\(\int (b \tan ^4(e+f x))^{5/2} \, dx\) [13]

Optimal result

Integrand size = 14, antiderivative size = 182 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\frac {b^2 \cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f}-b^2 x \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}-\frac {b^2 \tan (e+f x) \sqrt {b \tan ^4(e+f x)}}{3 f}+\frac {b^2 \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f}-\frac {b^2 \tan ^5(e+f x) \sqrt {b \tan ^4(e+f x)}}{7 f}+\frac {b^2 \tan ^7(e+f x) \sqrt {b \tan ^4(e+f x)}}{9 f} \]

[Out]

b^2*cot(f*x+e)*(b*tan(f*x+e)^4)^(1/2)/f-b^2*x*cot(f*x+e)^2*(b*tan(f*x+e)^4)^(1/2)-1/3*b^2*(b*tan(f*x+e)^4)^(1/
2)*tan(f*x+e)/f+1/5*b^2*(b*tan(f*x+e)^4)^(1/2)*tan(f*x+e)^3/f-1/7*b^2*(b*tan(f*x+e)^4)^(1/2)*tan(f*x+e)^5/f+1/
9*b^2*(b*tan(f*x+e)^4)^(1/2)*tan(f*x+e)^7/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=-\frac {b^2 \tan (e+f x) \sqrt {b \tan ^4(e+f x)}}{3 f}+\frac {b^2 \tan ^7(e+f x) \sqrt {b \tan ^4(e+f x)}}{9 f}-\frac {b^2 \tan ^5(e+f x) \sqrt {b \tan ^4(e+f x)}}{7 f}+\frac {b^2 \tan ^3(e+f x) \sqrt {b \tan ^4(e+f x)}}{5 f}-b^2 x \cot ^2(e+f x) \sqrt {b \tan ^4(e+f x)}+\frac {b^2 \cot (e+f x) \sqrt {b \tan ^4(e+f x)}}{f} \]

[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 8

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

Rule 3554

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 3739

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
]])

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.91 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.47 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\frac {\cot (e+f x) \left (35-45 \cot ^2(e+f x)+63 \cot ^4(e+f x)-105 \cot ^6(e+f x)+315 \cot ^8(e+f x)-315 \arctan (\tan (e+f x)) \cot ^9(e+f x)\right ) \left (b \tan ^4(e+f x)\right )^{5/2}}{315 f} \]

[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)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.46

method	result	size
derivativedivides	\(-\frac {\left (b \tan \left (f x +e \right )^{4}\right )^{\frac {5}{2}} \left (-35 \tan \left (f x +e \right )^{9}+45 \tan \left (f x +e \right )^{7}-63 \tan \left (f x +e \right )^{5}+105 \tan \left (f x +e \right )^{3}+315 \arctan \left (\tan \left (f x +e \right )\right )-315 \tan \left (f x +e \right )\right )}{315 f \tan \left (f x +e \right )^{10}}\)	\(84\)
default	\(-\frac {\left (b \tan \left (f x +e \right )^{4}\right )^{\frac {5}{2}} \left (-35 \tan \left (f x +e \right )^{9}+45 \tan \left (f x +e \right )^{7}-63 \tan \left (f x +e \right )^{5}+105 \tan \left (f x +e \right )^{3}+315 \arctan \left (\tan \left (f x +e \right )\right )-315 \tan \left (f x +e \right )\right )}{315 f \tan \left (f x +e \right )^{10}}\)	\(84\)
risch	\(\frac {b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, x}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {2 i b^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, \left (1575 \,{\mathrm e}^{16 i \left (f x +e \right )}+6300 \,{\mathrm e}^{14 i \left (f x +e \right )}+21000 \,{\mathrm e}^{12 i \left (f x +e \right )}+31500 \,{\mathrm e}^{10 i \left (f x +e \right )}+39438 \,{\mathrm e}^{8 i \left (f x +e \right )}+26292 \,{\mathrm e}^{6 i \left (f x +e \right )}+13968 \,{\mathrm e}^{4 i \left (f x +e \right )}+3492 \,{\mathrm e}^{2 i \left (f x +e \right )}+563\right )}{315 \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7} f}\)	\(218\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.53 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\frac {{\left (35 \, b^{2} \tan \left (f x + e\right )^{9} - 45 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, b^{2} \tan \left (f x + e\right )^{5} - 105 \, b^{2} \tan \left (f x + e\right )^{3} - 315 \, b^{2} f x + 315 \, b^{2} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{4}}}{315 \, f \tan \left (f x + e\right )^{2}} \]

[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)

Sympy [F]

\[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\int \left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.43 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\frac {35 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{9} - 45 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{7} + 63 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{5} - 105 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{3} - 315 \, {\left (f x + e\right )} b^{\frac {5}{2}} + 315 \, b^{\frac {5}{2}} \tan \left (f x + e\right )}{315 \, f} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 960 vs. \(2 (162) = 324\).

Time = 5.88 (sec) , antiderivative size = 960, normalized size of antiderivative = 5.27 \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\text {Too large to display} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \left (b \tan ^4(e+f x)\right )^{5/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}^{5/2} \,d x \]

[In]

int((b*tan(e + f*x)^4)^(5/2),x)

[Out]

int((b*tan(e + f*x)^4)^(5/2), x)