3.2.13 \(\int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx\) [113]

Optimal. Leaf size=160 \[ \frac {a \text {ArcTan}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}-\frac {2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}} \]

[Out]

a*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)+a*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)-2/3*a/d/e/(
e*sin(d*x+c))^(3/2)-2/3*a*cos(d*x+c)/d/e/(e*sin(d*x+c))^(3/2)-2/3*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)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/e^2/(e*sin(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3957, 2917, 2644, 331, 335, 218, 212, 209, 2716, 2721, 2720} \begin {gather*} \frac {a \text {ArcTan}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])/(e*Sin[c + d*x])^(5/2),x]

[Out]

(a*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(5/2)) + (a*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(5/2)) -
(2*a)/(3*d*e*(e*Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x])/(3*d*e*(e*Sin[c + d*x])^(3/2)) + (2*a*EllipticF[(c
- Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[e*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 218

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] &&  !Gt
Q[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 2720

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 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*} \int \frac {a+a \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx &=-\int \frac {(-a-a \cos (c+d x)) \sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx\\ &=a \int \frac {1}{(e \sin (c+d x))^{5/2}} \, dx+a \int \frac {\sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx\\ &=-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 e^2}+\frac {a \text {Subst}\left (\int \frac {1}{x^{5/2} \left (1-\frac {x^2}{e^2}\right )} \, dx,x,e \sin (c+d x)\right )}{d e}\\ &=-\frac {2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {a \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{e^2}\right )} \, dx,x,e \sin (c+d x)\right )}{d e^3}+\frac {\left (a \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 e^2 \sqrt {e \sin (c+d x)}}\\ &=-\frac {2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}}+\frac {(2 a) \text {Subst}\left (\int \frac {1}{1-\frac {x^4}{e^2}} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e^3}\\ &=-\frac {2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \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^2}+\frac {a \text {Subst}\left (\int \frac {1}{e+x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e^2}\\ &=\frac {a \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{5/2}}-\frac {2 a}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.24, size = 120, normalized size = 0.75 \begin {gather*} -\frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-3 \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right )-3 \tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )+2 F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+\csc ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sin (c+d x)}\right ) \sqrt {\sin (c+d x)}}{6 d e^2 \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])/(e*Sin[c + d*x])^(5/2),x]

[Out]

-1/6*(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-3*ArcTan[Sqrt[Sin[c + d*x]]] - 3*ArcTanh[Sqrt[Sin[c + d*x]]] +
2*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + Csc[(c + d*x)/2]^2*Sqrt[Sin[c + d*x]])*Sqrt[Sin[c + d*x]])/(d*e^2*Sqr
t[e*Sin[c + d*x]])

________________________________________________________________________________________

Maple [A]
time = 0.20, size = 164, normalized size = 1.02

method	result	size
default	\(\frac {-\frac {2 a}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a \arctanh \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}+\frac {a \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )}{e^{\frac {5}{2}}}-\frac {a \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(-2/3*a/e/(e*sin(d*x+c))^(3/2)+a/e^(5/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))+a/e^(5/2)*arctan((e*sin(d*x+c))
^(1/2)/e^(1/2))-1/3*a/e^2*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((-sin(d*x+c
)+1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^3+2*sin(d*x+c))/sin(d*x+c)^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

e^(-5/2)*integrate((a*sec(d*x + c) + a)/sin(d*x + c)^(5/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 4.33, size = 219, normalized size = 1.37 \begin {gather*} \frac {4 \, \sqrt {-i} {\left (\sqrt {2} a \cos \left (d x + c\right ) - \sqrt {2} a\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 4 \, \sqrt {i} {\left (\sqrt {2} a \cos \left (d x + c\right ) - \sqrt {2} a\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 6 \, {\left (a \cos \left (d x + c\right ) - a\right )} \arctan \left (\frac {\sin \left (d x + c\right ) - 1}{2 \, \sqrt {\sin \left (d x + c\right )}}\right ) + 3 \, {\left (a \cos \left (d x + c\right ) - a\right )} \log \left (\frac {\cos \left (d x + c\right )^{2} - 4 \, {\left (\sin \left (d x + c\right ) + 1\right )} \sqrt {\sin \left (d x + c\right )} - 6 \, \sin \left (d x + c\right ) - 2}{\cos \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) - 2}\right ) + 8 \, a \sqrt {\sin \left (d x + c\right )}}{12 \, {\left (d \cos \left (d x + c\right ) e^{\frac {5}{2}} - d e^{\frac {5}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/12*(4*sqrt(-I)*(sqrt(2)*a*cos(d*x + c) - sqrt(2)*a)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))
+ 4*sqrt(I)*(sqrt(2)*a*cos(d*x + c) - sqrt(2)*a)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 6
*(a*cos(d*x + c) - a)*arctan(1/2*(sin(d*x + c) - 1)/sqrt(sin(d*x + c))) + 3*(a*cos(d*x + c) - a)*log((cos(d*x
+ c)^2 - 4*(sin(d*x + c) + 1)*sqrt(sin(d*x + c)) - 6*sin(d*x + c) - 2)/(cos(d*x + c)^2 + 2*sin(d*x + c) - 2))
+ 8*a*sqrt(sin(d*x + c)))/(d*cos(d*x + c)*e^(5/2) - d*e^(5/2))

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3007 deep

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)/(e*sin(d*x + c))^(5/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+\frac {a}{\cos \left (c+d\,x\right )}}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))/(e*sin(c + d*x))^(5/2),x)

[Out]

int((a + a/cos(c + d*x))/(e*sin(c + d*x))^(5/2), x)

________________________________________________________________________________________