3.36 \(\int (b \tan ^4(c+d x))^{5/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 182, normalized size of antiderivative = 1.00, 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} \[ \frac {b^2 \tan ^7(c+d x) \sqrt {b \tan ^4(c+d x)}}{9 d}-\frac {b^2 \tan ^5(c+d x) \sqrt {b \tan ^4(c+d x)}}{7 d}+\frac {b^2 \tan ^3(c+d x) \sqrt {b \tan ^4(c+d x)}}{5 d}-\frac {b^2 \tan (c+d x) \sqrt {b \tan ^4(c+d x)}}{3 d}-b^2 x \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)}+\frac {b^2 \cot (c+d x) \sqrt {b \tan ^4(c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^4)^(5/2),x]

[Out]

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

Rule 8

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

Rubi steps

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

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Mathematica [A]  time = 0.76, size = 86, normalized size = 0.47 \[ \frac {\cot (c+d x) \left (b \tan ^4(c+d x)\right )^{5/2} \left (315 \cot ^8(c+d x)-105 \cot ^6(c+d x)+63 \cot ^4(c+d x)-45 \cot ^2(c+d x)-315 \tan ^{-1}(\tan (c+d x)) \cot ^9(c+d x)+35\right )}{315 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x]^4)^(5/2),x]

[Out]

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

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fricas [A]  time = 1.11, size = 96, normalized size = 0.53 \[ \frac {{\left (35 \, b^{2} \tan \left (d x + c\right )^{9} - 45 \, b^{2} \tan \left (d x + c\right )^{7} + 63 \, b^{2} \tan \left (d x + c\right )^{5} - 105 \, b^{2} \tan \left (d x + c\right )^{3} - 315 \, b^{2} d x + 315 \, b^{2} \tan \left (d x + c\right )\right )} \sqrt {b \tan \left (d x + c\right )^{4}}}{315 \, d \tan \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.12, size = 84, normalized size = 0.46 \[ -\frac {\left (b \left (\tan ^{4}\left (d x +c \right )\right )\right )^{\frac {5}{2}} \left (-35 \left (\tan ^{9}\left (d x +c \right )\right )+45 \left (\tan ^{7}\left (d x +c \right )\right )-63 \left (\tan ^{5}\left (d x +c \right )\right )+105 \left (\tan ^{3}\left (d x +c \right )\right )+315 \arctan \left (\tan \left (d x +c \right )\right )-315 \tan \left (d x +c \right )\right )}{315 d \tan \left (d x +c \right )^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(d*x+c)^4)^(5/2),x)

[Out]

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

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maxima [A]  time = 0.84, size = 79, normalized size = 0.43 \[ \frac {35 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{9} - 45 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{7} + 63 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{5} - 105 \, b^{\frac {5}{2}} \tan \left (d x + c\right )^{3} - 315 \, {\left (d x + c\right )} b^{\frac {5}{2}} + 315 \, b^{\frac {5}{2}} \tan \left (d x + c\right )}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(c + d*x)^4)^(5/2),x)

[Out]

int((b*tan(c + d*x)^4)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan ^{4}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tan(d*x+c)**4*b)**(5/2),x)

[Out]

Integral((b*tan(c + d*x)**4)**(5/2), x)

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