3.100 \(\int \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=79 \[ \frac {\sin (a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b d \sqrt {d \tan (a+b x)}} \]

[Out]

sin(b*x+a)/b/d/(d*tan(b*x+a))^(1/2)-1/2*(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))*sec(b*x+a)*sin(2*b*x+2*a)^(1/2)/b/d/(d*tan(b*x+a))^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 79, 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 = {2602, 2569, 2573, 2641} \[ \frac {\sin (a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {\sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b d \sqrt {d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/(d*Tan[a + b*x])^(3/2),x]

[Out]

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

Rule 2569

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(b*Sin[e +
f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Sin[e + f*x])^
n*(a*Cos[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m
, 2*n]

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 2602

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), 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]

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 \frac {\sin (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=\frac {\sqrt {\sin (a+b x)} \int \frac {\cos ^{\frac {3}{2}}(a+b x)}{\sqrt {\sin (a+b x)}} \, dx}{d \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {\sin (a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {\sqrt {\sin (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{2 d \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {\sin (a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {\left (\sec (a+b x) \sqrt {\sin (2 a+2 b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx}{2 d \sqrt {d \tan (a+b x)}}\\ &=\frac {\sin (a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b d \sqrt {d \tan (a+b x)}}\\ \end {align*}

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Mathematica [C]  time = 0.77, size = 126, normalized size = 1.59 \[ \frac {\cos (2 (a+b x)) \tan ^{\frac {3}{2}}(a+b x) \sec (a+b x) \left (-\sqrt {\tan (a+b x)} \sqrt {\sec ^2(a+b x)}+\sqrt [4]{-1} \sec ^2(a+b x) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right )\right |-1\right )\right )}{b \left (\tan ^2(a+b x)-1\right ) \sqrt {\sec ^2(a+b x)} (d \tan (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]/(d*Tan[a + b*x])^(3/2),x]

[Out]

(Cos[2*(a + b*x)]*Sec[a + b*x]*((-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Sec[a + b*x
]^2 - Sqrt[Sec[a + b*x]^2]*Sqrt[Tan[a + b*x]])*Tan[a + b*x]^(3/2))/(b*Sqrt[Sec[a + b*x]^2]*(d*Tan[a + b*x])^(3
/2)*(-1 + Tan[a + b*x]^2))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \tan \left (b x + a\right )} \sin \left (b x + a\right )}{d^{2} \tan \left (b x + a\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*sin(b*x + a)/(d^2*tan(b*x + a)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)/(d*tan(b*x + a))^(3/2), x)

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maple [B]  time = 0.42, size = 196, normalized size = 2.48 \[ -\frac {\left (\cos \left (b x +a \right )+1\right )^{2} \left (-1+\cos \left (b x +a \right )\right ) \left (\sin \left (b x +a \right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+\cos \left (b x +a \right ) \sqrt {2}\right ) \sqrt {2}}{2 b \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)/(d*tan(b*x+a))^(3/2),x)

[Out]

-1/2/b*(cos(b*x+a)+1)^2*(-1+cos(b*x+a))*(sin(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+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)-cos(b*x+a)^2*2^(1/2)+cos(b*x+a)*2^(1/2))/cos(b*x+a)^2/sin(b*x+a)^2/(d*sin(b*x+a)/cos
(b*x+a))^(3/2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)/(d*tan(b*x + a))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (a+b\,x\right )}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/(d*tan(a + b*x))^(3/2),x)

[Out]

int(sin(a + b*x)/(d*tan(a + b*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*tan(b*x+a))**(3/2),x)

[Out]

Integral(sin(a + b*x)/(d*tan(a + b*x))**(3/2), x)

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