3.561 \(\int \frac{a+b \tan (c+d x)}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=202 \[ -\frac{(a-b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{(a-b) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{(a+b) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a+b) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\tan (c+d x)}}-\frac{2 b}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

[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)/(5*d*Tan[c + d*x]^(5/2)) - (2*b)/(3*d*Tan
[c + d*x]^(3/2)) + (2*a)/(d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.152606, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{(a-b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{(a-b) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{(a+b) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a+b) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\tan (c+d x)}}-\frac{2 b}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])/Tan[c + d*x]^(7/2),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)/(5*d*Tan[c + d*x]^(5/2)) - (2*b)/(3*d*Tan
[c + d*x]^(3/2)) + (2*a)/(d*Sqrt[Tan[c + d*x]])

Rule 3529

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 3534

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]

Rule 1168

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{a+b \tan (c+d x)}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 a}{5 d \tan ^{\frac{5}{2}}(c+d x)}+\int \frac{b-a \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 b}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\int \frac{-a-b \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 b}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\tan (c+d x)}}+\int \frac{-b+a \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 b}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\tan (c+d x)}}+\frac{2 \operatorname{Subst}\left (\int \frac{-b+a x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 a}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 b}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\tan (c+d x)}}+\frac{(a-b) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}-\frac{(a+b) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 a}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 b}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\tan (c+d x)}}+\frac{(a-b) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}+\frac{(a-b) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}+\frac{(a+b) \operatorname{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) \operatorname{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) \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}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 b}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\tan (c+d x)}}+\frac{(a-b) \operatorname{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) \operatorname{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}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 b}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.101, size = 72, normalized size = 0.36 \[ \frac{-(a+i b) \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};-i \tan (c+d x)\right )-(a-i b) \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};i \tan (c+d x)\right )}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])/Tan[c + d*x]^(7/2),x]

[Out]

(-((a + I*b)*Hypergeometric2F1[-5/2, 1, -3/2, (-I)*Tan[c + d*x]]) - (a - I*b)*Hypergeometric2F1[-5/2, 1, -3/2,
I*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(5/2))

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Maple [A]  time = 0.012, size = 248, normalized size = 1.2 \begin{align*} -{\frac{2\,a}{5\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}-{\frac{2\,b}{3\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{a}{d\sqrt{\tan \left ( dx+c \right ) }}}-{\frac{b\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{b\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{b\sqrt{2}}{4\,d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{a\sqrt{2}}{4\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{a\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{a\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))/tan(d*x+c)^(7/2),x)

[Out]

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

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Maxima [A]  time = 1.81155, size = 215, normalized size = 1.06 \begin{align*} \frac{30 \, \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 ) + 30 \, \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 ) - 15 \, \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 ) + 15 \, \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 (15 \, a \tan \left (d x + c\right )^{2} - 5 \, b \tan \left (d x + c\right ) - 3 \, a\right )}}{\tan \left (d x + c\right )^{\frac{5}{2}}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/60*(30*sqrt(2)*(a - b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 30*sqrt(2)*(a - b)*arctan(-1/2
*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 15*sqrt(2)*(a + b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c)
+ 1) + 15*sqrt(2)*(a + b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 8*(15*a*tan(d*x + c)^2 - 5*b*t
an(d*x + c) - 3*a)/tan(d*x + c)^(5/2))/d

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Fricas [B]  time = 2.78161, size = 6718, normalized size = 33.26 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/60*(60*sqrt(2)*(d^5*cos(d*x + c)^4 - 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)*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)*(b*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^3 + a*b^2)*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^5 - 2*a^3*b^2 + a*b^4)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c
) + (a^6*b - a^4*b^3 - a^2*b^5 + b^7)*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^4*b
- b^5)*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^7 + a^5*b^2 - a^3*b^4 - a
*b^6)*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)) + 60*sqrt(2)*(d^5*cos(d*x + c)^4 - 2*d^5*c
os(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)*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)*(b*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^3 + a*b^2)*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^5 - 2*a
^3*b^2 + a*b^4)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + (a^6*b - a^4*b^3 - a^2*b^5 + b^7)*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))/co
s(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4) - sqrt(2)*((a^4*b - b^5)*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^7 + a^5*b^2 - a^3*b^4 - a*b^6)*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)) - 15*sqrt(2)*((a^4 + 2*a^2*b^2 + b^4)*d*cos(d*x + c)^4 - 2*(a^4 + 2*a^2*b^2 + b^4)*d*co
s(d*x + c)^2 + (a^4 + 2*a^2*b^2 + b^4)*d + 2*(a*b*d^3*cos(d*x + c)^4 - 2*a*b*d^3*cos(d*x + c)^2 + a*b*d^3)*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))*((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^5 - 2*a^3*b^2 + a*b^4)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4
)*cos(d*x + c) + (a^6*b - a^4*b^3 - a^2*b^5 + b^7)*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)) + 15*sqrt(2)*((a^4 + 2*a^2*b^2 + b^4)*d*
cos(d*x + c)^4 - 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)^4 - 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^5 - 2*a^3*b^
2 + a*b^4)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + (a^6*b - a^4*b^3 - a^2*b^5 + b^7)*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*(5*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^4 - 5*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^2 - 3*(6*(a^
5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c)^3 - 5*(a^5 + 2*a^3*b^2 + a*b^4)*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)^4 - 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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \tan{\left (c + d x \right )}}{\tan ^{\frac{7}{2}}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

Integral((a + b*tan(c + d*x))/tan(c + d*x)**(7/2), x)

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Giac [A]  time = 1.26983, size = 252, normalized size = 1.25 \begin{align*} \frac{{\left (\sqrt{2} a - \sqrt{2} b\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} + \frac{{\left (\sqrt{2} a - \sqrt{2} b\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} - \frac{{\left (\sqrt{2} a + \sqrt{2} b\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} + \frac{{\left (\sqrt{2} a + \sqrt{2} b\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} + \frac{2 \,{\left (15 \, a \tan \left (d x + c\right )^{2} - 5 \, b \tan \left (d x + c\right ) - 3 \, a\right )}}{15 \, d \tan \left (d x + c\right )^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

1/2*(sqrt(2)*a - sqrt(2)*b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c))))/d + 1/2*(sqrt(2)*a - sqrt(2)*
b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))))/d - 1/4*(sqrt(2)*a + sqrt(2)*b)*log(sqrt(2)*sqrt(tan(
d*x + c)) + tan(d*x + c) + 1)/d + 1/4*(sqrt(2)*a + sqrt(2)*b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) +
1)/d + 2/15*(15*a*tan(d*x + c)^2 - 5*b*tan(d*x + c) - 3*a)/(d*tan(d*x + c)^(5/2))