3.3.75 \(\int \frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx\) [275]

Optimal. Leaf size=161 \[ -\frac {2 \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \sqrt {e} (1+\cos (c+d x)+\sin (c+d x))}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d \sqrt {e} (1+\cos (c+d x)+\sin (c+d x))} \]

[Out]

-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
))/e^(1/2)+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*si
n(d*x+c))^(1/2)/d/(1+cos(d*x+c)+sin(d*x+c))/e^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2756, 2854, 209, 2912, 65, 221} \begin {gather*} \frac {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 )}{d \sqrt {e} (\sin (c+d x)+\cos (c+d x)+1)}-\frac {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 )}{d \sqrt {e} (\sin (c+d x)+\cos (c+d x)+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sin[c + d*x]]/Sqrt[e*Cos[c + d*x]],x]

[Out]

(-2*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*Sqrt[e]*(1 + Cos
[c + d*x] + Sin[c + d*x])) + (2*ArcTan[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])]*S
qrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*Sqrt[e]*(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 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 \frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx &=\frac {\left (a \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}{a+a \cos (c+d x)+a \sin (c+d x)}+\frac {\left (a \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}{a+a \cos (c+d x)+a \sin (c+d x)}\\ &=-\frac {\left (a \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 )}{d (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {\left (2 a \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 )}{d (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac {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)}}{d \sqrt {e} (1+\cos (c+d x)+\sin (c+d x))}-\frac {\left (2 a \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 )}{d e (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=-\frac {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)}}{d \sqrt {e} (1+\cos (c+d x)+\sin (c+d x))}+\frac {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)}}{d \sqrt {e} (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 = 0.48, size = 108, normalized size = 0.67 \begin {gather*} \frac {\sqrt {1+e^{2 i (c+d x)}} \left (d x-\sinh ^{-1}\left (e^{i (c+d x)}\right )+i \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {a (1+\sin (c+d x))}}{d \left (1-i e^{i (c+d x)}\right ) \sqrt {e \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sin[c + d*x]]/Sqrt[e*Cos[c + d*x]],x]

[Out]

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

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Maple [A]
time = 0.16, size = 142, normalized size = 0.88

method	result	size
default	\(-\frac {\left (\arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )-\arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )\right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, \left (\cos \left (d x +c \right )-1+\sin \left (d x +c \right )\right ) \sqrt {2}}{d \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {e \cos \left (d x +c \right )}}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*sin(d*x+c)/cos(d*x+c)*2^(1/2)))*(a*(1+sin(d*x+c)))^(1/2)*(cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)/(e*cos(d*x+c))^(1/2)*2^(1/2)

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 3105 vs. \(2 (126) = 252\).
time = 178.64, size = 3105, normalized size = 19.29 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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