3.302 \(\int \frac {(b \tan (e+f x))^{3/2}}{\sqrt {d \sec (e+f x)}} \, dx\)

Optimal. Leaf size=167 \[ \frac {b^{3/2} d (b \tan (e+f x))^{3/2} \tan ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}+\frac {b^{3/2} d (b \tan (e+f x))^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}-\frac {2 d \csc (e+f x) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2}} \]

[Out]

-2*d*csc(f*x+e)*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(3/2)+b^(3/2)*d*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*(b*
tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(3/2)/(b*sin(f*x+e))^(3/2)+b^(3/2)*d*arctanh((b*sin(f*x+e))^(1/2)/b^(1/2))*
(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+e))^(3/2)/(b*sin(f*x+e))^(3/2)

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Rubi [A]  time = 0.13, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2616, 2564, 321, 329, 212, 206, 203} \[ \frac {b^{3/2} d (b \tan (e+f x))^{3/2} \tan ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}+\frac {b^{3/2} d (b \tan (e+f x))^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{f (b \sin (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}-\frac {2 d \csc (e+f x) (b \tan (e+f x))^{3/2}}{f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(3/2)/Sqrt[d*Sec[e + f*x]],x]

[Out]

(-2*d*Csc[e + f*x]*(b*Tan[e + f*x])^(3/2))/(f*(d*Sec[e + f*x])^(3/2)) + (b^(3/2)*d*ArcTan[Sqrt[b*Sin[e + f*x]]
/Sqrt[b]]*(b*Tan[e + f*x])^(3/2))/(f*(d*Sec[e + f*x])^(3/2)*(b*Sin[e + f*x])^(3/2)) + (b^(3/2)*d*ArcTanh[Sqrt[
b*Sin[e + f*x]]/Sqrt[b]]*(b*Tan[e + f*x])^(3/2))/(f*(d*Sec[e + f*x])^(3/2)*(b*Sin[e + f*x])^(3/2))

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2616

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^(m + n)*(b
*Tan[e + f*x])^n)/((a*Sec[e + f*x])^n*(b*Sin[e + f*x])^n), Int[(b*Sin[e + f*x])^n/Cos[e + f*x]^(m + n), x], x]
/; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[n + 1/2] && IntegerQ[m + 1/2]

Rubi steps

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

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Mathematica [C]  time = 4.29, size = 64, normalized size = 0.38 \[ \frac {2 (b \tan (e+f x))^{5/2} \, _2F_1\left (-\frac {1}{4},-\frac {1}{4};\frac {3}{4};\sec ^2(e+f x)\right )}{b f \left (-\tan ^2(e+f x)\right )^{5/4} \sqrt {d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^(3/2)/Sqrt[d*Sec[e + f*x]],x]

[Out]

(2*Hypergeometric2F1[-1/4, -1/4, 3/4, Sec[e + f*x]^2]*(b*Tan[e + f*x])^(5/2))/(b*f*Sqrt[d*Sec[e + f*x]]*(-Tan[
e + f*x]^2)^(5/4))

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fricas [B]  time = 1.07, size = 741, normalized size = 4.44 \[ \left [-\frac {2 \, b d \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b \cos \left (f x + e\right )^{2} - {\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - b\right )}}\right ) - b d \sqrt {-\frac {b}{d}} \log \left (\frac {b \cos \left (f x + e\right )^{4} - 72 \, b \cos \left (f x + e\right )^{2} + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 28 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, b\right )} \sin \left (f x + e\right ) + 72 \, b}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) + 16 \, b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{8 \, d f}, \frac {2 \, b d \sqrt {\frac {b}{d}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b \cos \left (f x + e\right )^{2} + {\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - b\right )}}\right ) + b d \sqrt {\frac {b}{d}} \log \left (\frac {b \cos \left (f x + e\right )^{4} - 72 \, b \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {b}{d}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} - 28 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, b\right )} \sin \left (f x + e\right ) + 72 \, b}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) - 16 \, b \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{8 \, d f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(2*b*d*sqrt(-b/d)*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 - (cos(f*x + e)^2 + 6*cos(f*x + e) + 4)*
sin(f*x + e) - 2*cos(f*x + e) + 4)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-b/d)*sqrt(d/cos(f*x + e))/(b*cos(f*
x + e)^2 - (b*cos(f*x + e) + b)*sin(f*x + e) - b)) - b*d*sqrt(-b/d)*log((b*cos(f*x + e)^4 - 72*b*cos(f*x + e)^
2 + 8*(7*cos(f*x + e)^3 - (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos(f*x + e))*sqrt(b*sin(f*x + e)
/cos(f*x + e))*sqrt(-b/d)*sqrt(d/cos(f*x + e)) + 28*(b*cos(f*x + e)^2 - 2*b)*sin(f*x + e) + 72*b)/(cos(f*x + e
)^4 - 8*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)) + 16*b*sqrt(b*sin(f*x + e)/cos(f*x + e))*sq
rt(d/cos(f*x + e))*cos(f*x + e))/(d*f), 1/8*(2*b*d*sqrt(b/d)*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 + (
cos(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt
(b/d)*sqrt(d/cos(f*x + e))/(b*cos(f*x + e)^2 + (b*cos(f*x + e) + b)*sin(f*x + e) - b)) + b*d*sqrt(b/d)*log((b*
cos(f*x + e)^4 - 72*b*cos(f*x + e)^2 - 8*(7*cos(f*x + e)^3 + (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) -
8*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(b/d)*sqrt(d/cos(f*x + e)) - 28*(b*cos(f*x + e)^2 - 2*b)
*sin(f*x + e) + 72*b)/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)) - 16*b*sq
rt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e))/(d*f)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {d \sec \left (f x + e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e))^(3/2)/sqrt(d*sec(f*x + e)), x)

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maple [C]  time = 0.68, size = 719, normalized size = 4.31 \[ \frac {\left (2 i \sin \left (f x +e \right ) \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-i \sin \left (f x +e \right ) \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \sin \left (f x +e \right ) \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\sin \left (f x +e \right ) \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\sin \left (f x +e \right ) \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (f x +e \right ) \sqrt {2}+2 \sqrt {2}\right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \cos \left (f x +e \right ) \sqrt {2}}{2 f \left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(2*I*sin(f*x+e)*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(
I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1
/2))-I*sin(f*x+e)*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(I*co
s(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1
/2*2^(1/2))-I*sin(f*x+e)*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)*
(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-
1/2*I,1/2*2^(1/2))-sin(f*x+e)*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(
1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^(1/2)
,1/2+1/2*I,1/2*2^(1/2))+sin(f*x+e)*(-I*(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+
e))^(1/2)*(-(I*cos(f*x+e)-I-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((I*cos(f*x+e)-I+sin(f*x+e))/sin(f*x+e))^
(1/2),1/2-1/2*I,1/2*2^(1/2))-2*cos(f*x+e)*2^(1/2)+2*2^(1/2))*(b*sin(f*x+e)/cos(f*x+e))^(3/2)*cos(f*x+e)/(-1+co
s(f*x+e))/(d/cos(f*x+e))^(1/2)/sin(f*x+e)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {d \sec \left (f x + e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e))^(3/2)/sqrt(d*sec(f*x + e)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*tan(e + f*x))**(3/2)/sqrt(d*sec(e + f*x)), x)

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