3.4.36 \(\int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx\) [336]

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

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 115, 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 = {3609, 3613, 214} \begin {gather*} \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^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 {gather*}

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 214

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

Rule 3609

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 3613

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 (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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.85, size = 117, normalized size = 1.02 \begin {gather*} \frac {\left (\frac {1}{15}+\frac {i}{15}\right ) a (d \tan (e+f x))^{5/2} \left (-15 \sqrt [4]{-1} \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+15 (-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+(1-i) \sqrt {\tan (e+f x)} \left (-15+5 \tan (e+f x)+3 \tan ^2(e+f x)\right )\right )}{f \tan ^{\frac {5}{2}}(e+f x)} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(317\) vs. \(2(94)=188\).
time = 0.49, size = 318, normalized size = 2.77

method	result	size
derivativedivides	\(\frac {a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\)	\(318\)
default	\(\frac {a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\)	\(318\)

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,method=_RETURNVERBOSE)

[Out]

1/f*a*(2/5*(d*tan(f*x+e))^(5/2)+2/3*d*(d*tan(f*x+e))^(3/2)-2*d^2*(d*tan(f*x+e))^(1/2)+2*d^3*(1/8/d*(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)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)
/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/8/(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)))+2*arctan(2^(1/2)/
(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))

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Maxima [A]
time = 0.50, size = 141, normalized size = 1.23 \begin {gather*} \frac {15 \, a d^{4} {\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 )} + 12 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a d + 20 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d^{2} - 60 \, \sqrt {d \tan \left (f x + e\right )} a d^{3}}{30 \, d f} \end {gather*}

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]

1/30*(15*a*d^4*(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
*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) + 12*(d*tan(f*x + e))^(5/2)*a*d + 20*(d*tan
(f*x + e))^(3/2)*a*d^2 - 60*sqrt(d*tan(f*x + e))*a*d^3)/(d*f)

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Fricas [A]
time = 1.49, size = 241, normalized size = 2.10 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (99) = 198\).
time = 0.73, size = 314, normalized size = 2.73 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 5.47, size = 144, normalized size = 1.25 \begin {gather*} \frac {2\,a\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,f}+\frac {2\,a\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}-\frac {2\,a\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}-\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (-1-\mathrm {i}\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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