3.644 \(\int \frac{1}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=159 \[ \frac{2 b^2 \sqrt{\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{i \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]

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

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Rubi [A]  time = 0.465684, antiderivative size = 159, 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 = {3569, 3616, 3615, 93, 203, 206} \[ \frac{2 b^2 \sqrt{\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt{a+b \tan (c+d x)}}+\frac{i \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}+\frac{i \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

Rule 3569

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

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

Mathematica [A]  time = 0.893916, size = 183, normalized size = 1.15 \[ -\frac{-\frac{2 b^2 \sqrt{\tan (c+d x)}}{a \sqrt{a+b \tan (c+d x)}}+\frac{\sqrt [4]{-1} (b+i a) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}+\frac{(-1)^{3/4} (a+i b) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.678, size = 761738, normalized size = 4790.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/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{1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sqrt{\tan \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*tan(d*x + c) + a)^(3/2)*sqrt(tan(d*x + c))), 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(1/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{1}{\left (a + b \tan{\left (c + d x \right )}\right )^{\frac{3}{2}} \sqrt{\tan{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError