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

Optimal. Leaf size=78 \[ -\frac{3 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos (a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

[Out]

(-2*Cos[a + b*x])/(b*d*Sqrt[d*Tan[a + b*x]]) - (3*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a + b*x
]])/(b*d^2*Sqrt[Sin[2*a + 2*b*x]])

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Rubi [A]  time = 0.0982776, antiderivative size = 78, 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 = {2609, 2615, 2572, 2639} \[ -\frac{3 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos (a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x])/(b*d*Sqrt[d*Tan[a + b*x]]) - (3*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a + b*x
]])/(b*d^2*Sqrt[Sin[2*a + 2*b*x]])

Rule 2609

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

Rule 2615

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[(Sqrt[Cos[e + f*x]]*Sqrt[b*
Tan[e + f*x]])/Sqrt[Sin[e + f*x]], Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]

Rule 2572

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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

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

Mathematica [C]  time = 0.395016, size = 66, normalized size = 0.85 \[ -\frac{2 \sin (a+b x) \left (\tan ^2(a+b x) \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )+1\right )}{b (d \tan (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.143, size = 501, normalized size = 6.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*2^(1/2)*(6*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))*((cos(b*x+a)-1
)/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)-
3*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))*((cos(b*x+a)-1)/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)+6*EllipticE(((1
-cos(b*x+a)+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)*((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)-3*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/si
n(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/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)-3*cos(b*x+a)*2^(1/2))*sin(b*x+a)/cos(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{\cos \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(cos(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(sqrt(d*tan(b*x + a))*cos(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{\cos{\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(cos(b*x+a)/(d*tan(b*x+a))**(3/2),x)

[Out]

Integral(cos(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{\cos \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(cos(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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