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

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

[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)

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Rubi [A]  time = 0.0628983, 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} \[ \frac{b^2 \tan ^7(c+d x) \sqrt{b \tan ^4(c+d x)}}{9 d}-\frac{b^2 \tan ^5(c+d x) \sqrt{b \tan ^4(c+d x)}}{7 d}+\frac{b^2 \tan ^3(c+d x) \sqrt{b \tan ^4(c+d x)}}{5 d}-\frac{b^2 \tan (c+d x) \sqrt{b \tan ^4(c+d x)}}{3 d}-b^2 x \cot ^2(c+d x) \sqrt{b \tan ^4(c+d x)}+\frac{b^2 \cot (c+d x) \sqrt{b \tan ^4(c+d x)}}{d} \]

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 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(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*}

Mathematica [A]  time = 0.766847, 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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Maple [A]  time = 0.036, size = 84, normalized size = 0.5 \begin{align*} -{\frac{-35\, \left ( \tan \left ( dx+c \right ) \right ) ^{9}+45\, \left ( \tan \left ( dx+c \right ) \right ) ^{7}-63\, \left ( \tan \left ( dx+c \right ) \right ) ^{5}+105\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}+315\,\arctan \left ( \tan \left ( dx+c \right ) \right ) -315\,\tan \left ( dx+c \right ) }{315\,d \left ( \tan \left ( dx+c \right ) \right ) ^{10}} \left ( b \left ( \tan \left ( dx+c \right ) \right ) ^{4} \right ) ^{{\frac{5}{2}}}} \end{align*}

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 = 1.38655, size = 107, normalized size = 0.59 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.55799, size = 247, normalized size = 1.36 \begin{align*} \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}} \end{align*}

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

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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Giac [B]  time = 10.3889, size = 1296, normalized size = 7.12 \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)^4*b)^(5/2),x, algorithm="giac")

[Out]

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