3.342 \(\int \frac{a+a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=121 \[ \frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac{\sqrt{2} a \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}} \]

[Out]

-((Sqrt[2]*a*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(d^(7/2)*f)) - (2*a)/(5
*d*f*(d*Tan[e + f*x])^(5/2)) - (2*a)/(3*d^2*f*(d*Tan[e + f*x])^(3/2)) + (2*a)/(d^3*f*Sqrt[d*Tan[e + f*x]])

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Rubi [A]  time = 0.165502, antiderivative size = 121, 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 = {3529, 3532, 208} \[ \frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac{\sqrt{2} a \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tan[e + f*x])/(d*Tan[e + f*x])^(7/2),x]

[Out]

-((Sqrt[2]*a*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(d^(7/2)*f)) - (2*a)/(5
*d*f*(d*Tan[e + f*x])^(5/2)) - (2*a)/(3*d^2*f*(d*Tan[e + f*x])^(3/2)) + (2*a)/(d^3*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 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 \frac{a+a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx &=-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac{\int \frac{a d-a d \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{d^2}\\ &=-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac{2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{\int \frac{-a d^2-a d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{d^4}\\ &=-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac{2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}+\frac{\int \frac{-a d^3+a d^3 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{d^6}\\ &=-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac{2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}-\frac{\left (2 a^2\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 \tanh ^{-1}\left (\frac{\sqrt{d}+\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac{2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac{2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac{2 a}{d^3 f \sqrt{d \tan (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.209191, size = 68, normalized size = 0.56 \[ -\frac{\left (\frac{1}{5}+\frac{i}{5}\right ) a \left (\, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};-i \tan (e+f x)\right )-i \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};i \tan (e+f x)\right )\right )}{d f (d \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tan[e + f*x])/(d*Tan[e + f*x])^(7/2),x]

[Out]

((-1/5 - I/5)*a*(Hypergeometric2F1[-5/2, 1, -3/2, (-I)*Tan[e + f*x]] - I*Hypergeometric2F1[-5/2, 1, -3/2, I*Ta
n[e + f*x]]))/(d*f*(d*Tan[e + f*x])^(5/2))

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Maple [B]  time = 0.02, size = 393, normalized size = 3.3 \begin{align*} -{\frac{a\sqrt{2}}{4\,f{d}^{4}}\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}^{4}}\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}^{4}}\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}^{3}}\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}^{3}}\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}^{3}}\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{2\,a}{3\,{d}^{2}f} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,a}{5\,df} \left ( d\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}+2\,{\frac{a}{f{d}^{3}\sqrt{d\tan \left ( fx+e \right ) }}} \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))^(7/2),x)

[Out]

-1/4/f*a/d^4*(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^4*(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^4*(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*ta
n(f*x+e))^(1/2)+1)-2/3*a/d^2/f/(d*tan(f*x+e))^(3/2)-2/5*a/d/f/(d*tan(f*x+e))^(5/2)+2*a/d^3/f/(d*tan(f*x+e))^(1
/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))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.87575, size = 668, normalized size = 5.52 \begin{align*} \left [\frac{15 \, \sqrt{2} a \sqrt{d} \log \left (\frac{\tan \left (f x + e\right )^{2} - \frac{2 \, \sqrt{2} \sqrt{d \tan \left (f x + e\right )}{\left (\tan \left (f x + e\right ) + 1\right )}}{\sqrt{d}} + 4 \, \tan \left (f x + e\right ) + 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{3} + 4 \,{\left (15 \, a \tan \left (f x + e\right )^{2} - 5 \, a \tan \left (f x + e\right ) - 3 \, a\right )} \sqrt{d \tan \left (f x + e\right )}}{30 \, d^{4} f \tan \left (f x + e\right )^{3}}, \frac{15 \, \sqrt{2} a d \sqrt{-\frac{1}{d}} \arctan \left (\frac{\sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{-\frac{1}{d}}{\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{3} + 2 \,{\left (15 \, a \tan \left (f x + e\right )^{2} - 5 \, a \tan \left (f x + e\right ) - 3 \, a\right )} \sqrt{d \tan \left (f x + e\right )}}{15 \, d^{4} f \tan \left (f x + e\right )^{3}}\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))^(7/2),x, algorithm="fricas")

[Out]

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

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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{7}{2}}}\, dx + \int \frac{\tan{\left (e + f x \right )}}{\left (d \tan{\left (e + f x \right )}\right )^{\frac{7}{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))**(7/2),x)

[Out]

a*(Integral((d*tan(e + f*x))**(-7/2), x) + Integral(tan(e + f*x)/(d*tan(e + f*x))**(7/2), x))

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Giac [B]  time = 1.24128, size = 398, normalized size = 3.29 \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^{5} 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^{5} 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^{5} 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^{5} f} + \frac{2 \,{\left (15 \, a d^{2} \tan \left (f x + e\right )^{2} - 5 \, a d^{2} \tan \left (f x + e\right ) - 3 \, a d^{2}\right )}}{15 \, \sqrt{d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}} \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))^(7/2),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^5*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^5*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^5*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^5*f) + 2/15*(15
*a*d^2*tan(f*x + e)^2 - 5*a*d^2*tan(f*x + e) - 3*a*d^2)/(sqrt(d*tan(f*x + e))*d^5*f*tan(f*x + e)^2)