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

Optimal. Leaf size=410 \[ -\frac {a e^2 \left (a^2-b^2\right ) \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{b^2 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}-\frac {a e^2 \left (a^2-b^2\right ) \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{b^2 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {e^{3/2} \sqrt [4]{b^2-a^2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}+\frac {e^{3/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}+\frac {2 a e^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{b^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d} \]

[Out]

(-a^2+b^2)^(1/4)*e^(3/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/d+(-a^2+b^2)^(1
/4)*e^(3/2)*arctanh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/b^(3/2)/d-2*a*e^2*(sin(1/2*c+1/4*Pi
+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/b^2
/d/(e*sin(d*x+c))^(1/2)+a*(a^2-b^2)*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Elliptic
Pi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b^2/d/(a^2-b*(b-(-a^2+b^2)^(1/
2)))/(e*sin(d*x+c))^(1/2)+a*(a^2-b^2)*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Ellipt
icPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b^2/d/(a^2-b*(b+(-a^2+b^2)^(
1/2)))/(e*sin(d*x+c))^(1/2)-2*e*(e*sin(d*x+c))^(1/2)/b/d

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Rubi [A]  time = 0.90, antiderivative size = 410, 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 = {2695, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac {e^{3/2} \sqrt [4]{b^2-a^2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}+\frac {e^{3/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac {a e^2 \left (a^2-b^2\right ) \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{b^2 d \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}-\frac {a e^2 \left (a^2-b^2\right ) \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{b^2 d \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {2 a e^2 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{b^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e \sqrt {e \sin (c+d x)}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2695

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

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Mathematica [C]  time = 5.82, size = 434, normalized size = 1.06 \[ -\frac {\left (\frac {1}{20}-\frac {i}{20}\right ) \cos (c+d x) (e \sin (c+d x))^{3/2} \left (a+b \sqrt {\cos ^2(c+d x)}\right ) \left ((4+4 i) a b^{3/2} \sin ^{\frac {5}{2}}(c+d x) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right )-5 \left (a^2-b^2\right ) \left (\sqrt [4]{b^2-a^2} \log \left (-(1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}+i b \sin (c+d x)\right )-\sqrt [4]{b^2-a^2} \log \left ((1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}+i b \sin (c+d x)\right )+2 \sqrt [4]{b^2-a^2} \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \sqrt [4]{b^2-a^2} \tan ^{-1}\left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )+(4+4 i) \sqrt {b} \sqrt {\sin (c+d x)}\right )\right )}{b^{3/2} d \left (b^2-a^2\right ) \sin ^{\frac {3}{2}}(c+d x) \sqrt {\cos ^2(c+d x)} (a+b \cos (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/20 + I/20)*Cos[c + d*x]*(a + b*Sqrt[Cos[c + d*x]^2])*(e*Sin[c + d*x])^(3/2)*(-5*(a^2 - b^2)*(2*(-a^2 + b^
2)^(1/4)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - 2*(-a^2 + b^2)^(1/4)*ArcTan[1 +
((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] + (-a^2 + b^2)^(1/4)*Log[Sqrt[-a^2 + b^2] - (1 + I)*
Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[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[Sin[c + d*x]] + I*b*Sin[c + d*x]] + (4 + 4*I)*Sqrt[b]*Sqrt[Sin[c + d*x
]]) + (4 + 4*I)*a*b^(3/2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c
+ d*x]^(5/2)))/(b^(3/2)*(-a^2 + b^2)*d*Sqrt[Cos[c + d*x]^2]*(a + b*Cos[c + d*x])*Sin[c + d*x]^(3/2))

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.87, size = 1314, normalized size = 3.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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