\(\int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx\) [122]

Optimal result

Integrand size = 25, antiderivative size = 199 \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}+\sqrt {e} \tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {e^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{a d \sqrt {e \tan (c+d x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.64 (sec) , antiderivative size = 1211, normalized size of antiderivative = 6.09 \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(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[Sqr 
t[-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])*T 
an[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)*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)))]*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))]*Hypergeometric2...
 

Rubi [A] (warning: unable to verify)

Time = 0.85 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.20, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {3042, 4376, 25, 3042, 4372, 3042, 3094, 3042, 3053, 3042, 3120, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\)

\(\Big \downarrow \) 4376

\(\displaystyle \frac {e^2 \int -\frac {a-a \sec (c+d x)}{\sqrt {e \tan (c+d x)}}dx}{a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {e^2 \int \frac {a-a \sec (c+d x)}{\sqrt {e \tan (c+d x)}}dx}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \int \frac {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}\)

\(\Big \downarrow \) 4372

\(\displaystyle -\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-a \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}}dx\right )}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-a \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}}dx\right )}{a^2}\)

\(\Big \downarrow \) 3094

\(\displaystyle -\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}}dx}{\sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}}dx}{\sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 3053

\(\displaystyle -\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}}dx}{\sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}}dx}{\sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {e^2 \left (a \int \frac {1}{\sqrt {e \tan (c+d x)}}dx-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle -\frac {e^2 \left (\frac {a e \int \frac {1}{\sqrt {e \tan (c+d x)} \left (\tan ^2(c+d x) e^2+e^2\right )}d(e \tan (c+d x))}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {e^2 \left (\frac {2 a e \int \frac {1}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 755

\(\displaystyle -\frac {e^2 \left (\frac {2 a e \left (\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\int \frac {e^2 \tan ^2(c+d x)+e}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle -\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}+\frac {1}{2} \int \frac {1}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}-\frac {\int \frac {1}{-e^2 \tan ^2(c+d x)-1}d\left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}\right )}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {e^2 \left (\frac {2 a e \left (\frac {\int \frac {e-e^2 \tan ^2(c+d x)}{e^4 \tan ^4(c+d x)+e^2}d\sqrt {e \tan (c+d x)}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle -\frac {e^2 \left (\frac {2 a e \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {e}-2 \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)-\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {2} \sqrt {e}}+\frac {\int \frac {\sqrt {e}+\sqrt {2} \sqrt {e \tan (c+d x)}}{e^2 \tan ^2(c+d x)+\sqrt {2} e^{3/2} \tan (c+d x)+e}d\sqrt {e \tan (c+d x)}}{2 \sqrt {e}}}{2 e}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {e^2 \left (\frac {2 a e \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {e} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {e}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {e} \tan (c+d x)\right )}{\sqrt {2} \sqrt {e}}}{2 e}+\frac {\frac {\log \left (\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\log \left (-\sqrt {2} e^{3/2} \tan (c+d x)+e^2 \tan ^2(c+d x)+e\right )}{2 \sqrt {2} \sqrt {e}}}{2 e}\right )}{d}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\right )}{a^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
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 1479

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[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 3094

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] 
:> Simp[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[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 3120

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[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 4372

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

rule 4376

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Simp[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] && EqQ[a 
^2 - b^2, 0] && ILtQ[n, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.36 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (i \operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-2 i \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticPi}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+2 \operatorname {EllipticF}\left (\sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {e \tan \left (d x +c \right )}\, e \sqrt {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}\, \sqrt {2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )+2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \left (\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{a d}\)	\(203\)

Input:

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

Output:

(1/2+1/2*I)/a/d*(I*EllipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2+1/2*I,1 
/2*2^(1/2))-2*I*EllipticF((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2))-El 
lipticPi((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))+2*Ellipti 
cF((-cot(d*x+c)+csc(d*x+c)+1)^(1/2),1/2*2^(1/2)))*(e*tan(d*x+c))^(1/2)*e*( 
-cot(d*x+c)+csc(d*x+c)+1)^(1/2)*(2*cot(d*x+c)-2*csc(d*x+c)+2)^(1/2)*(-csc( 
d*x+c)+cot(d*x+c))^(1/2)*(csc(d*x+c)+cot(d*x+c))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) e}{a} \] Input:

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

Output:

(sqrt(e)*int((sqrt(tan(c + d*x))*tan(c + d*x))/(sec(c + d*x) + 1),x)*e)/a