3.3.84 \(\int \frac {a+a \sec (c+d x)}{\sqrt {e \csc (c+d x)}} \, dx\) [284]

Optimal. Leaf size=122 \[ -\frac {a \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}} \]

[Out]

-a*arctan(sin(d*x+c)^(1/2))/d/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)+a*arctanh(sin(d*x+c)^(1/2))/d/(e*csc(d*x+c
))^(1/2)/sin(d*x+c)^(1/2)-2*a*(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)*EllipticE(cos(1/2*
c+1/4*Pi+1/2*d*x),2^(1/2))/d/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3963, 3957, 2917, 2644, 335, 304, 209, 212, 2719} \begin {gather*} -\frac {a \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {a \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {2 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])/Sqrt[e*Csc[c + d*x]],x]

[Out]

-((a*ArcTan[Sqrt[Sin[c + d*x]]])/(d*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]])) + (a*ArcTanh[Sqrt[Sin[c + d*x]]]
)/(d*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (2*a*EllipticE[(c - Pi/2 + d*x)/2, 2])/(d*Sqrt[e*Csc[c + d*x]]
*Sqrt[Sin[c + d*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 212

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

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

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 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 3963

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*((g_.)*sec[(e_.) + (f_.)*(x_)])^(p_), x_Symbol] :> Dist[g^Int
Part[p]*(g*Sec[e + f*x])^FracPart[p]*Cos[e + f*x]^FracPart[p], Int[(a + b*Csc[e + f*x])^m/Cos[e + f*x]^p, x],
x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {a+a \sec (c+d x)}{\sqrt {e \csc (c+d x)}} \, dx &=\frac {\int (a+a \sec (c+d x)) \sqrt {\sin (c+d x)} \, dx}{\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=-\frac {\int (-a-a \cos (c+d x)) \sec (c+d x) \sqrt {\sin (c+d x)} \, dx}{\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {a \int \sqrt {\sin (c+d x)} \, dx}{\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a \int \sec (c+d x) \sqrt {\sin (c+d x)} \, dx}{\sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a \text {Subst}\left (\int \frac {\sqrt {x}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {(2 a) \text {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {a \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=-\frac {a \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {a \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.57, size = 130, normalized size = 1.07 \begin {gather*} \frac {a \left (-4 \cot (c+d x) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};\csc ^2(c+d x)\right )+\sqrt {-\cot ^2(c+d x)} \sqrt {\csc (c+d x)} \left (2 \text {ArcTan}\left (\sqrt {\csc (c+d x)}\right )-\log \left (1-\sqrt {\csc (c+d x)}\right )+\log \left (1+\sqrt {\csc (c+d x)}\right )\right )\right )}{2 d \sqrt {-\cot ^2(c+d x)} \sqrt {e \csc (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])/Sqrt[e*Csc[c + d*x]],x]

[Out]

(a*(-4*Cot[c + d*x]*Hypergeometric2F1[-1/4, 1/2, 3/4, Csc[c + d*x]^2] + Sqrt[-Cot[c + d*x]^2]*Sqrt[Csc[c + d*x
]]*(2*ArcTan[Sqrt[Csc[c + d*x]]] - Log[1 - Sqrt[Csc[c + d*x]]] + Log[1 + Sqrt[Csc[c + d*x]]])))/(2*d*Sqrt[-Cot
[c + d*x]^2]*Sqrt[e*Csc[c + d*x]])

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Maple [C] Result contains complex when optimal does not.
time = 0.21, size = 1535, normalized size = 12.58

method	result	size
default	\(\text {Expression too large to display}\)	\(1535\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a/d*(I*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(
I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2+1/2
*I,1/2*2^(1/2))-I*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1
/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),
1/2-1/2*I,1/2*2^(1/2))+cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c
))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(
1/2),1/2+1/2*I,1/2*2^(1/2))+cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(
d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+
c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-4*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-
I)/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticE(((I*cos(d*x+c)+sin(d*x+c)-I)/si
n(d*x+c))^(1/2),1/2*2^(1/2))+2*cos(d*x+c)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((I*cos(d*x+c)+sin(d*x+c)-I)/s
in(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*
x+c))^(1/2),1/2*2^(1/2))+I*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*
x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-I*(-1+cos(d*x+c
))/sin(d*x+c))^(1/2)-I*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c)
)^(1/2)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/s
in(d*x+c))^(1/2)+((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2
)*EllipticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x
+c))^(1/2)+((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*Elli
pticPi(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(
1/2)-4*((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*Elliptic
E(((I*cos(d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)+2*((I*cos(
d*x+c)+sin(d*x+c)-I)/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-sin(d*x+c)-I)/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c
)+sin(d*x+c)-I)/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)-2*2^(1/2)*cos(d*x+c)+2*2^
(1/2))/(e/sin(d*x+c))^(1/2)/sin(d*x+c)*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*sec(d*x+c))/(e*csc(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.42, size = 206, normalized size = 1.69 \begin {gather*} \frac {{\left (4 \, \sqrt {2 i} a {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 4 \, \sqrt {-2 i} a {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, a \arctan \left (\frac {76 \, \cos \left (d x + c\right )^{2} + \frac {425 \, {\left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 1\right )}}{\sqrt {\sin \left (d x + c\right )}} - 152 \, \sin \left (d x + c\right ) - 152}{2 \, {\left (16 \, \cos \left (d x + c\right )^{2} + 393 \, \sin \left (d x + c\right ) - 32\right )}}\right ) + a \log \left (\frac {\cos \left (d x + c\right )^{2} + \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right ) - 1\right )}}{\sqrt {\sin \left (d x + c\right )}} - 6 \, \sin \left (d x + c\right ) - 2}{\cos \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) - 2}\right )\right )} e^{\left (-\frac {1}{2}\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*sqrt(2*I)*a*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 4*sqrt(-2
*I)*a*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*a*arctan(1/2*(76*cos
(d*x + c)^2 + 425*(cos(d*x + c)^2 + sin(d*x + c) - 1)/sqrt(sin(d*x + c)) - 152*sin(d*x + c) - 152)/(16*cos(d*x
+ c)^2 + 393*sin(d*x + c) - 32)) + a*log((cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - sin(d*x + c) - 1)/sqrt(sin(d*x
+ c)) - 6*sin(d*x + c) - 2)/(cos(d*x + c)^2 + 2*sin(d*x + c) - 2)))*e^(-1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*csc(d*x+c))**(1/2),x)

[Out]

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

[Out]

integrate((a*sec(d*x + c) + a)/sqrt(e*csc(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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