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[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]])

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Rubi [A]  time = 0.0937112, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, 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 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 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 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*}

Mathematica [C]  time = 0.740658, 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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Maple [B]  time = 0.125, size = 199, normalized size = 2.5 \begin{align*}{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{2\,b \left ( \cos \left ( bx+a \right ) \right ) ^{2} \left ( \sin \left ( bx+a \right ) \right ) ^{2}} \left ( -\sin \left ( bx+a \right ){\it EllipticF} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}+ \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}-\cos \left ( bx+a \right ) \sqrt{2} \right ) \left ({\frac{\sin \left ( bx+a \right ) d}{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (a + b x \right )}}{\left (d \tan{\left (a + b x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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)