3.3.73 \(\int (e \cos (c+d x))^{3/2} \sqrt {a+a \sin (c+d x)} \, dx\) [273]

Optimal. Leaf size=236 \[ -\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {3 e^{3/2} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d (1+\cos (c+d x)+\sin (c+d x))}+\frac {3 e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d (1+\cos (c+d x)+\sin (c+d x))} \]

[Out]

-1/2*a*(e*cos(d*x+c))^(5/2)/d/e/(a+a*sin(d*x+c))^(1/2)+3/4*e*(e*cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d-3/4
*e^(3/2)*arcsinh((e*cos(d*x+c))^(1/2)/e^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d/(1+cos(d*x+c)+sin
(d*x+c))+3/4*e^(3/2)*arctan(sin(d*x+c)*e^(1/2)/(e*cos(d*x+c))^(1/2)/(1+cos(d*x+c))^(1/2))*(1+cos(d*x+c))^(1/2)
*(a+a*sin(d*x+c))^(1/2)/d/(1+cos(d*x+c)+sin(d*x+c))

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Rubi [A]
time = 0.25, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2757, 2764, 2756, 2854, 209, 2912, 65, 221} \begin {gather*} \frac {3 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \text {ArcTan}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{4 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {3 e^{3/2} \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{4 d (\sin (c+d x)+\cos (c+d x)+1)}-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a \sin (c+d x)+a}}+\frac {3 e \sqrt {a \sin (c+d x)+a} \sqrt {e \cos (c+d x)}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*(e*Cos[c + d*x])^(5/2))/(d*e*Sqrt[a + a*Sin[c + d*x]]) + (3*e*Sqrt[e*Cos[c + d*x]]*Sqrt[a + a*Sin[c +
d*x]])/(4*d) - (3*e^(3/2)*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]
])/(4*d*(1 + Cos[c + d*x] + Sin[c + d*x])) + (3*e^(3/2)*ArcTan[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sq
rt[1 + Cos[c + d*x]])]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(4*d*(1 + Cos[c + d*x] + Sin[c + d*x])
)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 2756

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

Rule 2757

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

Rule 2764

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

Rule 2854

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-2*(b/f), Subst[Int[1/(b + d*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]))
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{3/2} \sqrt {a+a \sin (c+d x)} \, dx &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {1}{4} (3 a) \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {1}{8} \left (3 e^2\right ) \int \frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {\left (3 a e^2 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\sqrt {1+\cos (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx}{8 (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (3 a e^2 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}} \, dx}{8 (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {\left (3 a e^2 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {e x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{8 d (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {\left (3 a e^2 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1+e x^2} \, dx,x,-\frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right )}{4 d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {3 e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d (1+\cos (c+d x)+\sin (c+d x))}-\frac {\left (3 a e \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{e}}} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{4 d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {a (e \cos (c+d x))^{5/2}}{2 d e \sqrt {a+a \sin (c+d x)}}+\frac {3 e \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {3 e^{3/2} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d (1+\cos (c+d x)+\sin (c+d x))}+\frac {3 e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{4 d (1+\cos (c+d x)+\sin (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.99, size = 269, normalized size = 1.14 \begin {gather*} -\frac {i e e^{-i (c+d x)} \sqrt {e \cos (c+d x)} \left (-i \sqrt {1+e^{2 i (c+d x)}}-2 e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}}+2 i e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}}+e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}}-3 d e^{2 i (c+d x)} x+3 e^{2 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )-3 i e^{2 i (c+d x)} \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {a (1+\sin (c+d x))}}{4 d \left (i+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/4*I)*e*Sqrt[e*Cos[c + d*x]]*((-I)*Sqrt[1 + E^((2*I)*(c + d*x))] - 2*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c
+ d*x))] + (2*I)*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c
+ d*x))] - 3*d*E^((2*I)*(c + d*x))*x + 3*E^((2*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))] - (3*I)*E^((2*I)*(c + d*
x))*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]])*Sqrt[a*(1 + Sin[c + d*x])])/(d*E^(I*(c + d*x))*(I + E^(I*(c + d*x)
))*Sqrt[1 + E^((2*I)*(c + d*x))])

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Maple [A]
time = 6.41, size = 241, normalized size = 1.02

method	result	size
default	\(\frac {\left (3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sin \left (d x +c \right )-3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )-4 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 \left (\cos ^{3}\left (d x +c \right )\right )+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+6 \cos \left (d x +c \right )\right ) \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a \left (1+\sin \left (d x +c \right )\right )}}{8 d \left (\sin \left (d x +c \right )-\cos \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8/d*(3*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))
*sin(d*x+c)-3*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*si
n(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)-4*cos(d*x+c)^2*sin(d*x+c)-4*cos(d*x+c)^3+6*cos(d*x+c)*sin(d*x+c)-2*cos
(d*x+c)^2+6*cos(d*x+c))*(e*cos(d*x+c))^(3/2)*(a*(1+sin(d*x+c)))^(1/2)/(sin(d*x+c)-cos(d*x+c)+1)/cos(d*x+c)^2

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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))^(3/2)*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 3208 vs. \(2 (181) = 362\).
time = 191.68, size = 3208, normalized size = 13.59 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/32*(12*sqrt(2)*d*(a^2/d^4)^(1/4)*arctan(-1/4*(2*sqrt(1/2)*((sqrt(2)*d^3*cos(d*x + c)^6 + 5*sqrt(2)*d^3*cos(d
*x + c)^5 - 8*sqrt(2)*d^3*cos(d*x + c)^4 - 20*sqrt(2)*d^3*cos(d*x + c)^3 + 8*sqrt(2)*d^3*cos(d*x + c)^2 + 16*s
qrt(2)*d^3*cos(d*x + c) + (sqrt(2)*d^3*cos(d*x + c)^5 - 4*sqrt(2)*d^3*cos(d*x + c)^4 - 12*sqrt(2)*d^3*cos(d*x
+ c)^3 + 8*sqrt(2)*d^3*cos(d*x + c)^2 + 16*sqrt(2)*d^3*cos(d*x + c))*sin(d*x + c))*(a^2/d^4)^(3/4)*e^(9/2) + (
sqrt(2)*a*d*cos(d*x + c)^6*e^3 - 3*sqrt(2)*a*d*cos(d*x + c)^5*e^3 - 8*sqrt(2)*a*d*cos(d*x + c)^4*e^3 + 4*sqrt(
2)*a*d*cos(d*x + c)^3*e^3 + 8*sqrt(2)*a*d*cos(d*x + c)^2*e^3 - (sqrt(2)*a*d*cos(d*x + c)^5*e^3 + 4*sqrt(2)*a*d
*cos(d*x + c)^4*e^3 - 4*sqrt(2)*a*d*cos(d*x + c)^3*e^3 - 8*sqrt(2)*a*d*cos(d*x + c)^2*e^3)*sin(d*x + c))*(a^2/
d^4)^(1/4)*e^(3/2) - (a*cos(d*x + c)^4*e^(9/2) - 3*a*cos(d*x + c)^3*e^(9/2) - 8*a*cos(d*x + c)^2*e^(9/2) + 4*a
*cos(d*x + c)*e^(9/2) + (2*d^2*cos(d*x + c)^5*e^(3/2) - 5*d^2*cos(d*x + c)^4*e^(3/2) - 19*d^2*cos(d*x + c)^3*e
^(3/2) + 20*d^2*cos(d*x + c)*e^(3/2) + 8*d^2*e^(3/2) - (2*d^2*cos(d*x + c)^4*e^(3/2) + 9*d^2*cos(d*x + c)^3*e^
(3/2) - 4*d^2*cos(d*x + c)^2*e^(3/2) - 20*d^2*cos(d*x + c)*e^(3/2) - 8*d^2*e^(3/2))*sin(d*x + c))*sqrt(a^2/d^4
)*e^3 + 8*a*e^(9/2) - (a*cos(d*x + c)^3*e^(9/2) + 4*a*cos(d*x + c)^2*e^(9/2) - 4*a*cos(d*x + c)*e^(9/2) - 8*a*
e^(9/2))*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))*sqrt((2*a^3*cos(d*x + c)*e^9*sin(d*x + c)
+ 2*a^3*cos(d*x + c)*e^9 + (a^2*d^2*e^6*sin(d*x + c) + a^2*d^2*e^6)*sqrt(a^2/d^4)*e^3 + (sqrt(2)*a*d^3*(a^2/d^
4)^(3/4)*cos(d*x + c)*e^9 + (sqrt(2)*a^2*d*e^(15/2)*sin(d*x + c) + sqrt(2)*a^2*d*e^(15/2))*(a^2/d^4)^(1/4)*e^(
3/2))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))/(sin(d*x + c) + 1)) - ((7*sqrt(2)*a*d^3*cos(d*x + c)^4*e^(9
/2) + 3*sqrt(2)*a*d^3*cos(d*x + c)^3*e^(9/2) - 16*sqrt(2)*a*d^3*cos(d*x + c)^2*e^(9/2) - 4*sqrt(2)*a*d^3*cos(d
*x + c)*e^(9/2) + 8*sqrt(2)*a*d^3*e^(9/2) + (2*sqrt(2)*a*d^3*cos(d*x + c)^4*e^(9/2) + sqrt(2)*a*d^3*cos(d*x +
c)^3*e^(9/2) - 12*sqrt(2)*a*d^3*cos(d*x + c)^2*e^(9/2) - 4*sqrt(2)*a*d^3*cos(d*x + c)*e^(9/2) + 8*sqrt(2)*a*d^
3*e^(9/2))*sin(d*x + c))*(a^2/d^4)^(3/4)*e^(9/2) + (2*sqrt(2)*a^2*d*cos(d*x + c)^5*e^(15/2) + sqrt(2)*a^2*d*co
s(d*x + c)^4*e^(15/2) - 13*sqrt(2)*a^2*d*cos(d*x + c)^3*e^(15/2) - 8*sqrt(2)*a^2*d*cos(d*x + c)^2*e^(15/2) + 1
2*sqrt(2)*a^2*d*cos(d*x + c)*e^(15/2) + 8*sqrt(2)*a^2*d*e^(15/2) - (7*sqrt(2)*a^2*d*cos(d*x + c)^3*e^(15/2) +
4*sqrt(2)*a^2*d*cos(d*x + c)^2*e^(15/2) - 12*sqrt(2)*a^2*d*cos(d*x + c)*e^(15/2) - 8*sqrt(2)*a^2*d*e^(15/2))*s
in(d*x + c))*(a^2/d^4)^(1/4)*e^(3/2))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))/(a^3*cos(d*x + c)^6*e^9 + a
^3*cos(d*x + c)^5*e^9 - 8*a^3*cos(d*x + c)^4*e^9 - 8*a^3*cos(d*x + c)^3*e^9 + 8*a^3*cos(d*x + c)^2*e^9 + 8*a^3
*cos(d*x + c)*e^9 - 4*(a^3*cos(d*x + c)^4*e^9 + a^3*cos(d*x + c)^3*e^9 - 2*a^3*cos(d*x + c)^2*e^9 - 2*a^3*cos(
d*x + c)*e^9)*sin(d*x + c)))*e^(3/2) - 12*sqrt(2)*d*(a^2/d^4)^(1/4)*arctan(1/4*(2*sqrt(1/2)*((sqrt(2)*d^3*cos(
d*x + c)^6 + 5*sqrt(2)*d^3*cos(d*x + c)^5 - 8*sqrt(2)*d^3*cos(d*x + c)^4 - 20*sqrt(2)*d^3*cos(d*x + c)^3 + 8*s
qrt(2)*d^3*cos(d*x + c)^2 + 16*sqrt(2)*d^3*cos(d*x + c) + (sqrt(2)*d^3*cos(d*x + c)^5 - 4*sqrt(2)*d^3*cos(d*x
+ c)^4 - 12*sqrt(2)*d^3*cos(d*x + c)^3 + 8*sqrt(2)*d^3*cos(d*x + c)^2 + 16*sqrt(2)*d^3*cos(d*x + c))*sin(d*x +
c))*(a^2/d^4)^(3/4)*e^(9/2) + (sqrt(2)*a*d*cos(d*x + c)^6*e^3 - 3*sqrt(2)*a*d*cos(d*x + c)^5*e^3 - 8*sqrt(2)*
a*d*cos(d*x + c)^4*e^3 + 4*sqrt(2)*a*d*cos(d*x + c)^3*e^3 + 8*sqrt(2)*a*d*cos(d*x + c)^2*e^3 - (sqrt(2)*a*d*co
s(d*x + c)^5*e^3 + 4*sqrt(2)*a*d*cos(d*x + c)^4*e^3 - 4*sqrt(2)*a*d*cos(d*x + c)^3*e^3 - 8*sqrt(2)*a*d*cos(d*x
+ c)^2*e^3)*sin(d*x + c))*(a^2/d^4)^(1/4)*e^(3/2) + (a*cos(d*x + c)^4*e^(9/2) - 3*a*cos(d*x + c)^3*e^(9/2) -
8*a*cos(d*x + c)^2*e^(9/2) + 4*a*cos(d*x + c)*e^(9/2) + (2*d^2*cos(d*x + c)^5*e^(3/2) - 5*d^2*cos(d*x + c)^4*e
^(3/2) - 19*d^2*cos(d*x + c)^3*e^(3/2) + 20*d^2*cos(d*x + c)*e^(3/2) + 8*d^2*e^(3/2) - (2*d^2*cos(d*x + c)^4*e
^(3/2) + 9*d^2*cos(d*x + c)^3*e^(3/2) - 4*d^2*cos(d*x + c)^2*e^(3/2) - 20*d^2*cos(d*x + c)*e^(3/2) - 8*d^2*e^(
3/2))*sin(d*x + c))*sqrt(a^2/d^4)*e^3 + 8*a*e^(9/2) - (a*cos(d*x + c)^3*e^(9/2) + 4*a*cos(d*x + c)^2*e^(9/2) -
4*a*cos(d*x + c)*e^(9/2) - 8*a*e^(9/2))*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))*sqrt((2*a^
3*cos(d*x + c)*e^9*sin(d*x + c) + 2*a^3*cos(d*x + c)*e^9 + (a^2*d^2*e^6*sin(d*x + c) + a^2*d^2*e^6)*sqrt(a^2/d
^4)*e^3 - (sqrt(2)*a*d^3*(a^2/d^4)^(3/4)*cos(d*x + c)*e^9 + (sqrt(2)*a^2*d*e^(15/2)*sin(d*x + c) + sqrt(2)*a^2
*d*e^(15/2))*(a^2/d^4)^(1/4)*e^(3/2))*sqrt(a*sin(d*x + c) + a)*sqrt(cos(d*x + c)))/(sin(d*x + c) + 1)) - ((7*s
qrt(2)*a*d^3*cos(d*x + c)^4*e^(9/2) + 3*sqrt(2)*a*d^3*cos(d*x + c)^3*e^(9/2) - 16*sqrt(2)*a*d^3*cos(d*x + c)^2
*e^(9/2) - 4*sqrt(2)*a*d^3*cos(d*x + c)*e^(9/2) + 8*sqrt(2)*a*d^3*e^(9/2) + (2*sqrt(2)*a*d^3*cos(d*x + c)^4*e^
(9/2) + sqrt(2)*a*d^3*cos(d*x + c)^3*e^(9/2) - 12*sqrt(2)*a*d^3*cos(d*x + c)^2*e^(9/2) - 4*sqrt(2)*a*d^3*cos(d
*x + c)*e^(9/2) + 8*sqrt(2)*a*d^3*e^(9/2))*sin(...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(3/2)*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*(e*cos(c + d*x))**(3/2), x)

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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))^(3/2)*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\sqrt {a+a\,\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))^(3/2)*(a + a*sin(c + d*x))^(1/2),x)

[Out]

int((e*cos(c + d*x))^(3/2)*(a + a*sin(c + d*x))^(1/2), x)

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