3.1.91 \(\int \frac {\sin (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\) [91]

Optimal. Leaf size=47 \[ \frac {E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}} \]

[Out]

-(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 = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2681, 2652, 2719} \begin {gather*} \frac {\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[Sin[a + b*x]/Sqrt[d*Tan[a + b*x]],x]

[Out]

(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 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 {\sin (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx &=\frac {\sqrt {\sin (a+b x)} \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)} \, dx}{\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {\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 {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 = 0.14, size = 60, normalized size = 1.28 \begin {gather*} \frac {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)} \sin (a+b x) \sqrt {d \tan (a+b x)}}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(522\) vs. \(2(69)=138\).
time = 0.35, size = 523, normalized size = 11.13

method	result	size
default	\(-\frac {\left (-1+\cos \left (b x +a \right )\right )^{2} \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 )}}+\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-\cos \left (b x +a \right ) \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \sqrt {2}}{2 b \sin \left (b x +a \right )^{4} \sqrt {\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}}\, \cos \left (b x +a \right )}\)	\(523\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b*(-1+cos(b*x+a))^2*(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))/si
n(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)+2*El
lipticE(((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*2^(1/2)-cos(b*x+a)*2^(1/2))*(cos(b*x+a)+1)^
2/sin(b*x+a)^4/(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(sin(b*x+a)/(d*tan(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [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(sin(b*x+a)/(d*tan(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*sin(b*x + a)/(d*tan(b*x + a)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\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(sin(b*x+a)/(d*tan(b*x+a))**(1/2),x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\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(sin(a + b*x)/(d*tan(a + b*x))^(1/2),x)

[Out]

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

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