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

Optimal. Leaf size=108 \[ \frac{d^2 \sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{12 b \sqrt{d \tan (a+b x)}}-\frac{d \cos ^3(a+b x) \sqrt{d \tan (a+b x)}}{3 b}+\frac{d \cos (a+b x) \sqrt{d \tan (a+b x)}}{6 b} \]

[Out]

(d^2*EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(12*b*Sqrt[d*Tan[a + b*x]]) + (d*Cos[a
+ b*x]*Sqrt[d*Tan[a + b*x]])/(6*b) - (d*Cos[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/(3*b)

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Rubi [A]  time = 0.133509, antiderivative size = 108, 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 = {2610, 2612, 2614, 2573, 2641} \[ \frac{d^2 \sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{12 b \sqrt{d \tan (a+b x)}}-\frac{d \cos ^3(a+b x) \sqrt{d \tan (a+b x)}}{3 b}+\frac{d \cos (a+b x) \sqrt{d \tan (a+b x)}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*(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]])/(12*b*Sqrt[d*Tan[a + b*x]]) + (d*Cos[a
+ b*x]*Sqrt[d*Tan[a + b*x]])/(6*b) - (d*Cos[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/(3*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 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 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 ^3(a+b x) (d \tan (a+b x))^{3/2} \, dx &=-\frac{d \cos ^3(a+b x) \sqrt{d \tan (a+b x)}}{3 b}+\frac{1}{6} d^2 \int \frac{\cos (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=\frac{d \cos (a+b x) \sqrt{d \tan (a+b x)}}{6 b}-\frac{d \cos ^3(a+b x) \sqrt{d \tan (a+b x)}}{3 b}+\frac{1}{12} 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)}}{6 b}-\frac{d \cos ^3(a+b x) \sqrt{d \tan (a+b x)}}{3 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}{12 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=\frac{d \cos (a+b x) \sqrt{d \tan (a+b x)}}{6 b}-\frac{d \cos ^3(a+b x) \sqrt{d \tan (a+b x)}}{3 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}{12 \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)}}{12 b \sqrt{d \tan (a+b x)}}+\frac{d \cos (a+b x) \sqrt{d \tan (a+b x)}}{6 b}-\frac{d \cos ^3(a+b x) \sqrt{d \tan (a+b x)}}{3 b}\\ \end{align*}

Mathematica [C]  time = 1.08639, size = 96, normalized size = 0.89 \[ -\frac{\cos (a+b x) (d \tan (a+b x))^{3/2} \left (\cos (2 (a+b x)) \sqrt{\tan (a+b x)}+\sqrt [4]{-1} \sqrt{\sec ^2(a+b x)} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (a+b x)}\right )\right |-1\right )\right )}{6 b \tan ^{\frac{3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[a + b*x]*((-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Sqrt[Sec[a + b*x]^2] + Cos[
2*(a + b*x)]*Sqrt[Tan[a + b*x]])*(d*Tan[a + b*x])^(3/2))/(6*b*Tan[a + b*x]^(3/2))

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Maple [A]  time = 0.172, size = 220, normalized size = 2. \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}}{12\,b \left ( \sin \left ( bx+a \right ) \right ) ^{5}} \left ( \sin \left ( bx+a \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{1-\cos \left ( bx+a \right ) +\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 ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{\sqrt{2}}{2}} \right ) +2\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}\sqrt{2}-2\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sqrt{2}- \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)^3*(d*tan(b*x+a))^(3/2),x)

[Out]

-1/12/b*2^(1/2)*(cos(b*x+a)-1)*(sin(b*x+a)*((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)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2
),1/2*2^(1/2))+2*cos(b*x+a)^4*2^(1/2)-2*cos(b*x+a)^3*2^(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 )^{3}\,{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((d*tan(b*x + a))^(3/2)*cos(b*x + a)^3, 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 )^{3} \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)^3*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: TypeError