3.416 \(\int \frac{\tan ^{\frac{5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=156 \[ \frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}-\frac{B \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{B \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d} \]

[Out]

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

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Rubi [A]  time = 0.10962, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {21, 3473, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{B \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}-\frac{B \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{B \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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{\tan ^{\frac{5}{2}}(c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx &=B \int \tan ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{3 d}-B \int \sqrt{\tan (c+d x)} \, dx\\ &=\frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{B \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{(2 B) \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{B \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}-\frac{B \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{B \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}-\frac{B \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}-\frac{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{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{B \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}+\frac{B \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}+\frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{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{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{B \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{B \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{B \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}+\frac{B \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}+\frac{2 B \tan ^{\frac{3}{2}}(c+d x)}{3 d}\\ \end{align*}

Mathematica [C]  time = 0.0466169, size = 38, normalized size = 0.24 \[ -\frac{2 B \tan ^{\frac{3}{2}}(c+d x) \left (\text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},-\tan ^2(c+d x)\right )-1\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*B*(-1 + Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2])*Tan[c + d*x]^(3/2))/(3*d)

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Maple [A]  time = 0.023, size = 118, normalized size = 0.8 \begin{align*}{\frac{2\,B}{3\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*B*tan(d*x+c)^(3/2)/d-1/2*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/2*B*arctan(-1+2^(1/2)*tan(d*x+c)
^(1/2))/d*2^(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)))

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Maxima [A]  time = 1.54973, size = 166, normalized size = 1.06 \begin{align*} \frac{8 \, B \tan \left (d x + c\right )^{\frac{3}{2}} - 3 \,{\left (2 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) - \sqrt{2} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt{2} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} B}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.54263, size = 1466, normalized size = 9.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*sqrt(2)*d*(B^4/d^4)^(1/4)*arctan(-(sqrt(2)*B^3*d*(B^4/d^4)^(1/4)*sqrt(sin(d*x + c)/cos(d*x + c)) + B^
4 - sqrt(2)*d*(B^4/d^4)^(1/4)*sqrt((sqrt(2)*B^3*d^3*(B^4/d^4)^(3/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x +
c) + B^4*d^2*sqrt(B^4/d^4)*cos(d*x + c) + B^6*sin(d*x + c))/cos(d*x + c)))/B^4)*cos(d*x + c) + 12*sqrt(2)*d*(B
^4/d^4)^(1/4)*arctan(-(sqrt(2)*B^3*d*(B^4/d^4)^(1/4)*sqrt(sin(d*x + c)/cos(d*x + c)) - B^4 - sqrt(2)*d*(B^4/d^
4)^(1/4)*sqrt(-(sqrt(2)*B^3*d^3*(B^4/d^4)^(3/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c) - B^4*d^2*sqrt(B^
4/d^4)*cos(d*x + c) - B^6*sin(d*x + c))/cos(d*x + c)))/B^4)*cos(d*x + c) + 3*sqrt(2)*d*(B^4/d^4)^(1/4)*cos(d*x
+ c)*log((sqrt(2)*B^3*d^3*(B^4/d^4)^(3/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c) + B^4*d^2*sqrt(B^4/d^4
)*cos(d*x + c) + B^6*sin(d*x + c))/cos(d*x + c)) - 3*sqrt(2)*d*(B^4/d^4)^(1/4)*cos(d*x + c)*log(-(sqrt(2)*B^3*
d^3*(B^4/d^4)^(3/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c) - B^4*d^2*sqrt(B^4/d^4)*cos(d*x + c) - B^6*si
n(d*x + c))/cos(d*x + c)) + 8*B*sqrt(sin(d*x + c)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out