\(\int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx\) [112]

Optimal result

Integrand size = 23, antiderivative size = 155 \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {a \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {a \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}-\frac {2 a}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}} \]

[Out]

-a*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)+a*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)-2*a/d/e/(e
*sin(d*x+c))^(1/2)-2*a*cos(d*x+c)/d/e/(e*sin(d*x+c))^(1/2)+2*a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1
/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/d/e^2/sin(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3957, 2917, 2644, 331, 335, 304, 209, 212, 2716, 2721, 2719} \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {a \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {a \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}-\frac {2 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {2 a}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}} \]

[In]

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

[Out]

-((a*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(3/2))) + (a*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(3/2)
) - (2*a)/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a*Cos[c + d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a*EllipticE[(c - Pi/
2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*e^2*Sqrt[Sin[c + d*x]])

Rule 209

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

Rule 212

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

Rule 304

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

Rule 331

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

Rule 335

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 2644

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 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\int \frac {(-a-a \cos (c+d x)) \sec (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx \\ & = a \int \frac {1}{(e \sin (c+d x))^{3/2}} \, dx+a \int \frac {\sec (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx \\ & = -\frac {2 a \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {a \int \sqrt {e \sin (c+d x)} \, dx}{e^2}+\frac {a \text {Subst}\left (\int \frac {1}{x^{3/2} \left (1-\frac {x^2}{e^2}\right )} \, dx,x,e \sin (c+d x)\right )}{d e} \\ & = -\frac {2 a}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}+\frac {a \text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e^3}-\frac {\left (a \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{e^2 \sqrt {\sin (c+d x)}} \\ & = -\frac {2 a}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {(2 a) \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{e^2}} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e^3} \\ & = -\frac {2 a}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {a \text {Subst}\left (\int \frac {1}{e-x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e}-\frac {a \text {Subst}\left (\int \frac {1}{e+x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e} \\ & = -\frac {a \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {a \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}-\frac {2 a}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.92 \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=-\frac {a (1+\cos (c+d x)) \sec \left (\frac {1}{2} (c+d x)\right ) \left (\arctan \left (\sqrt {\sin (c+d x)}\right ) \sec \left (\frac {1}{2} (c+d x)\right )-\text {arctanh}\left (\sqrt {\sin (c+d x)}\right ) \sec \left (\frac {1}{2} (c+d x)\right )-2 E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sec \left (\frac {1}{2} (c+d x)\right )+2 \csc \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sin (c+d x)}\right ) \sin ^{\frac {3}{2}}(c+d x)}{2 d (e \sin (c+d x))^{3/2}} \]

[In]

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

[Out]

-1/2*(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]*(ArcTan[Sqrt[Sin[c + d*x]]]*Sec[(c + d*x)/2] - ArcTanh[Sqrt[Sin[c
+ d*x]]]*Sec[(c + d*x)/2] - 2*EllipticE[(-2*c + Pi - 2*d*x)/4, 2]*Sec[(c + d*x)/2] + 2*Csc[(c + d*x)/2]*Sqrt[S
in[c + d*x]])*Sin[c + d*x]^(3/2))/(d*(e*Sin[c + d*x])^(3/2))

Maple [A] (verified)

Time = 10.07 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {-\frac {a \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {3}{2}}}-\frac {2 a}{e \sqrt {e \sin \left (d x +c \right )}}+\frac {a \,\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {3}{2}}}+\frac {a \left (2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c \right )^{2}\right )}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(201\)
parts	\(\frac {a \left (2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c \right )^{2}\right )}{e \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a \left (-\frac {\arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {3}{2}}}-\frac {2}{e \sqrt {e \sin \left (d x +c \right )}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {3}{2}}}\right )}{d}\)	\(203\)

[In]

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

[Out]

(-a/e^(3/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))-2*a/e/(e*sin(d*x+c))^(1/2)+a/e^(3/2)*arctanh((e*sin(d*x+c))^(
1/2)/e^(1/2))+a/e*(2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(
1/2),1/2*2^(1/2))-(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2
),1/2*2^(1/2))-2*cos(d*x+c)^2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.20 (sec) , antiderivative size = 670, normalized size of antiderivative = 4.32 \[ \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx=\left [\frac {-8 i \, \sqrt {2} a \sqrt {-i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 8 i \, \sqrt {2} a \sqrt {i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, a \sqrt {-e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} - e \sin \left (d x + c\right ) - e\right )}}\right ) \sin \left (d x + c\right ) - a \sqrt {-e} \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} + 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {-e} + 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) \sin \left (d x + c\right ) - 16 \, {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}}{8 \, d e^{2} \sin \left (d x + c\right )}, \frac {-8 i \, \sqrt {2} a \sqrt {-i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 8 i \, \sqrt {2} a \sqrt {i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, a \sqrt {e} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e}}{4 \, {\left (e \cos \left (d x + c\right )^{2} + e \sin \left (d x + c\right ) - e\right )}}\right ) \sin \left (d x + c\right ) + a \sqrt {e} \log \left (\frac {e \cos \left (d x + c\right )^{4} - 72 \, e \cos \left (d x + c\right )^{2} - 8 \, {\left (7 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 8\right )} \sin \left (d x + c\right ) - 8\right )} \sqrt {e \sin \left (d x + c\right )} \sqrt {e} - 28 \, {\left (e \cos \left (d x + c\right )^{2} - 2 \, e\right )} \sin \left (d x + c\right ) + 72 \, e}{\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8}\right ) \sin \left (d x + c\right ) - 16 \, {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}}{8 \, d e^{2} \sin \left (d x + c\right )}\right ] \]

[In]

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

[Out]

[1/8*(-8*I*sqrt(2)*a*sqrt(-I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*
sin(d*x + c))) + 8*I*sqrt(2)*a*sqrt(I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x
+ c) - I*sin(d*x + c))) - 2*a*sqrt(-e)*arctan(1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 2)*sqrt(e*sin(d*x + c))*s
qrt(-e)/(e*cos(d*x + c)^2 - e*sin(d*x + c) - e))*sin(d*x + c) - a*sqrt(-e)*log((e*cos(d*x + c)^4 - 72*e*cos(d*
x + c)^2 + 8*(7*cos(d*x + c)^2 - (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e*sin(d*x + c))*sqrt(-e) + 28*(e*
cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x
+ c) + 8))*sin(d*x + c) - 16*(a*cos(d*x + c) + a)*sqrt(e*sin(d*x + c)))/(d*e^2*sin(d*x + c)), 1/8*(-8*I*sqrt(
2)*a*sqrt(-I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) +
8*I*sqrt(2)*a*sqrt(I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x
+ c))) - 2*a*sqrt(e)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 2)*sqrt(e*sin(d*x + c))*sqrt(e)/(e*cos(d*x
+ c)^2 + e*sin(d*x + c) - e))*sin(d*x + c) + a*sqrt(e)*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 - 8*(7*cos
(d*x + c)^2 + (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e*sin(d*x + c))*sqrt(e) - 28*(e*cos(d*x + c)^2 - 2*e
)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8))*sin(d*x
+ c) - 16*(a*cos(d*x + c) + a)*sqrt(e*sin(d*x + c)))/(d*e^2*sin(d*x + c))]

Sympy [F]

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

[In]

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

[Out]

a*(Integral((e*sin(c + d*x))**(-3/2), x) + Integral(sec(c + d*x)/(e*sin(c + d*x))**(3/2), x))

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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