3.109 \(\int \frac {a+a \sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=282 \[ \frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}-\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{5/2}}+\frac {a \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{3 d e^2 \sqrt {e \tan (c+d x)}}-\frac {2 (a \sec (c+d x)+a)}{3 d e (e \tan (c+d x))^{3/2}} \]

[Out]

1/2*a*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)*2^(1/2)-1/2*a*arctan(1+2^(1/2)*(e*tan(d*x+c))^(
1/2)/e^(1/2))/d/e^(5/2)*2^(1/2)+1/4*a*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/d/e^(5/2)*2^
(1/2)-1/4*a*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/d/e^(5/2)*2^(1/2)+1/3*a*(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)/d/e^2/(e*
tan(d*x+c))^(1/2)-2/3*(a+a*sec(d*x+c))/d/e/(e*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.28, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3882, 3884, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2614, 2573, 2641} \[ \frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}-\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{5/2}}+\frac {a \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {a \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {a \sqrt {\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{3 d e^2 \sqrt {e \tan (c+d x)}}-\frac {2 (a \sec (c+d x)+a)}{3 d e (e \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2)) - (a*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan
[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2)) + (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x
]]])/(2*Sqrt[2]*d*e^(5/2)) - (a*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]
*d*e^(5/2)) - (2*(a + a*Sec[c + d*x]))/(3*d*e*(e*Tan[c + d*x])^(3/2)) - (a*EllipticF[c - Pi/4 + d*x, 2]*Sec[c
+ d*x]*Sqrt[Sin[2*c + 2*d*x]])/(3*d*e^2*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 3882

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[((e*Cot[c
+ d*x])^(m + 1)*(a + b*Csc[c + d*x]))/(d*e*(m + 1)), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)*(
a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -1]

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]

Rubi steps

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

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Mathematica [C]  time = 1.53, size = 200, normalized size = 0.71 \[ -\frac {a \csc (c+d x) \sqrt {e \tan (c+d x)} \left (\sqrt {\sec ^2(c+d x)} \left (2 \cot \left (\frac {1}{2} (c+d x)\right )-3 \sqrt {\sin (2 (c+d x))} \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+2 \cos \left (\frac {3}{2} (c+d x)\right ) \csc \left (\frac {1}{2} (c+d x)\right )+3 \sqrt {\sin (2 (c+d x))} \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )-4 \sqrt [4]{-1} \sqrt {\tan (c+d x)} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right )\right |-1\right )\right )}{6 d e^3 \sqrt {\sec ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a*Csc[c + d*x]*(Sqrt[Sec[c + d*x]^2]*(2*Cot[(c + d*x)/2] + 2*Cos[(3*(c + d*x))/2]*Csc[(c + d*x)/2] - 3*A
rcSin[Cos[c + d*x] - Sin[c + d*x]]*Sqrt[Sin[2*(c + d*x)]] + 3*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c
+ d*x)]]]*Sqrt[Sin[2*(c + d*x)]]) - 4*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]], -1]*Sqrt[
Tan[c + d*x]])*Sqrt[e*Tan[c + d*x]])/(d*e^3*Sqrt[Sec[c + d*x]^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((a+a*sec(d*x+c))/(e*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sec \left (d x + c\right ) + a}{\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.70, size = 648, normalized size = 2.30 \[ \frac {a \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (3 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 ) \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 )}}\, \sin \left (d x +c \right )-3 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 ) \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 )}}\, \sin \left (d x +c \right )+3 \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 ) \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 )}}\, \sin \left (d x +c \right )+3 \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 ) \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 )}}\, \sin \left (d x +c \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 ) \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 )}}\, \sin \left (d x +c \right )+2 \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{6 d \cos \left (d x +c \right )^{3} \left (\frac {e \sin \left (d x +c \right )}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))/(e*tan(d*x+c))^(5/2),x)

[Out]

1/6*a/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(3*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))*((-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)*sin(d*x+c)-3*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))*((-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)*sin(d*x+c)+3*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)+s
in(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+3*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)*sin(d*x+c)-4*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),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)*sin(d*x+c)+2*cos(d*x+c)*2^(1/2))/cos(d*x+c)^3/(e*sin(d*x+c)/cos(d*x+c))^(5/2)/sin(d*x+c)*2^(1/2)

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maxima [A]  time = 0.43, size = 169, normalized size = 0.60 \[ -\frac {a {\left (\frac {6 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{e^{\frac {3}{2}}} + \frac {6 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {e \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {e}}\right )}{e^{\frac {3}{2}}} + \frac {3 \, \sqrt {2} \log \left (e \tan \left (d x + c\right ) + \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{e^{\frac {3}{2}}} - \frac {3 \, \sqrt {2} \log \left (e \tan \left (d x + c\right ) - \sqrt {2} \sqrt {e \tan \left (d x + c\right )} \sqrt {e} + e\right )}{e^{\frac {3}{2}}} + \frac {8}{\left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}}}\right )}}{12 \, d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*a*(6*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e*tan(d*x + c)))/sqrt(e))/e^(3/2) + 6*sqrt(2)*
arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e*tan(d*x + c)))/sqrt(e))/e^(3/2) + 3*sqrt(2)*log(e*tan(d*x + c)
+ sqrt(2)*sqrt(e*tan(d*x + c))*sqrt(e) + e)/e^(3/2) - 3*sqrt(2)*log(e*tan(d*x + c) - sqrt(2)*sqrt(e*tan(d*x +
c))*sqrt(e) + e)/e^(3/2) + 8/(e*tan(d*x + c))^(3/2))/(d*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+\frac {a}{\cos \left (c+d\,x\right )}}{{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \frac {1}{\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {\sec {\left (c + d x \right )}}{\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral((e*tan(c + d*x))**(-5/2), x) + Integral(sec(c + d*x)/(e*tan(c + d*x))**(5/2), x))

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