3.1.92 \(\int \frac {\csc (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\) [92]

Optimal. Leaf size=72 \[ -\frac {2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}} \]

[Out]

-2*cos(b*x+a)/b/(d*tan(b*x+a))^(1/2)+2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*
x),2^(1/2))*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)/(d*tan(b*x+a))^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2681, 2650, 2652, 2719} \begin {gather*} -\frac {2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {2 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]/Sqrt[d*Tan[a + b*x]],x]

[Out]

(-2*Cos[a + b*x])/(b*Sqrt[d*Tan[a + b*x]]) - (2*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(b*Sqrt[Sin[2*a + 2
*b*x]]*Sqrt[d*Tan[a + b*x]])

Rule 2650

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

Rule 2652

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

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]
^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^n), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

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]

Rubi steps

\begin {align*} \int \frac {\csc (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx &=\frac {\sqrt {\sin (a+b x)} \int \frac {\sqrt {\cos (a+b x)}}{\sin ^{\frac {3}{2}}(a+b x)} \, dx}{\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {\left (2 \sqrt {\sin (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)} \, dx}{\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {(2 \sin (a+b x)) \int \sqrt {\sin (2 a+2 b x)} \, dx}{\sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {2 \cos (a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b 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 = 0.33, size = 69, normalized size = 0.96 \begin {gather*} -\frac {2 \cos (a+b x) \left (3+2 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)} \tan ^2(a+b x)\right )}{3 b \sqrt {d \tan (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]/Sqrt[d*Tan[a + b*x]],x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(481\) vs. \(2(91)=182\).
time = 0.38, size = 482, normalized size = 6.69

method	result	size
default	\(\frac {\left (2 \cos \left (b x +a \right ) \EllipticE \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-\cos \left (b x +a \right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+2 \EllipticE \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-\EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-\cos \left (b x +a \right ) \sqrt {2}\right ) \sqrt {2}}{b \sqrt {\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}}\, \cos \left (b x +a \right )}\)	\(482\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)/(d*tan(b*x+a))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/b*(2*cos(b*x+a)*EllipticE(((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))/sin(b*x
+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)-cos(b*x+a
)*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))/sin(b*x+a))^(1/2)*((1-c
os(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)+2*EllipticE(((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))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin
(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(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))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)
-1+sin(b*x+a))/sin(b*x+a))^(1/2)-cos(b*x+a)*2^(1/2))/(d*sin(b*x+a)/cos(b*x+a))^(1/2)/cos(b*x+a)*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(csc(b*x+a)/(d*tan(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)/sqrt(d*tan(b*x + a)), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(a + b*x)/sqrt(d*tan(a + b*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(csc(b*x+a)/(d*tan(b*x+a))^(1/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)/sqrt(d*tan(b*x + a)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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