3.352 \(\int \sqrt{d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=138 \[ \frac{2 \sqrt{2} a^3 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}+\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac{4 a^3 \sqrt{d \tan (e+f x)}}{f}+\frac{2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f} \]

[Out]

(2*Sqrt[2]*a^3*Sqrt[d]*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f + (4*a^3*Sqr
t[d*Tan[e + f*x]])/f + (8*a^3*(d*Tan[e + f*x])^(3/2))/(5*d*f) + (2*(d*Tan[e + f*x])^(3/2)*(a^3 + a^3*Tan[e + f
*x]))/(5*d*f)

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Rubi [A]  time = 0.18343, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3566, 3630, 3528, 3532, 205} \[ \frac{2 \sqrt{2} a^3 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}+\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac{4 a^3 \sqrt{d \tan (e+f x)}}{f}+\frac{2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])^3,x]

[Out]

(2*Sqrt[2]*a^3*Sqrt[d]*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/f + (4*a^3*Sqr
t[d*Tan[e + f*x]])/f + (8*a^3*(d*Tan[e + f*x])^(3/2))/(5*d*f) + (2*(d*Tan[e + f*x])^(3/2)*(a^3 + a^3*Tan[e + f
*x]))/(5*d*f)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

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 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 \sqrt{d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx &=\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+a^3 \tan (e+f x)\right )}{5 d f}+\frac{2 \int \sqrt{d \tan (e+f x)} \left (a^3 d+5 a^3 d \tan (e+f x)+6 a^3 d \tan ^2(e+f x)\right ) \, dx}{5 d}\\ &=\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+a^3 \tan (e+f x)\right )}{5 d f}+\frac{2 \int \sqrt{d \tan (e+f x)} \left (-5 a^3 d+5 a^3 d \tan (e+f x)\right ) \, dx}{5 d}\\ &=\frac{4 a^3 \sqrt{d \tan (e+f x)}}{f}+\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+a^3 \tan (e+f x)\right )}{5 d f}+\frac{2 \int \frac{-5 a^3 d^2-5 a^3 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{5 d}\\ &=\frac{4 a^3 \sqrt{d \tan (e+f x)}}{f}+\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+a^3 \tan (e+f x)\right )}{5 d f}-\frac{\left (20 a^6 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{50 a^6 d^4+d x^2} \, dx,x,\frac{-5 a^3 d^2+5 a^3 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=\frac{2 \sqrt{2} a^3 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{f}+\frac{4 a^3 \sqrt{d \tan (e+f x)}}{f}+\frac{8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 (d \tan (e+f x))^{3/2} \left (a^3+a^3 \tan (e+f x)\right )}{5 d f}\\ \end{align*}

Mathematica [C]  time = 1.81962, size = 315, normalized size = 2.28 \[ \frac{a^3 \cos (e+f x) (\tan (e+f x)+1)^3 \sqrt{d \tan (e+f x)} \left (3 \left (4 \sin ^2(e+f x) \sqrt{\tan (e+f x)}+10 \sin (2 (e+f x)) \sqrt{\tan (e+f x)}+10 \sqrt{2} \cos ^2(e+f x) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )-10 \sqrt{2} \cos ^2(e+f x) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )+40 \cos ^2(e+f x) \sqrt{\tan (e+f x)}+5 \sqrt{2} \cos ^2(e+f x) \log \left (\tan (e+f x)-\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-5 \sqrt{2} \cos ^2(e+f x) \log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )\right )-20 \sin (2 (e+f x)) \sqrt{\tan (e+f x)} \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\tan ^2(e+f x)\right )\right )}{30 f \sqrt{\tan (e+f x)} (\sin (e+f x)+\cos (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])^3,x]

[Out]

(a^3*Cos[e + f*x]*(3*(10*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^2 - 10*Sqrt[2]*ArcTan[1 +
Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^2 + 5*Sqrt[2]*Cos[e + f*x]^2*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Ta
n[e + f*x]] - 5*Sqrt[2]*Cos[e + f*x]^2*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] + 40*Cos[e + f*x]^2*
Sqrt[Tan[e + f*x]] + 4*Sin[e + f*x]^2*Sqrt[Tan[e + f*x]] + 10*Sin[2*(e + f*x)]*Sqrt[Tan[e + f*x]]) - 20*Hyperg
eometric2F1[3/4, 1, 7/4, -Tan[e + f*x]^2]*Sin[2*(e + f*x)]*Sqrt[Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]]*(1 + Tan[e
+ f*x])^3)/(30*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[Tan[e + f*x]])

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Maple [B]  time = 0.02, size = 391, normalized size = 2.8 \begin{align*}{\frac{2\,{a}^{3}}{5\,f{d}^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}+2\,{\frac{{a}^{3} \left ( d\tan \left ( fx+e \right ) \right ) ^{3/2}}{df}}+4\,{\frac{{a}^{3}\sqrt{d\tan \left ( fx+e \right ) }}{f}}-{\frac{{a}^{3}\sqrt{2}}{2\,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}^{3}\sqrt{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}^{3}\sqrt{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}^{3}d\sqrt{2}}{2\,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}^{3}d\sqrt{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}^{3}d\sqrt{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))^(1/2)*(a+a*tan(f*x+e))^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71056, size = 581, normalized size = 4.21 \begin{align*} \left [\frac{5 \, \sqrt{2} a^{3} \sqrt{-d} \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 ) + 2 \,{\left (a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + 10 \, a^{3}\right )} \sqrt{d \tan \left (f x + e\right )}}{5 \, f}, -\frac{2 \,{\left (5 \, \sqrt{2} a^{3} \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 ) -{\left (a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + 10 \, a^{3}\right )} \sqrt{d \tan \left (f x + e\right )}\right )}}{5 \, f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(1/2)*(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/5*(5*sqrt(2)*a^3*sqrt(-d)*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)) + 2*(a^3*tan(f*x + e)^2 + 5*a^3*tan(f*x + e) + 10*a^3)*sqrt(d*
tan(f*x + e)))/f, -2/5*(5*sqrt(2)*a^3*sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) - 1)/(sqrt
(d)*tan(f*x + e))) - (a^3*tan(f*x + e)^2 + 5*a^3*tan(f*x + e) + 10*a^3)*sqrt(d*tan(f*x + e)))/f]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \sqrt{d \tan{\left (e + f x \right )}}\, dx + \int 3 \sqrt{d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx + \int 3 \sqrt{d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \sqrt{d \tan{\left (e + f x \right )}} \tan ^{3}{\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))**(1/2)*(a+a*tan(f*x+e))**3,x)

[Out]

a**3*(Integral(sqrt(d*tan(e + f*x)), x) + Integral(3*sqrt(d*tan(e + f*x))*tan(e + f*x), x) + Integral(3*sqrt(d
*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(sqrt(d*tan(e + f*x))*tan(e + f*x)**3, x))

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Giac [B]  time = 1.3908, size = 464, normalized size = 3.36 \begin{align*} -\frac{\sqrt{2}{\left (a^{3} d \sqrt{{\left | d \right |}} - a^{3}{\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 )}{2 \, d f} + \frac{\sqrt{2}{\left (a^{3} d \sqrt{{\left | d \right |}} - a^{3}{\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 )}{2 \, d f} - \frac{{\left (\sqrt{2} a^{3} d \sqrt{{\left | d \right |}} + \sqrt{2} a^{3}{\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 )}{d f} - \frac{{\left (\sqrt{2} a^{3} d \sqrt{{\left | d \right |}} + \sqrt{2} a^{3}{\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 )}{d f} + \frac{2 \,{\left (\sqrt{d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right )^{2} + 5 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{10} f^{4} \tan \left (f x + e\right ) + 10 \, \sqrt{d \tan \left (f x + e\right )} a^{3} d^{10} f^{4}\right )}}{5 \, d^{10} f^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(1/2)*(a+a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) - a^3*abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(ab
s(d)) + abs(d))/(d*f) + 1/2*sqrt(2)*(a^3*d*sqrt(abs(d)) - a^3*abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(
d*tan(f*x + e))*sqrt(abs(d)) + abs(d))/(d*f) - (sqrt(2)*a^3*d*sqrt(abs(d)) + sqrt(2)*a^3*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*f) - (sqrt(2)*a^3*d*sqrt(abs(d))
+ sqrt(2)*a^3*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*f) + 2/5*(sqrt(d*tan(f*x + e))*a^3*d^10*f^4*tan(f*x + e)^2 + 5*sqrt(d*tan(f*x + e))*a^3*d^10*f^4*tan(f*x +
e) + 10*sqrt(d*tan(f*x + e))*a^3*d^10*f^4)/(d^10*f^5)