3.643 \(\int \frac{\sqrt{\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=149 \[ -\frac{2 b \sqrt{\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.453228, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3568, 3616, 3615, 93, 203, 206} \[ -\frac{2 b \sqrt{\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

Rule 3568

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

Rule 3616

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]

Rule 3615

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 93

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 203

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

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 1.53408, size = 164, normalized size = 1.1 \[ -\frac{\frac{2 b \sqrt{\tan (c+d x)}}{\left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{\sqrt [4]{-1} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(-a-i b)^{3/2}}+\frac{\sqrt [4]{-1} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a-i b)^{3/2}}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.677, size = 797817, normalized size = 5354.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\tan \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(1/2)/(a+b*tan(d*x+c))**(3/2),x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError