3.1.37 \(\int (b \tan ^4(c+d x))^{3/2} \, dx\) [37]

Optimal. Leaf size=110 \[ \frac {b \cot (c+d x) \sqrt {b \tan ^4(c+d x)}}{d}-b x \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)}-\frac {b \tan (c+d x) \sqrt {b \tan ^4(c+d x)}}{3 d}+\frac {b \tan ^3(c+d x) \sqrt {b \tan ^4(c+d x)}}{5 d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \begin {gather*} -\frac {b \tan (c+d x) \sqrt {b \tan ^4(c+d x)}}{3 d}+\frac {b \tan ^3(c+d x) \sqrt {b \tan ^4(c+d x)}}{5 d}-b x \cot ^2(c+d x) \sqrt {b \tan ^4(c+d x)}+\frac {b \cot (c+d x) \sqrt {b \tan ^4(c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^4)^(3/2),x]

[Out]

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

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

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Mathematica [A]
time = 0.81, size = 66, normalized size = 0.60 \begin {gather*} \frac {\cot (c+d x) \left (3-5 \cot ^2(c+d x)+15 \cot ^4(c+d x)-15 \text {ArcTan}(\tan (c+d x)) \cot ^5(c+d x)\right ) \left (b \tan ^4(c+d x)\right )^{3/2}}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x]^4)^(3/2),x]

[Out]

(Cot[c + d*x]*(3 - 5*Cot[c + d*x]^2 + 15*Cot[c + d*x]^4 - 15*ArcTan[Tan[c + d*x]]*Cot[c + d*x]^5)*(b*Tan[c + d
*x]^4)^(3/2))/(15*d)

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Maple [A]
time = 0.06, size = 64, normalized size = 0.58

method	result	size
derivativedivides	\(-\frac {\left (b \left (\tan ^{4}\left (d x +c \right )\right )\right )^{\frac {3}{2}} \left (-3 \left (\tan ^{5}\left (d x +c \right )\right )+5 \left (\tan ^{3}\left (d x +c \right )\right )+15 \arctan \left (\tan \left (d x +c \right )\right )-15 \tan \left (d x +c \right )\right )}{15 d \tan \left (d x +c \right )^{6}}\)	\(64\)
default	\(-\frac {\left (b \left (\tan ^{4}\left (d x +c \right )\right )\right )^{\frac {3}{2}} \left (-3 \left (\tan ^{5}\left (d x +c \right )\right )+5 \left (\tan ^{3}\left (d x +c \right )\right )+15 \arctan \left (\tan \left (d x +c \right )\right )-15 \tan \left (d x +c \right )\right )}{15 d \tan \left (d x +c \right )^{6}}\)	\(64\)
risch	\(\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, x}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 i b \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, \left (45 \,{\mathrm e}^{8 i \left (d x +c \right )}+90 \,{\mathrm e}^{6 i \left (d x +c \right )}+140 \,{\mathrm e}^{4 i \left (d x +c \right )}+70 \,{\mathrm e}^{2 i \left (d x +c \right )}+23\right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} d}\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(d*x+c)^4)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/d*(b*tan(d*x+c)^4)^(3/2)*(-3*tan(d*x+c)^5+5*tan(d*x+c)^3+15*arctan(tan(d*x+c))-15*tan(d*x+c))/tan(d*x+c)
^6

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Maxima [A]
time = 0.50, size = 53, normalized size = 0.48 \begin {gather*} \frac {3 \, b^{\frac {3}{2}} \tan \left (d x + c\right )^{5} - 5 \, b^{\frac {3}{2}} \tan \left (d x + c\right )^{3} - 15 \, {\left (d x + c\right )} b^{\frac {3}{2}} + 15 \, b^{\frac {3}{2}} \tan \left (d x + c\right )}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tan(d*x+c)^4*b)^(3/2),x, algorithm="maxima")

[Out]

1/15*(3*b^(3/2)*tan(d*x + c)^5 - 5*b^(3/2)*tan(d*x + c)^3 - 15*(d*x + c)*b^(3/2) + 15*b^(3/2)*tan(d*x + c))/d

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Fricas [A]
time = 0.36, size = 62, normalized size = 0.56 \begin {gather*} \frac {{\left (3 \, b \tan \left (d x + c\right )^{5} - 5 \, b \tan \left (d x + c\right )^{3} - 15 \, b d x + 15 \, b \tan \left (d x + c\right )\right )} \sqrt {b \tan \left (d x + c\right )^{4}}}{15 \, d \tan \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tan(d*x+c)^4*b)^(3/2),x, algorithm="fricas")

[Out]

1/15*(3*b*tan(d*x + c)^5 - 5*b*tan(d*x + c)^3 - 15*b*d*x + 15*b*tan(d*x + c))*sqrt(b*tan(d*x + c)^4)/(d*tan(d*
x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \tan ^{4}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tan(d*x+c)**4*b)**(3/2),x)

[Out]

Integral((b*tan(c + d*x)**4)**(3/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 992 vs. \(2 (98) = 196\).
time = 2.74, size = 992, normalized size = 9.02 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tan(d*x+c)^4*b)^(3/2),x, algorithm="giac")

[Out]

1/60*(15*pi - 60*d*x*tan(d*x)^5*tan(c)^5 - 15*pi*sgn(2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 - 2*tan(d*x) -
2*tan(c))*tan(d*x)^5*tan(c)^5 - 15*pi*tan(d*x)^5*tan(c)^5 + 30*arctan((tan(d*x)*tan(c) - 1)/(tan(d*x) + tan(c)
))*tan(d*x)^5*tan(c)^5 + 30*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(d*x)^5*tan(c)^5 + 300*d*x*ta
n(d*x)^4*tan(c)^4 + 75*pi*sgn(2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 - 2*tan(d*x) - 2*tan(c))*tan(d*x)^4*ta
n(c)^4 + 75*pi*tan(d*x)^4*tan(c)^4 - 150*arctan((tan(d*x)*tan(c) - 1)/(tan(d*x) + tan(c)))*tan(d*x)^4*tan(c)^4
- 150*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(d*x)^4*tan(c)^4 - 60*tan(d*x)^5*tan(c)^4 - 60*tan
(d*x)^4*tan(c)^5 - 600*d*x*tan(d*x)^3*tan(c)^3 - 150*pi*sgn(2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 - 2*tan(
d*x) - 2*tan(c))*tan(d*x)^3*tan(c)^3 + 20*tan(d*x)^5*tan(c)^2 - 150*pi*tan(d*x)^3*tan(c)^3 + 300*arctan((tan(d
*x)*tan(c) - 1)/(tan(d*x) + tan(c)))*tan(d*x)^3*tan(c)^3 + 300*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1
))*tan(d*x)^3*tan(c)^3 + 300*tan(d*x)^4*tan(c)^3 + 300*tan(d*x)^3*tan(c)^4 + 20*tan(d*x)^2*tan(c)^5 + 600*d*x*
tan(d*x)^2*tan(c)^2 + 150*pi*sgn(2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 - 2*tan(d*x) - 2*tan(c))*tan(d*x)^2
*tan(c)^2 - 12*tan(d*x)^5 - 100*tan(d*x)^4*tan(c) + 150*pi*tan(d*x)^2*tan(c)^2 - 300*arctan((tan(d*x)*tan(c) -
1)/(tan(d*x) + tan(c)))*tan(d*x)^2*tan(c)^2 - 300*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(d*x)^
2*tan(c)^2 - 600*tan(d*x)^3*tan(c)^2 - 600*tan(d*x)^2*tan(c)^3 - 100*tan(d*x)*tan(c)^4 - 12*tan(c)^5 - 300*d*x
*tan(d*x)*tan(c) - 75*pi*sgn(2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 - 2*tan(d*x) - 2*tan(c))*tan(d*x)*tan(c
) + 20*tan(d*x)^3 - 75*pi*tan(d*x)*tan(c) + 150*arctan((tan(d*x)*tan(c) - 1)/(tan(d*x) + tan(c)))*tan(d*x)*tan
(c) + 150*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c) - 1))*tan(d*x)*tan(c) + 300*tan(d*x)^2*tan(c) + 300*tan(
d*x)*tan(c)^2 + 20*tan(c)^3 + 60*d*x + 15*pi*sgn(2*tan(d*x)^2*tan(c) + 2*tan(d*x)*tan(c)^2 - 2*tan(d*x) - 2*ta
n(c)) - 30*arctan((tan(d*x)*tan(c) - 1)/(tan(d*x) + tan(c))) - 30*arctan((tan(d*x) + tan(c))/(tan(d*x)*tan(c)
- 1)) - 60*tan(d*x) - 60*tan(c))*b^(3/2)/(d*tan(d*x)^5*tan(c)^5 - 5*d*tan(d*x)^4*tan(c)^4 + 10*d*tan(d*x)^3*ta
n(c)^3 - 10*d*tan(d*x)^2*tan(c)^2 + 5*d*tan(d*x)*tan(c) - d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(c + d*x)^4)^(3/2),x)

[Out]

int((b*tan(c + d*x)^4)^(3/2), x)

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