3.284 \(\int \frac {a+a \sec (c+d x)}{\sqrt {e \csc (c+d x)}} \, dx\)
Optimal. Leaf size=122 \[ -\frac {a \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {a \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {2 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]
[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)
________________________________________________________________________________________
Rubi [A] time = 0.15, 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 =
{3878, 3872, 2838, 2564, 329, 298, 203, 206, 2639} \[ -\frac {a \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {a \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}+\frac {2 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]
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 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 298
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 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 2564
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 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rule 2838
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 3872
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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 3878
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 \operatorname {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) \operatorname {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 \operatorname {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 \operatorname {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*}
________________________________________________________________________________________
Mathematica [C] time = 0.83, size = 130, normalized size = 1.07 \[ \frac {a \left (\sqrt {-\cot ^2(c+d x)} \sqrt {\csc (c+d x)} \left (-\log \left (1-\sqrt {\csc (c+d x)}\right )+\log \left (\sqrt {\csc (c+d x)}+1\right )+2 \tan ^{-1}\left (\sqrt {\csc (c+d x)}\right )\right )-4 \cot (c+d x) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};\csc ^2(c+d x)\right )\right )}{2 d \sqrt {-\cot ^2(c+d x)} \sqrt {e \csc (c+d x)}} \]
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]])
________________________________________________________________________________________
fricas [F] time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \csc \left (d x + c\right )} {\left (a \sec \left (d x + c\right ) + a\right )}}{e \csc \left (d x + c\right )}, x\right ) \]
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]
integral(sqrt(e*csc(d*x + c))*(a*sec(d*x + c) + a)/(e*csc(d*x + c)), x)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sec \left (d x + c\right ) + a}{\sqrt {e \csc \left (d x + c\right )}}\,{d x} \]
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)
________________________________________________________________________________________
maple [C] time = 1.44, size = 1535, normalized size = 12.58 \[ \text {result too large to display} \]
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)
[Out]
1/2*a/d*(I*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))
^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/si
n(d*x+c))^(1/2)-I*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(
d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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)+cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))
/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-I*(-1+co
s(d*x+c))/sin(d*x+c))^(1/2)+cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*
x+c))/sin(d*x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I
-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticE(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+c
os(d*x+c))/sin(d*x+c))^(1/2)+2*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin
(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d
*x+c))/sin(d*x+c))^(1/2)+I*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*
x+c))^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)
)^(1/2)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2
)*EllipticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*Elli
pticPi(((I*cos(d*x+c)-I+sin(d*x+c))/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)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*Elliptic
E(((I*cos(d*x+c)-I+sin(d*x+c))/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)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c
)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)-2*cos(d*x+c)*2^(1/2)+2*2^
(1/2))/(e/sin(d*x+c))^(1/2)/sin(d*x+c)*2^(1/2)
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sec \left (d x + c\right ) + a}{\sqrt {e \csc \left (d x + c\right )}}\,{d x} \]
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]
integrate((a*sec(d*x + c) + a)/sqrt(e*csc(d*x + c)), x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+\frac {a}{\cos \left (c+d\,x\right )}}{\sqrt {\frac {e}{\sin \left (c+d\,x\right )}}} \,d x \]
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)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \frac {1}{\sqrt {e \csc {\left (c + d x \right )}}}\, dx + \int \frac {\sec {\left (c + d x \right )}}{\sqrt {e \csc {\left (c + d x \right )}}}\, dx\right ) \]
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))
________________________________________________________________________________________