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

Optimal. Leaf size=53 \[ \frac {\sqrt {\sin (2 a+2 b x)} F\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b} \]

[Out]

-(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))*(d*csc(b*x+a))^(1/2)*(c*se
c(b*x+a))^(1/2)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]  time = 0.09, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2630, 2573, 2641} \[ \frac {\sqrt {\sin (2 a+2 b x)} F\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]],x]

[Out]

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

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2630

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Csc[e + f*
x])^m*(b*Sec[e + f*x])^n*(a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n, Int[1/((a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n),
x], x] /; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[m - 1/2] && IntegerQ[n - 1/2]

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]

Rubi steps

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

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Mathematica [C]  time = 0.67, size = 68, normalized size = 1.28 \[ \frac {\tan ^3(a+b x) \left (-\cot ^2(a+b x)\right )^{7/4} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\csc ^2(a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]],x]

[Out]

((-Cot[a + b*x]^2)^(7/4)*Sqrt[d*Csc[a + b*x]]*Hypergeometric2F1[1/2, 3/4, 3/2, Csc[a + b*x]^2]*Sqrt[c*Sec[a +
b*x]]*Tan[a + b*x]^3)/b

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d \csc \left (b x + a\right )} \sqrt {c \sec \left (b x + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.42, size = 155, normalized size = 2.92 \[ -\frac {\sqrt {\frac {c}{\cos \left (b x +a \right )}}\, \sqrt {\frac {d}{\sin \left (b x +a \right )}}\, \left (\sin ^{2}\left (b x +a \right )\right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {2}}{b \left (-1+\cos \left (b x +a \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*csc(b*x+a))^(1/2)*(c*sec(b*x+a))^(1/2),x)

[Out]

-1/b*(c/cos(b*x+a))^(1/2)*(d/sin(b*x+a))^(1/2)*sin(b*x+a)^2*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+
cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a
))/sin(b*x+a))^(1/2),1/2*2^(1/2))/(-1+cos(b*x+a))*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \sqrt {\frac {c}{\cos \left (a+b\,x\right )}}\,\sqrt {\frac {d}{\sin \left (a+b\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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