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

Optimal. Leaf size=112 \[ -\frac{7 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d^3}-\frac{7 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{2 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.149311, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2609, 2612, 2615, 2572, 2639} \[ -\frac{7 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d^3}-\frac{7 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{2 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \cos ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2612

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*m), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integer
sQ[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 ^3(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=-\frac{2 \cos ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{7 \int \cos ^3(a+b x) \sqrt{d \tan (a+b x)} \, dx}{d^2}\\ &=-\frac{2 \cos ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{7 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d^3}-\frac{7 \int \cos (a+b x) \sqrt{d \tan (a+b x)} \, dx}{2 d^2}\\ &=-\frac{2 \cos ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{7 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d^3}-\frac{\left (7 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)} \, dx}{2 d^2 \sqrt{\sin (a+b x)}}\\ &=-\frac{2 \cos ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{7 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d^3}-\frac{\left (7 \cos (a+b x) \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{2 d^2 \sqrt{\sin (2 a+2 b x)}}\\ &=-\frac{2 \cos ^3(a+b x)}{b d \sqrt{d \tan (a+b x)}}-\frac{7 \cos (a+b x) E\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{d \tan (a+b x)}}{2 b d^2 \sqrt{\sin (2 a+2 b x)}}-\frac{7 \cos ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b d^3}\\ \end{align*}

Mathematica [C]  time = 0.552562, size = 77, normalized size = 0.69 \[ \frac{\sin (a+b x) \left (-14 \tan ^2(a+b x) \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )+\cos (2 (a+b x))-13\right )}{6 b (d \tan (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[a + b*x]*(-13 + Cos[2*(a + b*x)] - 14*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]
^2]*Tan[a + b*x]^2))/(6*b*(d*Tan[a + b*x])^(3/2))

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Maple [B]  time = 0.142, size = 515, normalized size = 4.6 \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)^3/(d*tan(b*x+a))^(3/2),x)

[Out]

1/12/b*2^(1/2)*(2*cos(b*x+a)^4*2^(1/2)+42*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)-21*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))*((c
os(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)+42*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)-21*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)+7*cos(b*x+a)^2*2^(1/2)-21*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 )^{3}}{\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)^3/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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