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

Optimal. Leaf size=98 \[ \frac{\sqrt{2} a \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{5/2} f}-\frac{2 a}{d^2 f \sqrt{d \tan (e+f x)}}-\frac{2 a}{3 d f (d \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.122455, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3529, 3532, 205} \[ \frac{\sqrt{2} a \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{5/2} f}-\frac{2 a}{d^2 f \sqrt{d \tan (e+f x)}}-\frac{2 a}{3 d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3529

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

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 205

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

Rubi steps

\begin{align*} \int \frac{a+a \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 a}{3 d f (d \tan (e+f x))^{3/2}}+\frac{\int \frac{a d-a d \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{d^2}\\ &=-\frac{2 a}{3 d f (d \tan (e+f x))^{3/2}}-\frac{2 a}{d^2 f \sqrt{d \tan (e+f x)}}+\frac{\int \frac{-a d^2-a d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{d^4}\\ &=-\frac{2 a}{3 d f (d \tan (e+f x))^{3/2}}-\frac{2 a}{d^2 f \sqrt{d \tan (e+f x)}}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2 d^4+d x^2} \, dx,x,\frac{-a d^2+a d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=\frac{\sqrt{2} a \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{5/2} f}-\frac{2 a}{3 d f (d \tan (e+f x))^{3/2}}-\frac{2 a}{d^2 f \sqrt{d \tan (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.145888, size = 68, normalized size = 0.69 \[ -\frac{\left (\frac{1}{3}+\frac{i}{3}\right ) a \left (\, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-i \tan (e+f x)\right )-i \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};i \tan (e+f x)\right )\right )}{d f (d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/3 - I/3)*a*(Hypergeometric2F1[-3/2, 1, -1/2, (-I)*Tan[e + f*x]] - I*Hypergeometric2F1[-3/2, 1, -1/2, I*Ta
n[e + f*x]]))/(d*f*(d*Tan[e + f*x])^(3/2))

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Maple [B]  time = 0.021, size = 374, normalized size = 3.8 \begin{align*} -{\frac{a\sqrt{2}}{4\,f{d}^{3}}\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\sqrt{2}}{2\,f{d}^{3}}\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\sqrt{2}}{2\,f{d}^{3}}\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\sqrt{2}}{4\,f{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{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{a\sqrt{2}}{2\,f{d}^{2}}\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\sqrt{2}}{2\,f{d}^{2}}\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}}}}}-2\,{\frac{a}{f{d}^{2}\sqrt{d\tan \left ( fx+e \right ) }}}-{\frac{2\,a}{3\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)-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*ta
n(f*x+e))^(1/2)+1)-2*a/d^2/f/(d*tan(f*x+e))^(1/2)-2/3*a/d/f/(d*tan(f*x+e))^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83769, size = 603, normalized size = 6.15 \begin{align*} \left [\frac{3 \, \sqrt{2} a d \sqrt{-\frac{1}{d}} \log \left (-\frac{2 \, \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{-\frac{1}{d}}{\left (\tan \left (f x + e\right ) - 1\right )} - \tan \left (f x + e\right )^{2} + 4 \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} - 4 \,{\left (3 \, a \tan \left (f x + e\right ) + a\right )} \sqrt{d \tan \left (f x + e\right )}}{6 \, d^{3} f \tan \left (f x + e\right )^{2}}, -\frac{3 \, \sqrt{2} a \sqrt{d} \arctan \left (\frac{\sqrt{2} \sqrt{d \tan \left (f x + e\right )}{\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt{d} \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{2} + 2 \,{\left (3 \, a \tan \left (f x + e\right ) + a\right )} \sqrt{d \tan \left (f x + e\right )}}{3 \, d^{3} f \tan \left (f x + e\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(2)*a*d*sqrt(-1/d)*log(-(2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-1/d)*(tan(f*x + e) - 1) - tan(f*x +
e)^2 + 4*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1))*tan(f*x + e)^2 - 4*(3*a*tan(f*x + e) + a)*sqrt(d*tan(f*x + e)
))/(d^3*f*tan(f*x + e)^2), -1/3*(3*sqrt(2)*a*sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) - 1
)/(sqrt(d)*tan(f*x + e)))*tan(f*x + e)^2 + 2*(3*a*tan(f*x + e) + a)*sqrt(d*tan(f*x + e)))/(d^3*f*tan(f*x + e)^
2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \frac{1}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx + \int \frac{\tan{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.34785, size = 371, normalized size = 3.79 \begin{align*} -\frac{\sqrt{2}{\left (a d \sqrt{{\left | d \right |}} + a{\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 \, d^{4} f} - \frac{\sqrt{2}{\left (a d \sqrt{{\left | d \right |}} + a{\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 \, d^{4} f} - \frac{\sqrt{2}{\left (a d \sqrt{{\left | d \right |}} - a{\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 \, d^{4} f} + \frac{\sqrt{2}{\left (a d \sqrt{{\left | d \right |}} - a{\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 \, d^{4} f} - \frac{2 \,{\left (3 \, a d \tan \left (f x + e\right ) + a d\right )}}{3 \, \sqrt{d \tan \left (f x + e\right )} d^{3} f \tan \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*(a*d*sqrt(abs(d)) + a*abs(d)^(3/2))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*x +
e)))/sqrt(abs(d)))/(d^4*f) - 1/2*sqrt(2)*(a*d*sqrt(abs(d)) + a*abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqr
t(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(d^4*f) - 1/4*sqrt(2)*(a*d*sqrt(abs(d)) - a*abs(d)^(3/2))*lo
g(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d^4*f) + 1/4*sqrt(2)*(a*d*sqrt(abs(d))
- a*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d^4*f) - 2/3*(3*a
*d*tan(f*x + e) + a*d)/(sqrt(d*tan(f*x + e))*d^3*f*tan(f*x + e))