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

Optimal. Leaf size=249 \[ -\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (a^2-b^2\right )}{d \sqrt{\tan (c+d x)}}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{4 a b}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

[Out]

-(((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d)) + ((a^2 - 2*a*b - b^2)*ArcTan[1 +
Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) + ((a^2 + 2*a*b - b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d
*x]])/(2*Sqrt[2]*d) - ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) -
(2*a^2)/(5*d*Tan[c + d*x]^(5/2)) - (4*a*b)/(3*d*Tan[c + d*x]^(3/2)) + (2*(a^2 - b^2))/(d*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.211728, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3542, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 \left (a^2-b^2\right )}{d \sqrt{\tan (c+d x)}}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{4 a b}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d)) + ((a^2 - 2*a*b - b^2)*ArcTan[1 +
Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) + ((a^2 + 2*a*b - b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d
*x]])/(2*Sqrt[2]*d) - ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) -
(2*a^2)/(5*d*Tan[c + d*x]^(5/2)) - (4*a*b)/(3*d*Tan[c + d*x]^(3/2)) + (2*(a^2 - b^2))/(d*Sqrt[Tan[c + d*x]])

Rule 3542

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

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

Mathematica [C]  time = 0.218838, size = 81, normalized size = 0.33 \[ \frac{-6 \left (a^2-b^2\right ) \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\tan ^2(c+d x)\right )-2 b \left (10 a \tan (c+d x) \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};-\tan ^2(c+d x)\right )+3 b\right )}{15 d \tan ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(a^2 - b^2)*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[c + d*x]^2] - 2*b*(3*b + 10*a*Hypergeometric2F1[-3/4, 1,
1/4, -Tan[c + d*x]^2]*Tan[c + d*x]))/(15*d*Tan[c + d*x]^(5/2))

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Maple [A]  time = 0.014, size = 386, normalized size = 1.6 \begin{align*} -{\frac{2\,{a}^{2}}{5\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}+2\,{\frac{{a}^{2}}{d\sqrt{\tan \left ( dx+c \right ) }}}-2\,{\frac{{b}^{2}}{d\sqrt{\tan \left ( dx+c \right ) }}}-{\frac{4\,ab}{3\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{ab\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{ab\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{ab\sqrt{2}}{2\,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{\sqrt{2}{a}^{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{\sqrt{2}{b}^{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{\sqrt{2}{a}^{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{\sqrt{2}{b}^{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{\sqrt{2}{a}^{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{\sqrt{2}{b}^{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))^2/tan(d*x+c)^(7/2),x)

[Out]

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

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Maxima [A]  time = 2.05114, size = 278, normalized size = 1.12 \begin{align*} \frac{30 \, \sqrt{2}{\left (a^{2} - 2 \, a b - b^{2}\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^{2} - 2 \, a b - b^{2}\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^{2} + 2 \, a b - b^{2}\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^{2} + 2 \, a b - b^{2}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \frac{8 \,{\left (10 \, a b \tan \left (d x + c\right ) - 15 \,{\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2}\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))^2/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/60*(30*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 30*sqrt(2)*(a^2 -
2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 15*sqrt(2)*(a^2 + 2*a*b - b^2)*log(sqrt(2
)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 15*sqrt(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + ta
n(d*x + c) + 1) - 8*(10*a*b*tan(d*x + c) - 15*(a^2 - b^2)*tan(d*x + c)^2 + 3*a^2)/tan(d*x + c)^(5/2))/d

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Fricas [B]  time = 6.45345, size = 11985, normalized size = 48.13 \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))^2/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((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*
b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2
+ 38*a^4*b^4 - 12*a^2*b^6 + b^8))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4)*sqrt((a^8 - 12*
a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4)*arctan(-((a^16 - 20*a^12*b^4 - 64*a^10*b^6 - 90*a^8*b^8 - 64*a^6
*b^10 - 20*a^4*b^12 + b^16)*d^4*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b
^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) - sqrt(2)*(2*a*b*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 +
b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + (a^6 + a^4*b^2 - a^2*b^4 - b^6)*d^5*s
qrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b
^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a
^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(((a^12 - 10*a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 + b
^12)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + sqrt(2)*((a^10 - 13*a^8*b^2
+ 50*a^6*b^4 - 50*a^4*b^6 + 13*a^2*b^8 - b^10)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*c
os(d*x + c) + 2*(a^13*b - 10*a^11*b^3 + 15*a^9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d*x
+ c))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a
^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x
+ c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^10*
b^6 + 134*a^8*b^8 + 72*a^6*b^10 - 4*a^4*b^12 - 8*a^2*b^14 + b^16)*sin(d*x + c))/cos(d*x + c))*((a^8 + 4*a^6*b^
2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4) + sqrt(2)*(2*(a^9*b - 4*a^7*b^3 - 10*a^5*b^5 - 4*a^3*b^7 + a*b^9)*
d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6
+ b^8)/d^4) + (a^14 - 3*a^12*b^2 - 15*a^10*b^4 - 11*a^8*b^6 + 11*a^6*b^8 + 15*a^4*b^10 + 3*a^2*b^12 - b^14)*d
^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6
+ b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 +
38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^
8)/d^4)^(3/4))/(a^24 - 4*a^22*b^2 - 30*a^20*b^4 + 12*a^18*b^6 + 367*a^16*b^8 + 1016*a^14*b^10 + 1372*a^12*b^12
+ 1016*a^10*b^14 + 367*a^8*b^16 + 12*a^6*b^18 - 30*a^4*b^20 - 4*a^2*b^22 + b^24)) + 60*sqrt(2)*(d^5*cos(d*x +
c)^4 - 2*d^5*cos(d*x + c)^2 + d^5)*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^
2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8
))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^
6 + b^8)/d^4)*arctan(((a^16 - 20*a^12*b^4 - 64*a^10*b^6 - 90*a^8*b^8 - 64*a^6*b^10 - 20*a^4*b^12 + b^16)*d^4*s
qrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^
8)/d^4) + sqrt(2)*(2*a*b*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2
+ 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + (a^6 + a^4*b^2 - a^2*b^4 - b^6)*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^
4 - 12*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((
a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(
((a^12 - 10*a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 + b^12)*d^2*sqrt((a^8 + 4*a^6*b^2 +
6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) - sqrt(2)*((a^10 - 13*a^8*b^2 + 50*a^6*b^4 - 50*a^4*b^6 + 13*a^
2*b^8 - b^10)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + 2*(a^13*b - 10*a^11
*b^3 + 15*a^9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d*x + c))*sqrt((a^8 + 4*a^6*b^2 + 6*
a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(
a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b
^4 + 4*a^2*b^6 + b^8)/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^10*b^6 + 134*a^8*b^8 + 72*a^6*b^10 -
4*a^4*b^12 - 8*a^2*b^14 + b^16)*sin(d*x + c))/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/
d^4)^(3/4) - sqrt(2)*(2*(a^9*b - 4*a^7*b^3 - 10*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4
*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + (a^14 - 3*a^12*b^2
- 15*a^10*b^4 - 11*a^8*b^6 + 11*a^6*b^8 + 15*a^4*b^10 + 3*a^2*b^12 - b^14)*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^
4*b^4 - 12*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sq
rt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*s
qrt(sin(d*x + c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4))/(a^24 - 4*a^22*b^2
- 30*a^20*b^4 + 12*a^18*b^6 + 367*a^16*b^8 + 1016*a^14*b^10 + 1372*a^12*b^12 + 1016*a^10*b^14 + 367*a^8*b^16
+ 12*a^6*b^18 - 30*a^4*b^20 - 4*a^2*b^22 + b^24)) + 15*sqrt(2)*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8
)*d*cos(d*x + c)^4 - 2*(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d*cos(d*x + c)^2 + (a^8 + 4*a^6*b^2 + 6
*a^4*b^4 + 4*a^2*b^6 + b^8)*d + 4*((a^3*b - a*b^3)*d^3*cos(d*x + c)^4 - 2*(a^3*b - a*b^3)*d^3*cos(d*x + c)^2 +
(a^3*b - a*b^3)*d^3)*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4
*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8
- 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(1/4)*lo
g(((a^12 - 10*a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 + b^12)*d^2*sqrt((a^8 + 4*a^6*b^2
+ 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + sqrt(2)*((a^10 - 13*a^8*b^2 + 50*a^6*b^4 - 50*a^4*b^6 + 13*
a^2*b^8 - b^10)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + 2*(a^13*b - 10*a^
11*b^3 + 15*a^9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d*x + c))*sqrt((a^8 + 4*a^6*b^2 +
6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))
/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4
*b^4 + 4*a^2*b^6 + b^8)/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^10*b^6 + 134*a^8*b^8 + 72*a^6*b^10
- 4*a^4*b^12 - 8*a^2*b^14 + b^16)*sin(d*x + c))/cos(d*x + c)) - 15*sqrt(2)*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*
a^2*b^6 + b^8)*d*cos(d*x + c)^4 - 2*(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d*cos(d*x + c)^2 + (a^8 +
4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d + 4*((a^3*b - a*b^3)*d^3*cos(d*x + c)^4 - 2*(a^3*b - a*b^3)*d^3*cos
(d*x + c)^2 + (a^3*b - a*b^3)*d^3)*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^
6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^
8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/
d^4)^(1/4)*log(((a^12 - 10*a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 + b^12)*d^2*sqrt((a^8
+ 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) - sqrt(2)*((a^10 - 13*a^8*b^2 + 50*a^6*b^4 - 50*
a^4*b^6 + 13*a^2*b^8 - b^10)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + 2*(a
^13*b - 10*a^11*b^3 + 15*a^9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d*x + c))*sqrt((a^8 +
4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6
+ b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^8 + 4*a^
6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^10*b^6 + 134*a^8*b^8
+ 72*a^6*b^10 - 4*a^4*b^12 - 8*a^2*b^14 + b^16)*sin(d*x + c))/cos(d*x + c)) - 8*(10*(a^9*b + 4*a^7*b^3 + 6*a^5
*b^5 + 4*a^3*b^7 + a*b^9)*cos(d*x + c)^4 - 10*(a^9*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*cos(d*x + c)
^2 - 3*((6*a^10 + 19*a^8*b^2 + 16*a^6*b^4 - 6*a^4*b^6 - 14*a^2*b^8 - 5*b^10)*cos(d*x + c)^3 - 5*(a^10 + 3*a^8*
b^2 + 2*a^6*b^4 - 2*a^4*b^6 - 3*a^2*b^8 - b^10)*cos(d*x + c))*sin(d*x + c))*sqrt(sin(d*x + c)/cos(d*x + c)))/(
(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d*cos(d*x + c)^4 - 2*(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6
+ b^8)*d*cos(d*x + c)^2 + (a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d)

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

[Out]

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

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Giac [A]  time = 1.5266, size = 339, normalized size = 1.36 \begin{align*} \frac{{\left (\sqrt{2} a^{2} - 2 \, \sqrt{2} a b - \sqrt{2} b^{2}\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^{2} - 2 \, \sqrt{2} a b - \sqrt{2} b^{2}\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^{2} + 2 \, \sqrt{2} a b - \sqrt{2} b^{2}\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^{2} + 2 \, \sqrt{2} a b - \sqrt{2} b^{2}\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^{2} \tan \left (d x + c\right )^{2} - 15 \, b^{2} \tan \left (d x + c\right )^{2} - 10 \, a b \tan \left (d x + c\right ) - 3 \, a^{2}\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))^2/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

1/2*(sqrt(2)*a^2 - 2*sqrt(2)*a*b - sqrt(2)*b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c))))/d + 1/2*(
sqrt(2)*a^2 - 2*sqrt(2)*a*b - sqrt(2)*b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))))/d - 1/4*(sqrt
(2)*a^2 + 2*sqrt(2)*a*b - sqrt(2)*b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/d + 1/4*(sqrt(2)*a^2
+ 2*sqrt(2)*a*b - sqrt(2)*b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/d + 2/15*(15*a^2*tan(d*x +
c)^2 - 15*b^2*tan(d*x + c)^2 - 10*a*b*tan(d*x + c) - 3*a^2)/(d*tan(d*x + c)^(5/2))