3.336 \(\int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx\)

Optimal. Leaf size=115 \[ -\frac{2 a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{\sqrt{2} a d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 a (d \tan (e+f x))^{5/2}}{5 f} \]

[Out]

(Sqrt[2]*a*d^(5/2)*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f - (2*a*d^2*Sqrt
[d*Tan[e + f*x]])/f + (2*a*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (2*a*(d*Tan[e + f*x])^(5/2))/(5*f)

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Rubi [A]  time = 0.148818, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3528, 3532, 208} \[ -\frac{2 a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{\sqrt{2} a d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 a (d \tan (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x]),x]

[Out]

(Sqrt[2]*a*d^(5/2)*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f - (2*a*d^2*Sqrt
[d*Tan[e + f*x]])/f + (2*a*d*(d*Tan[e + f*x])^(3/2))/(3*f) + (2*a*(d*Tan[e + f*x])^(5/2))/(5*f)

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3532

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*d^2)/f,
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx &=\frac{2 a (d \tan (e+f x))^{5/2}}{5 f}+\int (d \tan (e+f x))^{3/2} (-a d+a d \tan (e+f x)) \, dx\\ &=\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 a (d \tan (e+f x))^{5/2}}{5 f}+\int \sqrt{d \tan (e+f x)} \left (-a d^2-a d^2 \tan (e+f x)\right ) \, dx\\ &=-\frac{2 a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 a (d \tan (e+f x))^{5/2}}{5 f}+\int \frac{a d^3-a d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx\\ &=-\frac{2 a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 a (d \tan (e+f x))^{5/2}}{5 f}-\frac{\left (2 a^2 d^6\right ) \operatorname{Subst}\left (\int \frac{1}{-2 a^2 d^6+d x^2} \, dx,x,\frac{a d^3+a d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=\frac{\sqrt{2} a d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d}+\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}-\frac{2 a d^2 \sqrt{d \tan (e+f x)}}{f}+\frac{2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac{2 a (d \tan (e+f x))^{5/2}}{5 f}\\ \end{align*}

Mathematica [C]  time = 0.771061, size = 117, normalized size = 1.02 \[ \frac{\left (\frac{1}{15}+\frac{i}{15}\right ) a (d \tan (e+f x))^{5/2} \left (-15 \sqrt [4]{-1} \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (e+f x)}\right )+(1-i) \sqrt{\tan (e+f x)} \left (3 \tan ^2(e+f x)+5 \tan (e+f x)-15\right )+15 (-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (e+f x)}\right )\right )}{f \tan ^{\frac{5}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x]),x]

[Out]

((1/15 + I/15)*a*(d*Tan[e + f*x])^(5/2)*(-15*(-1)^(1/4)*ArcTan[(-1)^(3/4)*Sqrt[Tan[e + f*x]]] + 15*(-1)^(3/4)*
ArcTanh[(-1)^(3/4)*Sqrt[Tan[e + f*x]]] + (1 - I)*Sqrt[Tan[e + f*x]]*(-15 + 5*Tan[e + f*x] + 3*Tan[e + f*x]^2))
)/(f*Tan[e + f*x]^(5/2))

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Maple [B]  time = 0.024, size = 388, normalized size = 3.4 \begin{align*}{\frac{2\,a}{5\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,ad}{3\,f} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-2\,{\frac{a{d}^{2}\sqrt{d\tan \left ( fx+e \right ) }}{f}}+{\frac{a{d}^{2}\sqrt{2}}{4\,f}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }+{\frac{a{d}^{2}\sqrt{2}}{2\,f}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{a{d}^{2}\sqrt{2}}{2\,f}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{a{d}^{3}\sqrt{2}}{4\,f}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{a{d}^{3}\sqrt{2}}{2\,f}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{a{d}^{3}\sqrt{2}}{2\,f}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(5/2)*(a+a*tan(f*x+e)),x)

[Out]

2/5*a*(d*tan(f*x+e))^(5/2)/f+2/3*a*d*(d*tan(f*x+e))^(3/2)/f-2*a*d^2*(d*tan(f*x+e))^(1/2)/f+1/4/f*a*d^2*(d^2)^(
1/4)*2^(1/2)*ln((d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*
(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+1/2/f*a*d^2*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f
*x+e))^(1/2)+1)-1/2/f*a*d^2*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/4/f*a*d^
3/(d^2)^(1/4)*2^(1/2)*ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^
2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))-1/2/f*a*d^3/(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)
*(d*tan(f*x+e))^(1/2)+1)+1/2/f*a*d^3/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+a*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71939, size = 613, normalized size = 5.33 \begin{align*} \left [\frac{15 \, \sqrt{2} a d^{\frac{5}{2}} \log \left (\frac{d \tan \left (f x + e\right )^{2} + 2 \, \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{d}{\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 4 \,{\left (3 \, a d^{2} \tan \left (f x + e\right )^{2} + 5 \, a d^{2} \tan \left (f x + e\right ) - 15 \, a d^{2}\right )} \sqrt{d \tan \left (f x + e\right )}}{30 \, f}, -\frac{15 \, \sqrt{2} a \sqrt{-d} d^{2} \arctan \left (\frac{\sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{-d}{\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) - 2 \,{\left (3 \, a d^{2} \tan \left (f x + e\right )^{2} + 5 \, a d^{2} \tan \left (f x + e\right ) - 15 \, a d^{2}\right )} \sqrt{d \tan \left (f x + e\right )}}{15 \, f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+a*tan(f*x+e)),x, algorithm="fricas")

[Out]

[1/30*(15*sqrt(2)*a*d^(5/2)*log((d*tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d)*(tan(f*x + e) + 1)
+ 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 4*(3*a*d^2*tan(f*x + e)^2 + 5*a*d^2*tan(f*x + e) - 15*a*d^2)*s
qrt(d*tan(f*x + e)))/f, -1/15*(15*sqrt(2)*a*sqrt(-d)*d^2*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan
(f*x + e) + 1)/(d*tan(f*x + e))) - 2*(3*a*d^2*tan(f*x + e)^2 + 5*a*d^2*tan(f*x + e) - 15*a*d^2)*sqrt(d*tan(f*x
+ e)))/f]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}\, dx + \int \left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}} \tan{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(5/2)*(a+a*tan(f*x+e)),x)

[Out]

a*(Integral((d*tan(e + f*x))**(5/2), x) + Integral((d*tan(e + f*x))**(5/2)*tan(e + f*x), x))

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Giac [B]  time = 1.36987, size = 424, normalized size = 3.69 \begin{align*} \frac{\sqrt{2}{\left (a d^{2} \sqrt{{\left | d \right |}} - a d{\left | d \right |}^{\frac{3}{2}}\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} + 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{2 \, f} + \frac{\sqrt{2}{\left (a d^{2} \sqrt{{\left | d \right |}} - a d{\left | d \right |}^{\frac{3}{2}}\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} - 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{2 \, f} + \frac{\sqrt{2}{\left (a d^{2} \sqrt{{\left | d \right |}} + a d{\left | d \right |}^{\frac{3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{4 \, f} - \frac{\sqrt{2}{\left (a d^{2} \sqrt{{\left | d \right |}} + a d{\left | d \right |}^{\frac{3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{4 \, f} + \frac{2 \,{\left (3 \, \sqrt{d \tan \left (f x + e\right )} a d^{2} f^{4} \tan \left (f x + e\right )^{2} + 5 \, \sqrt{d \tan \left (f x + e\right )} a d^{2} f^{4} \tan \left (f x + e\right ) - 15 \, \sqrt{d \tan \left (f x + e\right )} a d^{2} f^{4}\right )}}{15 \, f^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(5/2)*(a+a*tan(f*x+e)),x, algorithm="giac")

[Out]

1/2*sqrt(2)*(a*d^2*sqrt(abs(d)) - a*d*abs(d)^(3/2))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*
x + e)))/sqrt(abs(d)))/f + 1/2*sqrt(2)*(a*d^2*sqrt(abs(d)) - a*d*abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sq
rt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/f + 1/4*sqrt(2)*(a*d^2*sqrt(abs(d)) + a*d*abs(d)^(3/2))*log
(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/f - 1/4*sqrt(2)*(a*d^2*sqrt(abs(d)) + a*
d*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/f + 2/15*(3*sqrt(d*ta
n(f*x + e))*a*d^2*f^4*tan(f*x + e)^2 + 5*sqrt(d*tan(f*x + e))*a*d^2*f^4*tan(f*x + e) - 15*sqrt(d*tan(f*x + e))
*a*d^2*f^4)/f^5