3.122 \(\int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=257 \[ \frac {e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}+\frac {e^{3/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {e^{3/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {e^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{a d \sqrt {e \tan (c+d x)}} \]

[Out]

1/2*e^(3/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/d*2^(1/2)-1/2*e^(3/2)*arctan(1+2^(1/2)*(e*tan(d*x
+c))^(1/2)/e^(1/2))/a/d*2^(1/2)+1/4*e^(3/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d*2^
(1/2)-1/4*e^(3/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d*2^(1/2)-e^2*(sin(c+1/4*Pi+d*
x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*EllipticF(cos(c+1/4*Pi+d*x),2^(1/2))*sec(d*x+c)*sin(2*d*x+2*c)^(1/2)/a/d/(e*tan(
d*x+c))^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3888, 3884, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2614, 2573, 2641} \[ \frac {e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}+\frac {e^{3/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {e^{3/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {e^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{a d \sqrt {e \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*d) - (e^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt
[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*d) + (e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c
+ d*x]]])/(2*Sqrt[2]*a*d) - (e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*S
qrt[2]*a*d) + (e^2*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(a*d*Sqrt[e*Tan[c + d*x]]
)

Rule 204

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

Rule 211

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

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3884

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx &=\frac {e^2 \int \frac {-a+a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{a^2}\\ &=-\frac {e^2 \int \frac {1}{\sqrt {e \tan (c+d x)}} \, dx}{a}+\frac {e^2 \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{a}\\ &=-\frac {e^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{a d}+\frac {\left (e^2 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}} \, dx}{a \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}}\\ &=-\frac {\left (2 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}+\frac {\left (e^2 \sec (c+d x) \sqrt {\sin (2 c+2 d x)}\right ) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}} \, dx}{a \sqrt {e \tan (c+d x)}}\\ &=\frac {e^2 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}}-\frac {e^2 \operatorname {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {e^2 \operatorname {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}\\ &=\frac {e^2 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}}+\frac {e^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^2 \operatorname {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d}-\frac {e^2 \operatorname {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d}\\ &=\frac {e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^2 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}}-\frac {e^{3/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{3/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}\\ &=\frac {e^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {e^2 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 13.21, size = 1211, normalized size = 4.71 \[ -\frac {4 \sqrt [4]{-1} \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right )\right |-1\right ) (e \tan (c+d x))^{3/2} \sec ^4(c+d x)}{d (\sec (c+d x) a+a) \tan ^{\frac {3}{2}}(c+d x) \left (\tan ^2(c+d x)+1\right )^{3/2}}-\frac {2 e^{-i (c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{d (\sec (c+d x) a+a) \tan ^{\frac {3}{2}}(c+d x)}-\frac {e^{-2 i c} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (e^{4 i c} \sqrt {-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt {-1+e^{4 i (c+d x)}}\right )+2 \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{2 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \tan ^{\frac {3}{2}}(c+d x)}-\frac {e^{-2 i c} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (\sqrt {-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt {-1+e^{4 i (c+d x)}}\right )+2 e^{4 i c} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{2 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \tan ^{\frac {3}{2}}(c+d x)}+\frac {e^{-i (2 c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (3 \left (-1+e^{4 i (c+d x)}\right )+e^{4 i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1-e^{4 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{4 i (c+d x)}\right )\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{3 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \tan ^{\frac {3}{2}}(c+d x)}-\frac {e^{-i d x} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (e^{2 i (c+2 d x)} \left (-1+e^{2 i c}\right ) \sqrt {1-e^{4 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{4 i (c+d x)}\right )-3 e^{4 i (c+d x)}+3\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{3 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \tan ^{\frac {3}{2}}(c+d x)}+\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc (c+d x) \left (\frac {8 \cos (c) \cos (d x) \sec (2 c) \sin ^2\left (\frac {c}{2}\right )}{d}-\frac {16 \cos \left (\frac {c}{2}\right ) \sec (2 c) \sin ^3\left (\frac {c}{2}\right ) \sin (d x)}{d}\right ) (e \tan (c+d x))^{3/2}}{\sec (c+d x) a+a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c/2 + (d*x)/2]^2*Csc[c + d*x]*((8*Cos[c]*Cos[d*x]*Sec[2*c]*Sin[c/2]^2)/d - (16*Cos[c/2]*Sec[2*c]*Sin[c/2]
^3*Sin[d*x])/d)*(e*Tan[c + d*x])^(3/2))/(a + a*Sec[c + d*x]) - (2*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 +
E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*(e*Tan[c + d*x])^(3
/2))/(d*E^(I*(c + d*x))*(a + a*Sec[c + d*x])*Tan[c + d*x]^(3/2)) - (Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1
+ E^((2*I)*(c + d*x)))]*(E^((4*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]] + 2
*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^(
(2*I)*(c + d*x)))]])*Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*(e*Tan[c + d*x])^(3/2))/(2*d*E^((2*I)*c)*(-1 +
E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])*Tan[c + d*x]^(3/2)) - (Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 +
E^((2*I)*(c + d*x)))]*(Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]] + 2*E^((4*I)*c)*S
qrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2
*I)*(c + d*x)))]])*Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*(e*Tan[c + d*x])^(3/2))/(2*d*E^((2*I)*c)*(-1 + E
^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])*Tan[c + d*x]^(3/2)) + (Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^
((2*I)*(c + d*x)))]*Cos[c/2 + (d*x)/2]^2*(3*(-1 + E^((4*I)*(c + d*x))) + E^((4*I)*(c + d*x))*(-1 + E^((2*I)*c)
)*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]*(
e*Tan[c + d*x])^(3/2))/(3*d*E^(I*(2*c + d*x))*(-1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])*Tan[c + d*x]^(3/
2)) - (Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*Cos[c/2 + (d*x)/2]^2*(3 - 3*E^((4*I)*
(c + d*x)) + E^((2*I)*(c + 2*d*x))*(-1 + E^((2*I)*c))*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4
, 7/4, E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]*(e*Tan[c + d*x])^(3/2))/(3*d*E^(I*d*x)*(-1 + E^((2*I)*(c +
d*x)))*(a + a*Sec[c + d*x])*Tan[c + d*x]^(3/2)) - (4*(-1)^(1/4)*Cos[c/2 + (d*x)/2]^2*EllipticF[I*ArcSinh[(-1)^
(1/4)*Sqrt[Tan[c + d*x]]], -1]*Sec[c + d*x]^4*(e*Tan[c + d*x])^(3/2))/(d*(a + a*Sec[c + d*x])*Tan[c + d*x]^(3/
2)*(1 + Tan[c + d*x]^2)^(3/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*tan(d*x + c))^(3/2)/(a*sec(d*x + c) + a), x)

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maple [C]  time = 1.83, size = 319, normalized size = 1.24 \[ \frac {\left (1+\cos \left (d x +c \right )\right )^{2} \left (i \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-4 \EllipticF \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (\frac {e \sin \left (d x +c \right )}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \sqrt {2}}{2 a d \sin \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x)

[Out]

1/2/a/d*(1+cos(d*x+c))^2*(I*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-I*E
llipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-4*EllipticF(((1-cos(d*x+c)+sin(d
*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^
(1/2))+EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2)))*((-1+cos(d*x+c))/sin(d*
x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-1+cos
(d*x+c))*(e*sin(d*x+c)/cos(d*x+c))^(3/2)*cos(d*x+c)/sin(d*x+c)^4*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*tan(d*x + c))^(3/2)/(a*sec(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(c + d*x))^(3/2)/(a + a/cos(c + d*x)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))**(3/2)/(a+a*sec(d*x+c)),x)

[Out]

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

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