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(b*x+a)^3/b/d/(d*tan(b*x+a))^(1/2)+7/2*cos(b*x+a)*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*Elliptic
E(cos(a+1/4*Pi+b*x),2^(1/2))*(d*tan(b*x+a))^(1/2)/b/d^2/sin(2*b*x+2*a)^(1/2)-7/3*cos(b*x+a)^3*(d*tan(b*x+a))^(
3/2)/b/d^3

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Rubi [A]  time = 0.15, antiderivative size = 112, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [C]  time = 0.57, 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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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \tan \left (b x + a\right )} \cos \left (b x + a\right )^{3}}{d^{2} \tan \left (b x + a\right )^{2}}, x\right ) \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)

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maple [B]  time = 0.52, size = 523, normalized size = 4.67 \[ \frac {\left (2 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}+42 \EllipticE \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \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 {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}-21 \EllipticF \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \cos \left (b x +a \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 {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}+42 \EllipticE \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \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 {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}-21 \EllipticF \left (\sqrt {-\frac {-\sin \left (b x +a \right )-1+\cos \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 {-\sin \left (b x +a \right )-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}+7 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-21 \cos \left (b x +a \right ) \sqrt {2}\right ) \sin \left (b x +a \right ) \sqrt {2}}{12 b \cos \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(cos(b*x+a)^3/(d*tan(b*x+a))^(3/2),x)

[Out]

1/12/b*(2*cos(b*x+a)^4*2^(1/2)+42*EllipticE((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*cos(b*
x+a)*((-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)*(-(-sin(b*x+a)-1+cos(b*x
+a))/sin(b*x+a))^(1/2)-21*EllipticF((-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*cos(b*x+a)*((-
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)*(-(-sin(b*x+a)-1+cos(b*x+a))/sin
(b*x+a))^(1/2)+42*EllipticE((-(-sin(b*x+a)-1+cos(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)*(-(-sin(b*x+a)-1+cos(b*x+a))/sin(b*x+a))^(1/2)-21*E
llipticF((-(-sin(b*x+a)-1+cos(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+c
os(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(-sin(b*x+a)-1+cos(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)*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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