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

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

[Out]

-a*arctanh(1/2*(d^(1/2)+d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/d^(7/2)/f+2*a/d^3/f/(d*tan(f
*x+e))^(1/2)-2/5*a/d/f/(d*tan(f*x+e))^(5/2)-2/3*a/d^2/f/(d*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.17, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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]

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]

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*}

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Mathematica [C]  time = 0.23, 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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fricas [A]  time = 0.45, size = 243, normalized size = 2.01 \[ \left [\frac {15 \, \sqrt {2} a \sqrt {d} \log \left (\frac {\tan \left (f x + e\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) + 1\right )}}{\sqrt {d}} + 4 \, \tan \left (f x + e\right ) + 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{3} + 4 \, {\left (15 \, a \tan \left (f x + e\right )^{2} - 5 \, a \tan \left (f x + e\right ) - 3 \, a\right )} \sqrt {d \tan \left (f x + e\right )}}{30 \, d^{4} f \tan \left (f x + e\right )^{3}}, \frac {15 \, \sqrt {2} a d \sqrt {-\frac {1}{d}} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-\frac {1}{d}} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{3} + 2 \, {\left (15 \, a \tan \left (f x + e\right )^{2} - 5 \, a \tan \left (f x + e\right ) - 3 \, a\right )} \sqrt {d \tan \left (f x + e\right )}}{15 \, d^{4} f \tan \left (f x + e\right )^{3}}\right ] \]

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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giac [B]  time = 1.80, size = 295, normalized size = 2.44 \[ -\frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} - a {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{2 \, d^{5} f} - \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} - a {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{2 \, d^{5} f} - \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} + a {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{4 \, d^{5} f} + \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} + a {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{4 \, d^{5} f} + \frac {2 \, {\left (15 \, a d^{2} \tan \left (f x + e\right )^{2} - 5 \, a d^{2} \tan \left (f x + e\right ) - 3 \, a d^{2}\right )}}{15 \, \sqrt {d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}} \]

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)

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maple [B]  time = 0.18, size = 393, normalized size = 3.25 \[ -\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \,d^{4}}-\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{4}}+\frac {a \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{4}}+\frac {a \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f \,d^{3} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{3} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {a \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f \,d^{3} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2 a}{3 d^{2} f \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2 a}{5 d f \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {2 a}{d^{3} f \sqrt {d \tan \left (f x +e \right )}} \]

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*a/f/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*a/f/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*a/f/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*a/f/d^3*2^(1/2)/(d^2)^(1/4)*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*a/f/d^3*2^(1/2)/(d^2)^(1/4)*arctan
(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/2*a/f/d^3*2^(1/2)/(d^2)^(1/4)*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 [A]  time = 0.58, size = 136, normalized size = 1.12 \[ -\frac {\frac {15 \, a {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d^{2}} - \frac {4 \, {\left (15 \, a d^{2} \tan \left (f x + e\right )^{2} - 5 \, a d^{2} \tan \left (f x + e\right ) - 3 \, a d^{2}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} d^{2}}}{30 \, d f} \]

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]

-1/30*(15*a*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - sqrt(2)*log(d*ta
n(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/d^2 - 4*(15*a*d^2*tan(f*x + e)^2 - 5*a*d^2*tan
(f*x + e) - 3*a*d^2)/((d*tan(f*x + e))^(5/2)*d^2))/(d*f)

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mupad [B]  time = 5.80, size = 120, normalized size = 0.99 \[ -\frac {\frac {2\,a}{5\,d}-\frac {2\,a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{d}}{f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}-\frac {2\,a}{3\,d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (1+1{}\mathrm {i}\right )}{d^{7/2}\,f}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (-1+1{}\mathrm {i}\right )}{d^{7/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x))/(d*tan(e + f*x))^(7/2),x)

[Out]

((-1)^(1/4)*a*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2))*(1 + 1i))/(d^(7/2)*f) - (2*a)/(3*d^2*f*(d*tan(
e + f*x))^(3/2)) - ((2*a)/(5*d) - (2*a*tan(e + f*x)^2)/d)/(f*(d*tan(e + f*x))^(5/2)) - ((-1)^(1/4)*a*atanh(((-
1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2))*(1 - 1i))/(d^(7/2)*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {\tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx\right ) \]

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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