3.13 \(\int (b \tan ^4(e+f x))^{5/2} \, dx\)

Optimal. Leaf size=182 \[ \frac{b^2 \tan ^7(e+f x) \sqrt{b \tan ^4(e+f x)}}{9 f}-\frac{b^2 \tan ^5(e+f x) \sqrt{b \tan ^4(e+f x)}}{7 f}+\frac{b^2 \tan ^3(e+f x) \sqrt{b \tan ^4(e+f x)}}{5 f}-\frac{b^2 \tan (e+f x) \sqrt{b \tan ^4(e+f x)}}{3 f}-b^2 x \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}+\frac{b^2 \cot (e+f x) \sqrt{b \tan ^4(e+f x)}}{f} \]

[Out]

(b^2*Cot[e + f*x]*Sqrt[b*Tan[e + f*x]^4])/f - b^2*x*Cot[e + f*x]^2*Sqrt[b*Tan[e + f*x]^4] - (b^2*Tan[e + f*x]*
Sqrt[b*Tan[e + f*x]^4])/(3*f) + (b^2*Tan[e + f*x]^3*Sqrt[b*Tan[e + f*x]^4])/(5*f) - (b^2*Tan[e + f*x]^5*Sqrt[b
*Tan[e + f*x]^4])/(7*f) + (b^2*Tan[e + f*x]^7*Sqrt[b*Tan[e + f*x]^4])/(9*f)

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Rubi [A]  time = 0.0635726, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 8} \[ \frac{b^2 \tan ^7(e+f x) \sqrt{b \tan ^4(e+f x)}}{9 f}-\frac{b^2 \tan ^5(e+f x) \sqrt{b \tan ^4(e+f x)}}{7 f}+\frac{b^2 \tan ^3(e+f x) \sqrt{b \tan ^4(e+f x)}}{5 f}-\frac{b^2 \tan (e+f x) \sqrt{b \tan ^4(e+f x)}}{3 f}-b^2 x \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}+\frac{b^2 \cot (e+f x) \sqrt{b \tan ^4(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^4)^(5/2),x]

[Out]

(b^2*Cot[e + f*x]*Sqrt[b*Tan[e + f*x]^4])/f - b^2*x*Cot[e + f*x]^2*Sqrt[b*Tan[e + f*x]^4] - (b^2*Tan[e + f*x]*
Sqrt[b*Tan[e + f*x]^4])/(3*f) + (b^2*Tan[e + f*x]^3*Sqrt[b*Tan[e + f*x]^4])/(5*f) - (b^2*Tan[e + f*x]^5*Sqrt[b
*Tan[e + f*x]^4])/(7*f) + (b^2*Tan[e + f*x]^7*Sqrt[b*Tan[e + f*x]^4])/(9*f)

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx &=\left (b^2 \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}\right ) \int \tan ^{10}(e+f x) \, dx\\ &=\frac{b^2 \tan ^7(e+f x) \sqrt{b \tan ^4(e+f x)}}{9 f}-\left (b^2 \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}\right ) \int \tan ^8(e+f x) \, dx\\ &=-\frac{b^2 \tan ^5(e+f x) \sqrt{b \tan ^4(e+f x)}}{7 f}+\frac{b^2 \tan ^7(e+f x) \sqrt{b \tan ^4(e+f x)}}{9 f}+\left (b^2 \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}\right ) \int \tan ^6(e+f x) \, dx\\ &=\frac{b^2 \tan ^3(e+f x) \sqrt{b \tan ^4(e+f x)}}{5 f}-\frac{b^2 \tan ^5(e+f x) \sqrt{b \tan ^4(e+f x)}}{7 f}+\frac{b^2 \tan ^7(e+f x) \sqrt{b \tan ^4(e+f x)}}{9 f}-\left (b^2 \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}\right ) \int \tan ^4(e+f x) \, dx\\ &=-\frac{b^2 \tan (e+f x) \sqrt{b \tan ^4(e+f x)}}{3 f}+\frac{b^2 \tan ^3(e+f x) \sqrt{b \tan ^4(e+f x)}}{5 f}-\frac{b^2 \tan ^5(e+f x) \sqrt{b \tan ^4(e+f x)}}{7 f}+\frac{b^2 \tan ^7(e+f x) \sqrt{b \tan ^4(e+f x)}}{9 f}+\left (b^2 \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}\right ) \int \tan ^2(e+f x) \, dx\\ &=\frac{b^2 \cot (e+f x) \sqrt{b \tan ^4(e+f x)}}{f}-\frac{b^2 \tan (e+f x) \sqrt{b \tan ^4(e+f x)}}{3 f}+\frac{b^2 \tan ^3(e+f x) \sqrt{b \tan ^4(e+f x)}}{5 f}-\frac{b^2 \tan ^5(e+f x) \sqrt{b \tan ^4(e+f x)}}{7 f}+\frac{b^2 \tan ^7(e+f x) \sqrt{b \tan ^4(e+f x)}}{9 f}-\left (b^2 \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}\right ) \int 1 \, dx\\ &=\frac{b^2 \cot (e+f x) \sqrt{b \tan ^4(e+f x)}}{f}-b^2 x \cot ^2(e+f x) \sqrt{b \tan ^4(e+f x)}-\frac{b^2 \tan (e+f x) \sqrt{b \tan ^4(e+f x)}}{3 f}+\frac{b^2 \tan ^3(e+f x) \sqrt{b \tan ^4(e+f x)}}{5 f}-\frac{b^2 \tan ^5(e+f x) \sqrt{b \tan ^4(e+f x)}}{7 f}+\frac{b^2 \tan ^7(e+f x) \sqrt{b \tan ^4(e+f x)}}{9 f}\\ \end{align*}

Mathematica [A]  time = 0.718507, size = 86, normalized size = 0.47 \[ \frac{\cot (e+f x) \left (b \tan ^4(e+f x)\right )^{5/2} \left (315 \cot ^8(e+f x)-105 \cot ^6(e+f x)+63 \cot ^4(e+f x)-45 \cot ^2(e+f x)-315 \tan ^{-1}(\tan (e+f x)) \cot ^9(e+f x)+35\right )}{315 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^4)^(5/2),x]

[Out]

(Cot[e + f*x]*(35 - 45*Cot[e + f*x]^2 + 63*Cot[e + f*x]^4 - 105*Cot[e + f*x]^6 + 315*Cot[e + f*x]^8 - 315*ArcT
an[Tan[e + f*x]]*Cot[e + f*x]^9)*(b*Tan[e + f*x]^4)^(5/2))/(315*f)

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Maple [A]  time = 0.037, size = 84, normalized size = 0.5 \begin{align*} -{\frac{-35\, \left ( \tan \left ( fx+e \right ) \right ) ^{9}+45\, \left ( \tan \left ( fx+e \right ) \right ) ^{7}-63\, \left ( \tan \left ( fx+e \right ) \right ) ^{5}+105\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}+315\,\arctan \left ( \tan \left ( fx+e \right ) \right ) -315\,\tan \left ( fx+e \right ) }{315\,f \left ( \tan \left ( fx+e \right ) \right ) ^{10}} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{4} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^4)^(5/2),x)

[Out]

-1/315/f*(b*tan(f*x+e)^4)^(5/2)*(-35*tan(f*x+e)^9+45*tan(f*x+e)^7-63*tan(f*x+e)^5+105*tan(f*x+e)^3+315*arctan(
tan(f*x+e))-315*tan(f*x+e))/tan(f*x+e)^10

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Maxima [A]  time = 1.66781, size = 107, normalized size = 0.59 \begin{align*} \frac{35 \, b^{\frac{5}{2}} \tan \left (f x + e\right )^{9} - 45 \, b^{\frac{5}{2}} \tan \left (f x + e\right )^{7} + 63 \, b^{\frac{5}{2}} \tan \left (f x + e\right )^{5} - 105 \, b^{\frac{5}{2}} \tan \left (f x + e\right )^{3} - 315 \,{\left (f x + e\right )} b^{\frac{5}{2}} + 315 \, b^{\frac{5}{2}} \tan \left (f x + e\right )}{315 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(5/2),x, algorithm="maxima")

[Out]

1/315*(35*b^(5/2)*tan(f*x + e)^9 - 45*b^(5/2)*tan(f*x + e)^7 + 63*b^(5/2)*tan(f*x + e)^5 - 105*b^(5/2)*tan(f*x
+ e)^3 - 315*(f*x + e)*b^(5/2) + 315*b^(5/2)*tan(f*x + e))/f

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Fricas [A]  time = 2.53901, size = 247, normalized size = 1.36 \begin{align*} \frac{{\left (35 \, b^{2} \tan \left (f x + e\right )^{9} - 45 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, b^{2} \tan \left (f x + e\right )^{5} - 105 \, b^{2} \tan \left (f x + e\right )^{3} - 315 \, b^{2} f x + 315 \, b^{2} \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{4}}}{315 \, f \tan \left (f x + e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(5/2),x, algorithm="fricas")

[Out]

1/315*(35*b^2*tan(f*x + e)^9 - 45*b^2*tan(f*x + e)^7 + 63*b^2*tan(f*x + e)^5 - 105*b^2*tan(f*x + e)^3 - 315*b^
2*f*x + 315*b^2*tan(f*x + e))*sqrt(b*tan(f*x + e)^4)/(f*tan(f*x + e)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**4)**(5/2),x)

[Out]

Integral((b*tan(e + f*x)**4)**(5/2), x)

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Giac [B]  time = 14.4079, size = 1381, normalized size = 7.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^4)^(5/2),x, algorithm="giac")

[Out]

-1/315*(315*b^2*f*x*tan(f*x)^9*tan(e)^9 - 2835*b^2*f*x*tan(f*x)^8*tan(e)^8 + 315*b^2*tan(f*x)^9*tan(e)^8 + 315
*b^2*tan(f*x)^8*tan(e)^9 + 11340*b^2*f*x*tan(f*x)^7*tan(e)^7 - 105*b^2*tan(f*x)^9*tan(e)^6 - 2835*b^2*tan(f*x)
^8*tan(e)^7 - 2835*b^2*tan(f*x)^7*tan(e)^8 - 105*b^2*tan(f*x)^6*tan(e)^9 - 26460*b^2*f*x*tan(f*x)^6*tan(e)^6 +
63*b^2*tan(f*x)^9*tan(e)^4 + 945*b^2*tan(f*x)^8*tan(e)^5 + 11340*b^2*tan(f*x)^7*tan(e)^6 + 11340*b^2*tan(f*x)
^6*tan(e)^7 + 945*b^2*tan(f*x)^5*tan(e)^8 + 63*b^2*tan(f*x)^4*tan(e)^9 + 39690*b^2*f*x*tan(f*x)^5*tan(e)^5 - 4
5*b^2*tan(f*x)^9*tan(e)^2 - 567*b^2*tan(f*x)^8*tan(e)^3 - 3780*b^2*tan(f*x)^7*tan(e)^4 - 26460*b^2*tan(f*x)^6*
tan(e)^5 - 26460*b^2*tan(f*x)^5*tan(e)^6 - 3780*b^2*tan(f*x)^4*tan(e)^7 - 567*b^2*tan(f*x)^3*tan(e)^8 - 45*b^2
*tan(f*x)^2*tan(e)^9 - 39690*b^2*f*x*tan(f*x)^4*tan(e)^4 + 35*b^2*tan(f*x)^9 + 405*b^2*tan(f*x)^8*tan(e) + 226
8*b^2*tan(f*x)^7*tan(e)^2 + 8820*b^2*tan(f*x)^6*tan(e)^3 + 39690*b^2*tan(f*x)^5*tan(e)^4 + 39690*b^2*tan(f*x)^
4*tan(e)^5 + 8820*b^2*tan(f*x)^3*tan(e)^6 + 2268*b^2*tan(f*x)^2*tan(e)^7 + 405*b^2*tan(f*x)*tan(e)^8 + 35*b^2*
tan(e)^9 + 26460*b^2*f*x*tan(f*x)^3*tan(e)^3 - 45*b^2*tan(f*x)^7 - 567*b^2*tan(f*x)^6*tan(e) - 3780*b^2*tan(f*
x)^5*tan(e)^2 - 26460*b^2*tan(f*x)^4*tan(e)^3 - 26460*b^2*tan(f*x)^3*tan(e)^4 - 3780*b^2*tan(f*x)^2*tan(e)^5 -
567*b^2*tan(f*x)*tan(e)^6 - 45*b^2*tan(e)^7 - 11340*b^2*f*x*tan(f*x)^2*tan(e)^2 + 63*b^2*tan(f*x)^5 + 945*b^2
*tan(f*x)^4*tan(e) + 11340*b^2*tan(f*x)^3*tan(e)^2 + 11340*b^2*tan(f*x)^2*tan(e)^3 + 945*b^2*tan(f*x)*tan(e)^4
+ 63*b^2*tan(e)^5 + 2835*b^2*f*x*tan(f*x)*tan(e) - 105*b^2*tan(f*x)^3 - 2835*b^2*tan(f*x)^2*tan(e) - 2835*b^2
*tan(f*x)*tan(e)^2 - 105*b^2*tan(e)^3 - 315*b^2*f*x + 315*b^2*tan(f*x) + 315*b^2*tan(e))*sqrt(b)/(f*tan(f*x)^9
*tan(e)^9 - 9*f*tan(f*x)^8*tan(e)^8 + 36*f*tan(f*x)^7*tan(e)^7 - 84*f*tan(f*x)^6*tan(e)^6 + 126*f*tan(f*x)^5*t
an(e)^5 - 126*f*tan(f*x)^4*tan(e)^4 + 84*f*tan(f*x)^3*tan(e)^3 - 36*f*tan(f*x)^2*tan(e)^2 + 9*f*tan(f*x)*tan(e
) - f)