3.7.42 \(\int \frac {\tan ^{\frac {3}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) [642]

Optimal. Leaf size=154 \[ -\frac {i \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}+\frac {2 a \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.31, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3648, 3697, 3696, 95, 209, 212} \begin {gather*} \frac {2 a \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {i \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 {i \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]^(3/2)/(a + b*Tan[c + d*x])^(3/2),x]

[Out]

((-I)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a - b)^(3/2)*d) - (I*ArcTanh[(S
qrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((I*a + b)^(3/2)*d) + (2*a*Sqrt[Tan[c + d*x]])/((a
^2 + b^2)*d*Sqrt[a + b*Tan[c + d*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 209

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

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 3648

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[
1/((m + 1)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*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] && LtQ[m, -1] && LtQ[1, n, 2] && IntegerQ[2*m]

Rule 3696

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

Rule 3697

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

Rubi steps

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

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Mathematica [A]
time = 1.29, size = 182, normalized size = 1.18 \begin {gather*} \frac {-\frac {(-1)^{3/4} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{3/2}}+\frac {\frac {\sqrt [4]{-1} (i a+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 )}{(a+i b)^{3/2}}+\frac {2 a \sqrt {\tan (c+d x)}}{(a+i b) \sqrt {a+b \tan (c+d x)}}}{a-i b}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 2.29, size = 762478, normalized size = 4951.16 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Integral(tan(c + d*x)**(3/2)/(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)^(3/2)/(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 )}^{3/2}}{{\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)^(3/2)/(a + b*tan(c + d*x))^(3/2),x)

[Out]

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

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