3.7.41 \(\int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) [641]

Optimal. Leaf size=195 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]

[Out]

arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a-b)^(3/2)/d+2*arctanh(b^(1/2)*tan(d*x+c)^(1/
2)/(a+b*tan(d*x+c))^(1/2))/b^(3/2)/d-arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a+b)^(3
/2)/d-2*a^2*tan(d*x+c)^(1/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)

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Rubi [A]
time = 0.89, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3646, 3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} -\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(5/2)/(a + b*Tan[c + d*x])^(3/2),x]

[Out]

ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/((I*a - b)^(3/2)*d) + (2*ArcTanh[(Sqrt[b]*
Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(b^(3/2)*d) - ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a
+ b*Tan[c + d*x]]]/((I*a + b)^(3/2)*d) - (2*a^2*Sqrt[Tan[c + d*x]])/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]
)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3736

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^2)/(1 + ff^2*x^2)), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx &=-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \int \frac {\frac {a^2}{2}-\frac {1}{2} a b \tan (c+d x)+\frac {1}{2} \left (a^2+b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {\frac {a^2}{2}-\frac {a b x}{2}+\frac {1}{2} \left (a^2+b^2\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \left (\frac {a^2+b^2}{2 \sqrt {x} \sqrt {a+b x}}-\frac {b^2+a b x}{2 \sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{b d}-\frac {\text {Subst}\left (\int \frac {b^2+a b x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b d}-\frac {\text {Subst}\left (\int \left (\frac {-a b+i b^2}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {a b+i b^2}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b) d}+\frac {\text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b) d}+\frac {2 \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b d}\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a-i b) d}+\frac {\text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b) d}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.26, size = 321, normalized size = 1.65 \begin {gather*} -\frac {2 \sqrt {a} \sqrt {-a+i b} \sqrt {a+i b} \left (a^2+b^2\right ) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}+\sqrt {b} \left (\sqrt [4]{-1} (a+i b)^{3/2} b \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {-a+i b} \left (-2 a^2 \sqrt {a+i b} \sqrt {\tan (c+d x)}-\sqrt [4]{-1} (a-i b) b \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}\right )\right )}{(-a+i b)^{3/2} (a+i b)^{3/2} b^{3/2} d \sqrt {a+b \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(5/2)/(a + b*Tan[c + d*x])^(3/2),x]

[Out]

-((2*Sqrt[a]*Sqrt[-a + I*b]*Sqrt[a + I*b]*(a^2 + b^2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (
b*Tan[c + d*x])/a] + Sqrt[b]*((-1)^(1/4)*(a + I*b)^(3/2)*b*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]
])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + Sqrt[-a + I*b]*(-2*a^2*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]
] - (-1)^(1/4)*(a - I*b)*b*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt
[a + b*Tan[c + d*x]])))/((-a + I*b)^(3/2)*(a + I*b)^(3/2)*b^(3/2)*d*Sqrt[a + b*Tan[c + d*x]]))

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.97, size = 798950, normalized size = 4097.18 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(3/2),x)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)^(5/2)/(b*tan(d*x + c) + a)^(3/2), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{\frac {5}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\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)**(5/2)/(a+b*tan(d*x+c))**(3/2),x)

[Out]

Integral(tan(c + d*x)**(5/2)/(a + b*tan(c + d*x))**(3/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(5/2)/(a + b*tan(c + d*x))^(3/2),x)

[Out]

int(tan(c + d*x)^(5/2)/(a + b*tan(c + d*x))^(3/2), x)

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