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

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

[Out]

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

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Rubi [A]  time = 0.0951669, 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 = {2610, 2614, 2573, 2641} \[ \frac{d^2 \sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b \sqrt{d \tan (a+b x)}}-\frac{d \cos (a+b x) \sqrt{d \tan (a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2610

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(n - 1))/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan
[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, 3/2]
)) && IntegersQ[2*m, 2*n]

Rule 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

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

Mathematica [C]  time = 0.128121, size = 58, normalized size = 0.74 \[ \frac{d \cos (a+b x) \sqrt{d \tan (a+b x)} \left (\sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2(a+b x)\right )-1\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Cos[a + b*x]*(-1 + Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*Sqrt[d*Tan[a + b
*x]])/b

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Maple [B]  time = 0.132, size = 196, normalized size = 2.5 \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) \cos \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{2\,b \left ( \sin \left ( bx+a \right ) \right ) ^{5}} \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{d\sin \left ( bx+a \right ) }{\cos \left ( bx+a \right ) }} \right ) ^{{\frac{3}{2}}}} \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)*(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)*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))
^(3/2)/sin(b*x+a)^5

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{d \tan \left (b x + a\right )} d \cos \left (b x + a\right ) \tan \left (b x + a\right ), 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))*d*cos(b*x + a)*tan(b*x + a), 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)*(d*tan(b*x+a))**(3/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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]

Exception raised: TypeError