3.283 \(\int \sqrt {e \csc (c+d x)} (a+a \sec (c+d x)) \, dx\)

Optimal. Leaf size=121 \[ \frac {a \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {a \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 121, 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, 212, 206, 203, 2641} \[ \frac {a \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {a \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \csc (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.85, size = 111, normalized size = 0.92 \[ -\frac {a \sqrt {e \csc (c+d x)} \left (\log \left (1-\sqrt {\csc (c+d x)}\right )-\log \left (\sqrt {\csc (c+d x)}+1\right )+2 \tan ^{-1}\left (\sqrt {\csc (c+d x)}\right )+4 \sqrt {\sin (c+d x)} \sqrt {\csc (c+d x)} F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )\right )}{2 d \sqrt {\csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*Sqrt[e*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]]] + 4*Sqrt[Csc[c + d*x]]*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sqrt[Sin[c + d*x]]))/(d*Sqrt[Csc[c + d*x]]
)

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {e \csc \left (d x + c\right )} {\left (a \sec \left (d x + c\right ) + a\right )}, x\right ) \]

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), x)

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

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

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maple [C]  time = 1.63, size = 291, normalized size = 2.40 \[ -\frac {a \sqrt {2}\, \sqrt {\frac {e}{\sin \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}\, \left (i \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+i \EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2}}{2 d \sin \left (d x +c \right )^{2}} \]

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*2^(1/2)*(e/sin(d*x+c))^(1/2)*(-1+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)*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(I*EllipticPi(((I*cos(d*x+c)-I+si
n(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+I*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))+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))-Ellip
ticPi(((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)))/sin(d*x+c)^2*(1+cos(d*x+c))^2

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

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

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

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)

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

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(sqrt(e*csc(c + d*x)), x) + Integral(sqrt(e*csc(c + d*x))*sec(c + d*x), x))

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