3.580 \(\int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x)) \, dx\)

Optimal. Leaf size=88 \[ -\frac{2 a d^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 a d \sin (e+f x) \sqrt{d \sec (e+f x)}}{f}+\frac{2 b (d \sec (e+f x))^{3/2}}{3 f} \]

[Out]

(-2*a*d^2*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*b*(d*Sec[e + f*x])^(3/2)
)/(3*f) + (2*a*d*Sqrt[d*Sec[e + f*x]]*Sin[e + f*x])/f

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Rubi [A]  time = 0.0694522, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3486, 3768, 3771, 2639} \[ -\frac{2 a d^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 a d \sin (e+f x) \sqrt{d \sec (e+f x)}}{f}+\frac{2 b (d \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*d^2*EllipticE[(e + f*x)/2, 2])/(f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*b*(d*Sec[e + f*x])^(3/2)
)/(3*f) + (2*a*d*Sqrt[d*Sec[e + f*x]]*Sin[e + f*x])/f

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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

Mathematica [A]  time = 0.287762, size = 58, normalized size = 0.66 \[ \frac{(d \sec (e+f x))^{3/2} \left (3 a \sin (2 (e+f x))-6 a \cos ^{\frac{3}{2}}(e+f x) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )+2 b\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*Sec[e + f*x])^(3/2)*(2*b - 6*a*Cos[e + f*x]^(3/2)*EllipticE[(e + f*x)/2, 2] + 3*a*Sin[2*(e + f*x)]))/(3*f)

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Maple [C]  time = 0.211, size = 355, normalized size = 4. \begin{align*}{\frac{2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) ^{2}}{3\,f \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ( 3\,i\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) a-3\,i\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) a+3\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) a\sin \left ( fx+e \right ) -3\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) a\sin \left ( fx+e \right ) -3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}a+b\sin \left ( fx+e \right ) +3\,a\cos \left ( fx+e \right ) \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e)),x)

[Out]

2/3/f*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(3*I*sin(f*x+e)*cos(f*x+e)^2*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos
(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a-3*I*sin(f*x+e)*cos(f*x+e)^2*(1/(cos(f*x+e)+1))^(1
/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a+3*I*(1/(cos(f*x+e)+1))^(1/2)*
(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*a*sin(f*x+e)-3*I*(1/(cos
(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*a*sin(
f*x+e)-3*cos(f*x+e)^2*a+b*sin(f*x+e)+3*a*cos(f*x+e))*(d/cos(f*x+e))^(3/2)/sin(f*x+e)^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(3/2)*(b*tan(f*x + e) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b d \sec \left (f x + e\right ) \tan \left (f x + e\right ) + a d \sec \left (f x + e\right )\right )} \sqrt{d \sec \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

integral((b*d*sec(f*x + e)*tan(f*x + e) + a*d*sec(f*x + e))*sqrt(d*sec(f*x + e)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec{\left (e + f x \right )}\right )^{\frac{3}{2}} \left (a + b \tan{\left (e + f x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(3/2)*(a+b*tan(f*x+e)),x)

[Out]

Integral((d*sec(e + f*x))**(3/2)*(a + b*tan(e + f*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}{\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(3/2)*(b*tan(f*x + e) + a), x)