3.6.60 \(\int \frac {a+b \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [560]

Optimal. Leaf size=184 \[ \frac {(a+b) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b}{d \sqrt {\tan (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3610, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {(a+b) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {(a-b) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a-b) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b}{d \sqrt {\tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a + b)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]
)/(Sqrt[2]*d) + ((a - b)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - ((a - b)*Log[1 +
Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (2*a)/(3*d*Tan[c + d*x]^(3/2)) - (2*b)/(d*Sqrt[Tan
[c + d*x]])

Rule 210

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 631

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 642

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 1176

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 1179

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 1182

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

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rubi steps

\begin {align*} \int \frac {a+b \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx &=-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\int \frac {b-a \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b}{d \sqrt {\tan (c+d x)}}+\int \frac {-a-b \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b}{d \sqrt {\tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {-a-b x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b}{d \sqrt {\tan (c+d x)}}-\frac {(a-b) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}-\frac {(a+b) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b}{d \sqrt {\tan (c+d x)}}+\frac {(a-b) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {(a-b) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {(a+b) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}-\frac {(a+b) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}\\ &=\frac {(a-b) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b}{d \sqrt {\tan (c+d x)}}-\frac {(a+b) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}\\ &=\frac {(a+b) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b}{d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.11, size = 72, normalized size = 0.39 \begin {gather*} \frac {-\left ((a+i b) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-i \tan (c+d x)\right )\right )-(a-i b) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};i \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + I*b)*Hypergeometric2F1[-3/2, 1, -1/2, (-I)*Tan[c + d*x]]) - (a - I*b)*Hypergeometric2F1[-3/2, 1, -1/2,
I*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2))

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Maple [A]
time = 0.05, size = 200, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 b}{\sqrt {\tan \left (d x +c \right )}}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(200\)
default	\(\frac {-\frac {2 a}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 b}{\sqrt {\tan \left (d x +c \right )}}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(200\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))/tan(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/3*a/tan(d*x+c)^(3/2)-2*b/tan(d*x+c)^(1/2)-1/4*a*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1
-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/
2)))-1/4*b*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arct
an(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))

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Maxima [A]
time = 0.49, size = 146, normalized size = 0.79 \begin {gather*} -\frac {6 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left (a - b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left (a - b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, {\left (3 \, b \tan \left (d x + c\right ) + a\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2861 vs. \(2 (150) = 300\).
time = 1.22, size = 2861, normalized size = 15.55 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(12*sqrt(2)*(d^5*cos(d*x + c)^2 - d^5)*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b
^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4)*arcta
n(-((a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^4*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4
) - sqrt(2)*(a*d^7*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) - (a^2*b + b^3)*d^5*sqr
t((a^4 - 2*a^2*b^2 + b^4)/d^4))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^
4 - 2*a^2*b^2 + b^4))*sqrt(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c)
+ sqrt(2)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - (a^7 - a^5*b^2 - a^
3*b^4 + a*b^6)*d*cos(d*x + c))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4
- 2*a^2*b^2 + b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) + (a^8 - 2*a^4*b^4 +
b^8)*sin(d*x + c))/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4) - sqrt(2)*((a^5 - a*b^4)*d^7*sqrt((a^4 +
2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) - (a^6*b + a^4*b^3 - a^2*b^5 - b^7)*d^5*sqrt((a^4 - 2*
a^2*b^2 + b^4)/d^4))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b
^2 + b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4))/(a^12 + 2*a^10*b^2 - a^8*b^4 -
4*a^6*b^6 - a^4*b^8 + 2*a^2*b^10 + b^12)) + 12*sqrt(2)*(d^5*cos(d*x + c)^2 - d^5)*sqrt((2*a*b*d^2*sqrt((a^4 +
2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4)*s
qrt((a^4 - 2*a^2*b^2 + b^4)/d^4)*arctan(((a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^4*sqrt((a^4 + 2*a^2*b^2 + b^4)/
d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) + sqrt(2)*(a*d^7*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^
2 + b^4)/d^4) - (a^2*b + b^3)*d^5*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b
^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*sqrt(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4
+ 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - sqrt(2)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^
4)*cos(d*x + c) - (a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*cos(d*x + c))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^
4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 +
b^4)/d^4)^(1/4) + (a^8 - 2*a^4*b^4 + b^8)*sin(d*x + c))/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4) + sq
rt(2)*((a^5 - a*b^4)*d^7*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) - (a^6*b + a^4*b^
3 - a^2*b^5 - b^7)*d^5*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) +
a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^
(3/4))/(a^12 + 2*a^10*b^2 - a^8*b^4 - 4*a^6*b^6 - a^4*b^8 + 2*a^2*b^10 + b^12)) - 3*sqrt(2)*((a^4 + 2*a^2*b^2
+ b^4)*d*cos(d*x + c)^2 - (a^4 + 2*a^2*b^2 + b^4)*d - 2*(a*b*d^3*cos(d*x + c)^2 - a*b*d^3)*sqrt((a^4 + 2*a^2*b
^2 + b^4)/d^4))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 +
b^4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4)*log(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4
)/d^4)*cos(d*x + c) + sqrt(2)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) -
(a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*cos(d*x + c))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*
a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) +
(a^8 - 2*a^4*b^4 + b^8)*sin(d*x + c))/cos(d*x + c)) + 3*sqrt(2)*((a^4 + 2*a^2*b^2 + b^4)*d*cos(d*x + c)^2 - (a
^4 + 2*a^2*b^2 + b^4)*d - 2*(a*b*d^3*cos(d*x + c)^2 - a*b*d^3)*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4))*sqrt((2*a*b*
d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*((a^4 + 2*a^2*b^2 + b^
4)/d^4)^(1/4)*log(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - sqrt(2
)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - (a^7 - a^5*b^2 - a^3*b^4 + a
*b^6)*d*cos(d*x + c))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*
b^2 + b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) + (a^8 - 2*a^4*b^4 + b^8)*sin(
d*x + c))/cos(d*x + c)) - 8*((a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c)^2 + 3*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x +
c)*sin(d*x + c))*sqrt(sin(d*x + c)/cos(d*x + c)))/((a^4 + 2*a^2*b^2 + b^4)*d*cos(d*x + c)^2 - (a^4 + 2*a^2*b^2
+ b^4)*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \tan {\left (c + d x \right )}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*tan(c + d*x))/tan(c + d*x)**(5/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [B]
time = 5.13, size = 114, normalized size = 0.62 \begin {gather*} \frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}-\frac {2\,b}{d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}-\frac {2\,a}{3\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-1)^(1/4)*a*atan((-1)^(1/4)*tan(c + d*x)^(1/2))*1i)/d - (2*b)/(d*tan(c + d*x)^(1/2)) - (2*a)/(3*d*tan(c + d*
x)^(3/2)) + ((-1)^(1/4)*a*atanh((-1)^(1/4)*tan(c + d*x)^(1/2))*1i)/d - ((-1)^(1/4)*b*atan((-1)^(1/4)*tan(c + d
*x)^(1/2)))/d + ((-1)^(1/4)*b*atanh((-1)^(1/4)*tan(c + d*x)^(1/2)))/d

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