3.582 \(\int \frac {1}{(e \cos (c+d x))^{5/2} (a+b \sin (c+d x))} \, dx\)

Optimal. Leaf size=434 \[ \frac {2 a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \left (a^2-b^2\right ) \sqrt {e \cos (c+d x)}}-\frac {a b^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {a b^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {2 (b-a \sin (c+d x))}{3 d e \left (a^2-b^2\right ) (e \cos (c+d x))^{3/2}}-\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}}-\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}} \]

[Out]

-b^(5/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/(-a^2+b^2)^(7/4)/d/e^(5/2)-b^(5/2)*arct
anh(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/(-a^2+b^2)^(7/4)/d/e^(5/2)-2/3*(b-a*sin(d*x+c))/(a^
2-b^2)/d/e/(e*cos(d*x+c))^(3/2)+2/3*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/
2*c),2^(1/2))*cos(d*x+c)^(1/2)/(a^2-b^2)/d/e^2/(e*cos(d*x+c))^(1/2)-a*b^2*(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)/(a^2-b^2)/d/e^2/(
a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)-a*b^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti
cPi(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)/(a^2-b^2)/d/e^2/(a^2-b*(b+(-a^2+b^2)
^(1/2)))/(e*cos(d*x+c))^(1/2)

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Rubi [A]  time = 1.00, antiderivative size = 434, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2696, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}}-\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {2 a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d e^2 \left (a^2-b^2\right ) \sqrt {e \cos (c+d x)}}-\frac {a b^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {a b^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {2 (b-a \sin (c+d x))}{3 d e \left (a^2-b^2\right ) (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

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 212

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

Rule 329

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 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2702

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

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 2807

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*
Sin[e + f*x])/(c + d)]), 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 2867

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

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Mathematica [C]  time = 24.60, size = 1192, normalized size = 2.75 \[ \frac {\left (-\frac {2 a b \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {1-\cos ^2(c+d x)} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )}{\left (2 \left (2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) b^2+\left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )\right ) \cos ^2(c+d x)-5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )\right ) \left (a^2+b^2 \left (\cos ^2(c+d x)-1\right )\right )}+\frac {a \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 (\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \cos (c+d x)-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+\sqrt {a^2-b^2}\right )+\log \left (b \cos (c+d x)+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+\sqrt {a^2-b^2}\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}\right ) \sin ^2(c+d x)}{\left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))}-\frac {2 \left (a^2-3 b^2\right ) \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) \sqrt {\cos (c+d x)}}{\sqrt {1-\cos ^2(c+d x)} \left (5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )-2 \left (2 F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right ) b^2+\left (b^2-a^2\right ) F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )\right ) \cos ^2(c+d x)\right ) \left (a^2+b^2 \left (\cos ^2(c+d x)-1\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {b} \left (2 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac {(1+i) \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )+\log \left (i b \cos (c+d x)-(1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\cos (c+d x)}+\sqrt {b^2-a^2}\right )-\log \left (i b \cos (c+d x)+(1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\cos (c+d x)}+\sqrt {b^2-a^2}\right )\right )}{\left (b^2-a^2\right )^{3/4}}\right ) \sin (c+d x)}{\sqrt {1-\cos ^2(c+d x)} (a+b \sin (c+d x))}\right ) \cos ^{\frac {5}{2}}(c+d x)}{3 (a-b) (a+b) d (e \cos (c+d x))^{5/2}}+\frac {2 (a \sin (c+d x)-b) \cos (c+d x)}{3 \left (a^2-b^2\right ) d (e \cos (c+d x))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])),x]

[Out]

(2*Cos[c + d*x]*(-b + a*Sin[c + d*x]))/(3*(a^2 - b^2)*d*(e*Cos[c + d*x])^(5/2)) + (Cos[c + d*x]^(5/2)*((-2*(a^
2 - 3*b^2)*(a + b*Sqrt[1 - Cos[c + d*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*
Cos[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[c + d*x]])/(Sqrt[1 - Cos[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2,
1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[c + d*x]^
2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d
*x]^2)/(-a^2 + b^2)])*Cos[c + d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x]^2))) - ((1/8 - I/8)*Sqrt[b]*(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]]))/(
-a^2 + b^2)^(3/4))*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*x]^2]*(a + b*Sin[c + d*x])) - (2*a*b*(a + b*Sqrt[1 - Cos[
c + d*x]^2])*((5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)]*
Sqrt[Cos[c + d*x]]*Sqrt[1 - Cos[c + d*x]^2])/((-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Cos[c + d*x]^2, (b^2
*Cos[c + d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a
^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Cos[c + d*x]^2, (b^2*Cos[c + d*x]^2)/(-a^2 + b^2)])*Cos[c
+ d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x]^2))) + (a*(-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]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)))*Sin[c + d*x]^2)/(
(1 - Cos[c + d*x]^2)*(a + b*Sin[c + d*x]))))/(3*(a - b)*(a + b)*d*(e*Cos[c + d*x])^(5/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 4.96, size = 1083, normalized size = 2.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/e*b^3/(a-b)/(a+b)*sum((_R^4+_R^2*e)/(_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))-1/12/d/e^3*b/(a^2-b^2)/(cos(1/2*d*x+1/2*c)+1/2*2^(1/2))^2*(-
2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)-1/12/d/e^3*b*2^(1/2)/(a^2-b^2)/(cos(1/2*d*x+1/2*c)+1/2*2^(1/2))*(-2*sin(1/2*
d*x+1/2*c)^2*e+e)^(1/2)-1/12/d/e^3*b/(a^2-b^2)/(cos(1/2*d*x+1/2*c)-1/2*2^(1/2))^2*(-2*sin(1/2*d*x+1/2*c)^2*e+e
)^(1/2)+1/12/d/e^3*b*2^(1/2)/(a^2-b^2)/(cos(1/2*d*x+1/2*c)-1/2*2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)+1/
8/d*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a/e^2/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c
)^2-1))^(1/2)/(a-b)/(a+b)*sum(1/_alpha/(2*_alpha^2-1)*(2^(1/2)/(e*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctan
h(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))+8*b^2/a^2*_alpha*(_alpha^2-1)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(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)*EllipticPi(cos(1/2*d*x+1/2*c),-4*b^2/a^2*(_alpha^2-1),2^(1/2)))
,_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2))+1/3/d*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a/e
^3/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-e*(2*sin(1/2*d*x+1/2
*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2-2/3/d*(e*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*
x+1/2*c)^2)^(1/2)*a/e^2/sin(1/2*d*x+1/2*c)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/(a^2-b^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)/(-e*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)*EllipticF(
cos(1/2*d*x+1/2*c),2^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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