3.586 \(\int \frac{(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=473 \[ -\frac{5 a e^{7/2} \sqrt [4]{b^2-a^2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{7/2} d}-\frac{5 a e^{7/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{7/2} d}+\frac{5 e^4 \left (3 a^2-b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d \sqrt{e \cos (c+d x)}}-\frac{5 a^2 e^4 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{2 b^4 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \cos (c+d x)}}-\frac{5 a^2 e^4 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{2 b^4 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac{e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.11844, antiderivative size = 473, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {2693, 2865, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac{5 a e^{7/2} \sqrt [4]{b^2-a^2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{7/2} d}-\frac{5 a e^{7/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{7/2} d}+\frac{5 e^4 \left (3 a^2-b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d \sqrt{e \cos (c+d x)}}-\frac{5 a^2 e^4 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{2 b^4 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \cos (c+d x)}}-\frac{5 a^2 e^4 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{2 b^4 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac{e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(7/2)/(a + b*Sin[c + d*x])^2,x]

[Out]

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

Rule 2693

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

Rule 2865

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 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]

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 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 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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^2} \, dx &=-\frac{e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac{\left (5 e^2\right ) \int \frac{(e \cos (c+d x))^{3/2} \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b}\\ &=\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac{e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac{\left (5 e^4\right ) \int \frac{-a b-\frac{1}{2} \left (3 a^2-b^2\right ) \sin (c+d x)}{\sqrt{e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{3 b^3}\\ &=\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac{e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac{\left (5 a \left (a^2-b^2\right ) e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{2 b^4}+\frac{\left (5 \left (3 a^2-b^2\right ) e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{6 b^4}\\ &=\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac{e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac{\left (5 a^2 \sqrt{-a^2+b^2} e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} \left (\sqrt{-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{4 b^4}-\frac{\left (5 a^2 \sqrt{-a^2+b^2} e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} \left (\sqrt{-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{4 b^4}-\frac{\left (5 a \left (a^2-b^2\right ) e^5\right ) \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 )}{2 b^3 d}+\frac{\left (5 \left (3 a^2-b^2\right ) e^4 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 b^4 \sqrt{e \cos (c+d x)}}\\ &=\frac{5 \left (3 a^2-b^2\right ) e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d \sqrt{e \cos (c+d x)}}+\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac{e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac{\left (5 a \left (a^2-b^2\right ) e^5\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 )}{b^3 d}-\frac{\left (5 a^2 \sqrt{-a^2+b^2} e^4 \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}{4 b^4 \sqrt{e \cos (c+d x)}}-\frac{\left (5 a^2 \sqrt{-a^2+b^2} e^4 \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}{4 b^4 \sqrt{e \cos (c+d x)}}\\ &=\frac{5 \left (3 a^2-b^2\right ) e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d \sqrt{e \cos (c+d x)}}+\frac{5 a^2 \sqrt{-a^2+b^2} e^4 \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 )}{2 b^4 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}-\frac{5 a^2 \sqrt{-a^2+b^2} e^4 \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 )}{2 b^4 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}+\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac{e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}-\frac{\left (5 a \sqrt{-a^2+b^2} e^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e-b x^2} \, dx,x,\sqrt{e \cos (c+d x)}\right )}{2 b^3 d}-\frac{\left (5 a \sqrt{-a^2+b^2} e^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e+b x^2} \, dx,x,\sqrt{e \cos (c+d x)}\right )}{2 b^3 d}\\ &=-\frac{5 a \sqrt [4]{-a^2+b^2} e^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{2 b^{7/2} d}-\frac{5 a \sqrt [4]{-a^2+b^2} e^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{2 b^{7/2} d}+\frac{5 \left (3 a^2-b^2\right ) e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^4 d \sqrt{e \cos (c+d x)}}+\frac{5 a^2 \sqrt{-a^2+b^2} e^4 \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 )}{2 b^4 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}-\frac{5 a^2 \sqrt{-a^2+b^2} e^4 \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 )}{2 b^4 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}+\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a-b \sin (c+d x))}{3 b^3 d}-\frac{e (e \cos (c+d x))^{5/2}}{b d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 27.3326, size = 1956, normalized size = 4.14 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Cos[c + d*x])^(7/2)/(a + b*Sin[c + d*x])^2,x]

[Out]

((e*Cos[c + d*x])^(7/2)*Sec[c + d*x]^3*((-2*Sin[c + d*x])/(3*b^2) + (a^2 - b^2)/(b^3*(a + b*Sin[c + d*x]))))/d
+ ((e*Cos[c + d*x])^(7/2)*((-8*a*b*(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[C
os[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])) + (6
*a*b*(a + b*Sqrt[1 - Cos[c + d*x]^2])*Cos[2*(c + d*x)]*(((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 - ((1 + I)*Sqrt[b
]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1
+ ((1 + I)*Sqrt[b]*Sqrt[Cos[c + d*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) + (4*Sqrt[Cos[c + d*
x]])/b - (4*a*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]^(5/2)
)/(5*(a^2 - b^2)) + (10*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)])*Co
s[c + d*x]^2)*(a^2 + b^2*(-1 + Cos[c + d*x]^2))) + ((1/4 - I/4)*(-2*a^2 + b^2)*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]])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/4 - I/4)
*(-2*a^2 + b^2)*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
]])/(b^(3/2)*(-a^2 + b^2)^(3/4)))*Sin[c + d*x])/(Sqrt[1 - Cos[c + d*x]^2]*(-1 + 2*Cos[c + d*x]^2)*(a + b*Sin[c
+ d*x])) - (2*(3*a^2 - 5*b^2)*(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*ArcTa
n[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*Sq
rt[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]))))/(6*b^3*d*Cos[c
+ d*x]^(7/2))

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Maple [C]  time = 8.384, size = 14392, normalized size = 30.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^2,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(7/2)/(b*sin(d*x + c) + a)^2, x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(7/2)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out