3.362 \(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+a \tan (e+f x))} \, dx\)

Optimal. Leaf size=111 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a d^{3/2} f}+\frac {\tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {2}{a d f \sqrt {d \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3569, 3653, 3532, 205, 3634, 63} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a d^{3/2} f}+\frac {\tan ^{-1}\left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {2}{a d f \sqrt {d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])),x]

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]

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 3569

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 + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*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] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3653

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

Rubi steps

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

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Mathematica [A]  time = 1.68, size = 122, normalized size = 1.10 \[ -\frac {-\sqrt {2} \sqrt {\tan (e+f x)} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )+\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right ) \sqrt {\tan (e+f x)}+2 \tan ^{-1}\left (\sqrt {\tan (e+f x)}\right ) \sqrt {\tan (e+f x)}+4}{2 a d f \sqrt {d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])),x]

[Out]

-1/2*(4 - Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*Sqrt[Tan[e + f*x]] + Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[
Tan[e + f*x]]]*Sqrt[Tan[e + f*x]] + 2*ArcTan[Sqrt[Tan[e + f*x]]]*Sqrt[Tan[e + f*x]])/(a*d*f*Sqrt[d*Tan[e + f*x
]])

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fricas [A]  time = 0.51, size = 271, normalized size = 2.44 \[ \left [-\frac {\sqrt {2} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right ) + 2 \, \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) \tan \left (f x + e\right ) + 8 \, \sqrt {d \tan \left (f x + e\right )}}{4 \, a d^{2} f \tan \left (f x + e\right )}, -\frac {\sqrt {2} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right ) + 2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) \tan \left (f x + e\right ) + 4 \, \sqrt {d \tan \left (f x + e\right )}}{2 \, a d^{2} f \tan \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(sqrt(2)*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))*tan(f*x + e) + 2*sqrt(-d)*log((d*tan(f*x + e) + 2*sqrt(d*tan(f*x +
e))*sqrt(-d) - d)/(tan(f*x + e) + 1))*tan(f*x + e) + 8*sqrt(d*tan(f*x + e)))/(a*d^2*f*tan(f*x + e)), -1/2*(sqr
t(2)*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*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))*tan(f*x + e) + 4*sqrt(d*tan(f*x + e)))/(a*d^2*f*tan(f*x + e))]

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giac [B]  time = 1.36, size = 284, normalized size = 2.56 \[ -\frac {\frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a d^{2} f} + \frac {2 \, \sqrt {2} {\left (d \sqrt {{\left | d \right |}} + {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{a d^{2} f} + \frac {8 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a \sqrt {d} f} + \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a d^{2} f} - \frac {\sqrt {2} {\left (d \sqrt {{\left | d \right |}} - {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{a d^{2} f} + \frac {16}{\sqrt {d \tan \left (f x + e\right )} a f}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*sqrt(2)*(d*sqrt(abs(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)))/(a*d^2*f) + 2*sqrt(2)*(d*sqrt(abs(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)))/(a*d^2*f) + 8*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a*sqrt(d)*f)
+ sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)) + abs
(d))/(a*d^2*f) - sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqr
t(abs(d)) + abs(d))/(a*d^2*f) + 16/(sqrt(d*tan(f*x + e))*a*f))/d

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maple [B]  time = 0.28, size = 395, normalized size = 3.56 \[ -\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{8 f a \,d^{2}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a \,d^{2}}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a \,d^{2}}-\frac {\sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{8 f a d \left (d^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a d \left (d^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{4 f a d \left (d^{2}\right )^{\frac {1}{4}}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{a \,d^{\frac {3}{2}} f}-\frac {2}{a d f \sqrt {d \tan \left (f x +e \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(3/2)/(a+a*tan(f*x+e)),x)

[Out]

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

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maxima [A]  time = 0.52, size = 124, normalized size = 1.12 \[ -\frac {\frac {\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}}{a} + \frac {2 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a \sqrt {d}} + \frac {4}{\sqrt {d \tan \left (f x + e\right )} a}}{2 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*arctan
(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/a + 2*arctan(sqrt(d*tan(f*x + e))/s
qrt(d))/(a*sqrt(d)) + 4/(sqrt(d*tan(f*x + e))*a))/(d*f)

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mupad [B]  time = 4.44, size = 124, normalized size = 1.12 \[ -\frac {2}{a\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,d^{3/2}\,f}-\frac {\sqrt {2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{4\,a\,d^{3/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d*tan(e + f*x))^(3/2)*(a + a*tan(e + f*x))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )} + \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))**(3/2)/(a+a*tan(f*x+e)),x)

[Out]

Integral(1/((d*tan(e + f*x))**(3/2)*tan(e + f*x) + (d*tan(e + f*x))**(3/2)), x)/a

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