3.111 \(\int \frac {a+a \sec (c+d x)}{\sqrt {e \sin (c+d x)}} \, dx\)

Optimal. Leaf size=103 \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \sqrt {e \sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 103, 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 = {3872, 2838, 2564, 329, 212, 206, 203, 2642, 2641} \[ \frac {a \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d \sqrt {e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 3.18, size = 193, normalized size = 1.87 \[ \frac {4 a \cos \left (\frac {1}{2} (c+d x)\right ) \left (4 F\left (\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{4} (c+d x)\right )}}\right )\right |-1\right )+\sqrt {2} \left (\Pi \left (-1-\sqrt {2};\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{4} (c+d x)\right )}}\right )\right |-1\right )-\Pi \left (1-\sqrt {2};\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{4} (c+d x)\right )}}\right )\right |-1\right )-\Pi \left (-1+\sqrt {2};\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{4} (c+d x)\right )}}\right )\right |-1\right )+\Pi \left (1+\sqrt {2};\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{4} (c+d x)\right )}}\right )\right |-1\right )\right )\right )}{d \sqrt {\tan \left (\frac {1}{4} (c+d x)\right )} \sqrt {1-\cot ^2\left (\frac {1}{4} (c+d x)\right )} \sqrt {e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*Cos[(c + d*x)/2]*(4*EllipticF[ArcSin[1/Sqrt[Tan[(c + d*x)/4]]], -1] + Sqrt[2]*(EllipticPi[-1 - Sqrt[2], A
rcSin[1/Sqrt[Tan[(c + d*x)/4]]], -1] - EllipticPi[1 - Sqrt[2], ArcSin[1/Sqrt[Tan[(c + d*x)/4]]], -1] - Ellipti
cPi[-1 + Sqrt[2], ArcSin[1/Sqrt[Tan[(c + d*x)/4]]], -1] + EllipticPi[1 + Sqrt[2], ArcSin[1/Sqrt[Tan[(c + d*x)/
4]]], -1])))/(d*Sqrt[1 - Cot[(c + d*x)/4]^2]*Sqrt[e*Sin[c + d*x]]*Sqrt[Tan[(c + d*x)/4]])

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}}{e \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sec(d*x + c) + a)*sqrt(e*sin(d*x + c))/(e*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sec \left (d x + c\right ) + a}{\sqrt {e \sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 2.54, size = 122, normalized size = 1.18 \[ \frac {a \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{d \sqrt {e}}+\frac {a \arctanh \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{d \sqrt {e}}-\frac {a \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{d \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sec \left (d x + c\right ) + a}{\sqrt {e \sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx + \int \frac {\sec {\left (c + d x \right )}}{\sqrt {e \sin {\left (c + d x \right )}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(1/sqrt(e*sin(c + d*x)), x) + Integral(sec(c + d*x)/sqrt(e*sin(c + d*x)), x))

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