3.16 \(\int \frac{1}{(b \tan (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=234 \[ -\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}\right )}{\sqrt{2} b^{7/2} d}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}+1\right )}{\sqrt{2} b^{7/2} d}+\frac{2}{b^3 d \sqrt{b \tan (c+d x)}}+\frac{\log \left (\sqrt{b} \tan (c+d x)-\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} b^{7/2} d}-\frac{\log \left (\sqrt{b} \tan (c+d x)+\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} b^{7/2} d}-\frac{2}{5 b d (b \tan (c+d x))^{5/2}} \]

[Out]

-(ArcTan[1 - (Sqrt[2]*Sqrt[b*Tan[c + d*x]])/Sqrt[b]]/(Sqrt[2]*b^(7/2)*d)) + ArcTan[1 + (Sqrt[2]*Sqrt[b*Tan[c +
d*x]])/Sqrt[b]]/(Sqrt[2]*b^(7/2)*d) + Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] - Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*S
qrt[2]*b^(7/2)*d) - Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] + Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*Sqrt[2]*b^(7/2)*d) -
2/(5*b*d*(b*Tan[c + d*x])^(5/2)) + 2/(b^3*d*Sqrt[b*Tan[c + d*x]])

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Rubi [A]  time = 0.180953, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {3474, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}\right )}{\sqrt{2} b^{7/2} d}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b \tan (c+d x)}}{\sqrt{b}}+1\right )}{\sqrt{2} b^{7/2} d}+\frac{2}{b^3 d \sqrt{b \tan (c+d x)}}+\frac{\log \left (\sqrt{b} \tan (c+d x)-\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} b^{7/2} d}-\frac{\log \left (\sqrt{b} \tan (c+d x)+\sqrt{2} \sqrt{b \tan (c+d x)}+\sqrt{b}\right )}{2 \sqrt{2} b^{7/2} d}-\frac{2}{5 b d (b \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x])^(-7/2),x]

[Out]

-(ArcTan[1 - (Sqrt[2]*Sqrt[b*Tan[c + d*x]])/Sqrt[b]]/(Sqrt[2]*b^(7/2)*d)) + ArcTan[1 + (Sqrt[2]*Sqrt[b*Tan[c +
d*x]])/Sqrt[b]]/(Sqrt[2]*b^(7/2)*d) + Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] - Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*S
qrt[2]*b^(7/2)*d) - Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] + Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*Sqrt[2]*b^(7/2)*d) -
2/(5*b*d*(b*Tan[c + d*x])^(5/2)) + 2/(b^3*d*Sqrt[b*Tan[c + d*x]])

Rule 3474

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]

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 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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

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

Mathematica [C]  time = 0.102988, size = 40, normalized size = 0.17 \[ -\frac{2 \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\tan ^2(c+d x)\right )}{5 b d (b \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x])^(-7/2),x]

[Out]

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

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Maple [A]  time = 0.019, size = 202, normalized size = 0.9 \begin{align*}{\frac{\sqrt{2}}{4\,d{b}^{3}}\ln \left ({ \left ( b\tan \left ( dx+c \right ) -\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( dx+c \right ) }\sqrt{2}+\sqrt{{b}^{2}} \right ) \left ( b\tan \left ( dx+c \right ) +\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( dx+c \right ) }\sqrt{2}+\sqrt{{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{b}^{2}}}}}+{\frac{\sqrt{2}}{2\,d{b}^{3}}\arctan \left ({\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{b}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{b}^{2}}}}}-{\frac{\sqrt{2}}{2\,d{b}^{3}}\arctan \left ( -{\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{b}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{b}^{2}}}}}-{\frac{2}{5\,bd} \left ( b\tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}+2\,{\frac{1}{d{b}^{3}\sqrt{b\tan \left ( dx+c \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(d*x+c))^(7/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.77854, size = 1985, normalized size = 8.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/20*(8*(6*cos(d*x + c)^3 - 5*cos(d*x + c))*sqrt(b*sin(d*x + c)/cos(d*x + c))*sin(d*x + c) + 20*(sqrt(2)*b^4*
d*cos(d*x + c)^4 - 2*sqrt(2)*b^4*d*cos(d*x + c)^2 + sqrt(2)*b^4*d)*(1/(b^14*d^4))^(1/4)*arctan(-sqrt(2)*b^3*d*
sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(b^14*d^4))^(1/4) + sqrt(2)*b^3*d*sqrt((sqrt(2)*b^11*d^3*sqrt(b*sin(d*x +
c)/cos(d*x + c))*(1/(b^14*d^4))^(3/4)*cos(d*x + c) + b^8*d^2*sqrt(1/(b^14*d^4))*cos(d*x + c) + b*sin(d*x + c)
)/cos(d*x + c))*(1/(b^14*d^4))^(1/4) - 1) + 20*(sqrt(2)*b^4*d*cos(d*x + c)^4 - 2*sqrt(2)*b^4*d*cos(d*x + c)^2
+ sqrt(2)*b^4*d)*(1/(b^14*d^4))^(1/4)*arctan(-sqrt(2)*b^3*d*sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(b^14*d^4))^(
1/4) + sqrt(2)*b^3*d*sqrt(-(sqrt(2)*b^11*d^3*sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(b^14*d^4))^(3/4)*cos(d*x +
c) - b^8*d^2*sqrt(1/(b^14*d^4))*cos(d*x + c) - b*sin(d*x + c))/cos(d*x + c))*(1/(b^14*d^4))^(1/4) + 1) + 5*(sq
rt(2)*b^4*d*cos(d*x + c)^4 - 2*sqrt(2)*b^4*d*cos(d*x + c)^2 + sqrt(2)*b^4*d)*(1/(b^14*d^4))^(1/4)*log((sqrt(2)
*b^11*d^3*sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(b^14*d^4))^(3/4)*cos(d*x + c) + b^8*d^2*sqrt(1/(b^14*d^4))*cos
(d*x + c) + b*sin(d*x + c))/cos(d*x + c)) - 5*(sqrt(2)*b^4*d*cos(d*x + c)^4 - 2*sqrt(2)*b^4*d*cos(d*x + c)^2 +
sqrt(2)*b^4*d)*(1/(b^14*d^4))^(1/4)*log(-(sqrt(2)*b^11*d^3*sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(b^14*d^4))^(
3/4)*cos(d*x + c) - b^8*d^2*sqrt(1/(b^14*d^4))*cos(d*x + c) - b*sin(d*x + c))/cos(d*x + c)))/(b^4*d*cos(d*x +
c)^4 - 2*b^4*d*cos(d*x + c)^2 + b^4*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tan{\left (c + d x \right )}\right )^{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c))**(7/2),x)

[Out]

Integral((b*tan(c + d*x))**(-7/2), x)

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Giac [A]  time = 1.50618, size = 313, normalized size = 1.34 \begin{align*} \frac{1}{20} \, b{\left (\frac{10 \, \sqrt{2}{\left | b \right |}^{\frac{3}{2}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | b \right |}} + 2 \, \sqrt{b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt{{\left | b \right |}}}\right )}{b^{6} d} + \frac{10 \, \sqrt{2}{\left | b \right |}^{\frac{3}{2}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | b \right |}} - 2 \, \sqrt{b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt{{\left | b \right |}}}\right )}{b^{6} d} - \frac{5 \, \sqrt{2}{\left | b \right |}^{\frac{3}{2}} \log \left (b \tan \left (d x + c\right ) + \sqrt{2} \sqrt{b \tan \left (d x + c\right )} \sqrt{{\left | b \right |}} +{\left | b \right |}\right )}{b^{6} d} + \frac{5 \, \sqrt{2}{\left | b \right |}^{\frac{3}{2}} \log \left (b \tan \left (d x + c\right ) - \sqrt{2} \sqrt{b \tan \left (d x + c\right )} \sqrt{{\left | b \right |}} +{\left | b \right |}\right )}{b^{6} d} + \frac{8 \,{\left (5 \, b^{2} \tan \left (d x + c\right )^{2} - b^{2}\right )}}{\sqrt{b \tan \left (d x + c\right )} b^{6} d \tan \left (d x + c\right )^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

1/20*b*(10*sqrt(2)*abs(b)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) + 2*sqrt(b*tan(d*x + c)))/sqrt(abs(b)
))/(b^6*d) + 10*sqrt(2)*abs(b)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sqrt(b*tan(d*x + c)))/sqrt(
abs(b)))/(b^6*d) - 5*sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs
(b))/(b^6*d) + 5*sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b))
/(b^6*d) + 8*(5*b^2*tan(d*x + c)^2 - b^2)/(sqrt(b*tan(d*x + c))*b^6*d*tan(d*x + c)^2))