3.588 \(\int \frac {(e \cos (c+d x))^{3/2}}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=404 \[ \frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {a e^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} d \left (b^2-a^2\right )^{3/4}}-\frac {a e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} d \left (b^2-a^2\right )^{3/4}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {e \cos (c+d x)}} \]

[Out]

-1/2*a*e^(3/2)*arctan(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/(-a^2+b^2)^(3/4)/d-1/2*a*
e^(3/2)*arctanh(b^(1/2)*(e*cos(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/(-a^2+b^2)^(3/4)/d-e^2*(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)/b^2/d/(e*cos(d*
x+c))^(1/2)+1/2*a^2*e^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)/b^2/d/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)+1/2*a^2*e^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)/b^2/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*cos(d*x+c))^(1/2)-e*(e*cos(d*x+c))^(1/2)/b/d/(a+b*si
n(d*x+c))

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Rubi [A]  time = 0.89, antiderivative size = 404, 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 = {2693, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac {a e^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} d \left (b^2-a^2\right )^{3/4}}-\frac {a e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \cos (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 b^{3/2} d \left (b^2-a^2\right )^{3/4}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \cos (c+d x)}}+\frac {a^2 e^2 \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (c+d x)\right |2\right )}{2 b^2 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \cos (c+d x)}}-\frac {e \sqrt {e \cos (c+d x)}}{b d (a+b \sin (c+d x))}-\frac {e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(3/2)/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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

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Mathematica [C]  time = 12.53, size = 614, normalized size = 1.52 \[ \frac {\sin ^2(c+d x) (e \cos (c+d x))^{3/2} \left (a+b \sqrt {1-\cos ^2(c+d x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {1-\cos ^2(c+d x)} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )}{\left (a^2+b^2 \left (\cos ^2(c+d x)-1\right )\right ) \left (2 \cos ^2(c+d x) \left (2 b^2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )+\left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )\right )-5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{b^2-a^2}\right )\right )}+\frac {a \left (-\log \left (-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+\sqrt {a^2-b^2}+b \cos (c+d x)\right )+\log \left (\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (c+d x)}+\sqrt {a^2-b^2}+b \cos (c+d x)\right )-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}\right )}{b d \cos ^{\frac {3}{2}}(c+d x) \left (1-\cos ^2(c+d x)\right ) (a+b \sin (c+d x))}-\frac {\sec (c+d x) (e \cos (c+d x))^{3/2}}{b d (a+b \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Cos[c + d*x])^(3/2)/(a + b*Sin[c + d*x])^2,x]

[Out]

-(((e*Cos[c + d*x])^(3/2)*Sec[c + d*x])/(b*d*(a + b*Sin[c + d*x]))) + ((e*Cos[c + d*x])^(3/2)*(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)])*Co
s[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)/(b*d*Cos[c + d*x]^(3/2)*(1 - Cos[c + d*x]^2)*(a + b*Sin[c + d*x]))

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fricas [F]  time = 123.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e \cos \left (d x + c\right )} e \cos \left (d x + c\right )}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))*e*cos(d*x + c)/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2), x)

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giac [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((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 8.24, size = 9301, normalized size = 23.02 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^2,x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(3/2)/(b*sin(d*x + c) + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(3/2)/(a + b*sin(c + d*x))^2,x)

[Out]

int((e*cos(c + d*x))^(3/2)/(a + b*sin(c + d*x))^2, 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((e*cos(d*x+c))**(3/2)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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