left (a \left (5 a^2-8 b^2\right ) e^4\right ) \int \sqrt {e \cos (c+d x)} \, dx}{5 b^4}+\frac {\left (\left (a^2-b^2\right )^2 e^4\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a+b \sin (c+d x)} \, dx}{b^4}\\ &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 b d}-\frac {2 e^3 (e \cos (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \sin (c+d x)\right )}{15 b^3 d}-\frac {\left (a \left (a^2-b^2\right )^2 e^5\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^5}+\frac {\left (a \left (a^2-b^2\right )^2 e^5\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^5}+\frac {\left (\left (a^2-b^2\right )^2 e^5\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^3 d}-\frac {\left (a \left (5 a^2-8 b^2\right ) e^4 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^4 \sqrt {\cos (c+d x)}}\\ &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 b d}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt {\cos (c+d x)}}-\frac {2 e^3 (e \cos (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \sin (c+d x)\right )}{15 b^3 d}+\frac {\left (2 \left (a^2-b^2\right )^2 e^5\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^3 d}-\frac {\left (a \left (a^2-b^2\right )^2 e^5 \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^5 \sqrt {e \cos (c+d x)}}+\frac {\left (a \left (a^2-b^2\right )^2 e^5 \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^5 \sqrt {e \cos (c+d x)}}\\ &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 b d}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt {\cos (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {2 e^3 (e \cos (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \sin (c+d x)\right )}{15 b^3 d}-\frac {\left (\left (a^2-b^2\right )^2 e^5\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^4 d}+\frac {\left (\left (a^2-b^2\right )^2 e^5\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^4 d}\\ &=\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}-\frac {\left (-a^2+b^2\right )^{7/4} e^{9/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{b^{9/2} d}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 b d}-\frac {2 a \left (5 a^2-8 b^2\right ) e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt {\cos (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}+\frac {a \left (a^2-b^2\right )^2 e^5 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \cos (c+d x)}}-\frac {2 e^3 (e \cos (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \sin (c+d x)\right )}{15 b^3 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 46.91, size = 834, normalized size = 1.87 \begin {gather*} -\frac {(e \cos (c+d x))^{9/2} \left (-\frac {2 \left (2 a^2 b-5 b^3\right ) \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {a F_1\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 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \tan ^{-1}\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)}{\sqrt {1-\cos ^2(c+d x)} (a+b \sin (c+d x))}-\frac {\left (5 a^3-8 a b^2\right ) \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (8 b^{5/2} F_1\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 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\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 ) \sin ^2(c+d x)}{12 b^{3/2} \left (-a^2+b^2\right ) \left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))}\right )}{5 b^3 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {(e \cos (c+d x))^{9/2} \sec ^4(c+d x) \left (\frac {\left (-28 a^2+37 b^2\right ) \cos (c+d x)}{42 b^3}+\frac {\cos (3 (c+d x))}{14 b}+\frac {a \sin (2 (c+d x))}{5 b^2}\right )}{d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Cos[c + d*x])^(9/2)/(a + b*Sin[c + d*x]),x]

[Out]

-1/5*((e*Cos[c + d*x])^(9/2)*((-2*(2*a^2*b - 5*b^3)*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((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[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]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)))*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*x]^2]*(a + b*Sin[c + d*x])) - ((5*a
^3 - 8*a*b^2)*(a + b*Sqrt[1 - Cos[c + d*x]^2])*(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]*S
qrt[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]]))*Sin[c + d*x]^2)/(12*b^(3/2)*(
-a^2 + b^2)*(1 - Cos[c + d*x]^2)*(a + b*Sin[c + d*x]))))/(b^3*d*Cos[c + d*x]^(9/2)) + ((e*Cos[c + d*x])^(9/2)*
Sec[c + d*x]^4*(((-28*a^2 + 37*b^2)*Cos[c + d*x])/(42*b^3) + Cos[3*(c + d*x)]/(14*b) + (a*Sin[2*(c + d*x)])/(5
*b^2)))/d

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 10.35, size = 1583, normalized size = 3.55

method	result	size
default	\(\text {Expression too large to display}\)	\(1583\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(9/2)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

(16/7*e^4/b*cos(1/2*d*x+1/2*c)^6*(2*cos(1/2*d*x+1/2*c)^2*e-e)^(1/2)-24/7*e^4/b*cos(1/2*d*x+1/2*c)^4*(2*cos(1/2
*d*x+1/2*c)^2*e-e)^(1/2)+64/21*e^4/b*cos(1/2*d*x+1/2*c)^2*(2*cos(1/2*d*x+1/2*c)^2*e-e)^(1/2)+64/21*e^4/b*(2*co
s(1/2*d*x+1/2*c)^2*e-e)^(1/2)-4/3*e^4/b^3*cos(1/2*d*x+1/2*c)^2*(2*cos(1/2*d*x+1/2*c)^2*e-e)^(1/2)*a^2-4/3*e^4/
b^3*(2*cos(1/2*d*x+1/2*c)^2*e-e)^(1/2)*a^2+2*e^4/b^3*(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)*a^2-4*e^4/b*(e*(2*co
s(1/2*d*x+1/2*c)^2-1))^(1/2)+1/2*e^5/b^3*sum((_R^6-_R^4*e-_R^2*e^2+e^3)/(_R^7*b^2-3*_R^5*b^2*e+8*_R^3*a^2*e^2-
5*_R^3*b^2*e^2-_R*b^2*e^3)*ln((-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-e^(1/2)*cos(1/2*d*x+1/2*c)*2^(1/2)-_R),_R=Ro
otOf(b^2*_Z^8-4*b^2*e*_Z^6+(16*a^2*e^2-10*b^2*e^2)*_Z^4-4*b^2*e^3*_Z^2+b^2*e^4))*a^4-e^5/b*sum((_R^6-_R^4*e-_R
^2*e^2+e^3)/(_R^7*b^2-3*_R^5*b^2*e+8*_R^3*a^2*e^2-5*_R^3*b^2*e^2-_R*b^2*e^3)*ln((-2*sin(1/2*d*x+1/2*c)^2*e+e)^
(1/2)-e^(1/2)*cos(1/2*d*x+1/2*c)*2^(1/2)-_R),_R=RootOf(b^2*_Z^8-4*b^2*e*_Z^6+(16*a^2*e^2-10*b^2*e^2)*_Z^4-4*b^
2*e^3*_Z^2+b^2*e^4))*a^2+1/2*e^5*b*sum((_R^6-_R^4*e-_R^2*e^2+e^3)/(_R^7*b^2-3*_R^5*b^2*e+8*_R^3*a^2*e^2-5*_R^3
*b^2*e^2-_R*b^2*e^3)*ln((-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-e^(1/2)*cos(1/2*d*x+1/2*c)*2^(1/2)-_R),_R=RootOf(b
^2*_Z^8-4*b^2*e*_Z^6+(16*a^2*e^2-10*b^2*e^2)*_Z^4-4*b^2*e^3*_Z^2+b^2*e^4))-1/40*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*
sin(1/2*d*x+1/2*c)^2)^(1/2)*e^5*(128*cos(1/2*d*x+1/2*c)^7*a^2*b^4-256*cos(1/2*d*x+1/2*c)^5*a^2*b^4+160*cos(1/2
*d*x+1/2*c)^3*a^2*b^4+80*(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))*a^4*b^2-128*(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))*a^2*b^4-32*cos(1/2*d*x+1/2*c)*a^2*b^4+5*sum((a^4-2*a^2*b^2+b^4)/_alpha*(8*(e*(2*_alpha^2*b^
2+a^2-2*b^2)/b^2)^(1/2)*(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))*_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*cos(1/2*d*x+1/2*c)^2*a^2-3*b^2*cos(1/2*
d*x+1/2*c)^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))*(-sin(1/2*d*x+1/2*c)^2*e*(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)/(-sin(1/2*d*x+1/2*c)^2*e*(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))/a/b^6/(-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

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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((e*cos(d*x+c))^(9/2)/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

e^(9/2)*integrate(cos(d*x + c)^(9/2)/(b*sin(d*x + c) + a), x)

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Fricas [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((e*cos(d*x+c))^(9/2)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(9/2)/(a+b*sin(d*x+c)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8857 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(9/2)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(9/2)*e^(9/2)/(b*sin(d*x + c) + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}}{a+b\,\sin \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(9/2)/(a + b*sin(c + d*x)),x)

[Out]

int((e*cos(c + d*x))^(9/2)/(a + b*sin(c + d*x)), x)

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