3.7.36 \(\int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\) [636]

Optimal. Leaf size=109 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3656, 926, 95, 211, 214} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]),x]

[Out]

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

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 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 926

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 3656

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

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 124, normalized size = 1.14 \begin {gather*} \frac {(-1)^{3/4} \left (-\frac {\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+i b}}-\frac {\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+i b}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.70, size = 940263, normalized size = 8626.27 \[\text {output too large to display}\]

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

[Out]

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

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))**(1/2),x)

[Out]

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

[Out]

Timed out

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Mupad [B]
time = 13.30, size = 1716, normalized size = 15.74 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(((400*b^6*d^3*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(3/2))/((a + b*tan(c + d*x))^(1/2)
- a^(1/2)) - (48*b^5*d*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2))/((a + b*tan(c + d*x))^(1
/2) - a^(1/2)) - (a^7*d^5*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(5/2)*4096i)/((a + b*tan(c +
d*x))^(1/2) - a^(1/2)) - (a^5*d*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*256i)/((a + b*t
an(c + d*x))^(1/2) - a^(1/2)) - (832*b^7*d^5*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(5/2))/((
a + b*tan(c + d*x))^(1/2) - a^(1/2)) - (112*a^2*b^3*d*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^
(1/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (a^3*b^2*d*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d
^2))^(1/2)*720i)/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) - (a*b^5*d^3*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2
+ 4*b^2*d^2))^(3/2)*1200i)/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (a^5*b*d^3*tan(c + d*x)^(1/2)*((a*1i + b)
/(4*a^2*d^2 + 4*b^2*d^2))^(3/2)*2048i)/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (a*b^6*d^5*tan(c + d*x)^(1/2)*
((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(5/2)*2496i)/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (12288*a^6*b*d^5*ta
n(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(5/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (848*a^2
*b^4*d^3*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(3/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))
- (a^3*b^3*d^3*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(3/2)*5744i)/((a + b*tan(c + d*x))^(1/
2) - a^(1/2)) - (6144*a^4*b^2*d^3*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(3/2))/((a + b*tan(c
+ d*x))^(1/2) - a^(1/2)) - (2368*a^2*b^5*d^5*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(5/2))/(
(a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (a^3*b^4*d^5*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(
5/2)*13760i)/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (10496*a^4*b^3*d^5*tan(c + d*x)^(1/2)*((a*1i + b)/(4*a^2
*d^2 + 4*b^2*d^2))^(5/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (a^5*b^2*d^5*tan(c + d*x)^(1/2)*((a*1i + b)
/(4*a^2*d^2 + 4*b^2*d^2))^(5/2)*7424i)/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (a*b^4*d*tan(c + d*x)^(1/2)*((
a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*144i)/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (768*a^4*b*d*tan(c + d
*x)^(1/2)*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2)))/(a*b^3 + (a*b^4*
tan(c + d*x))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2))*((a*1i + b)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*2i - atan(
((2*a^3*d^5*tan(c + d*x)^(1/2)*(-(a*1i - b)/(a^2*d^2 + b^2*d^2))^(5/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))
+ (b^3*d^5*tan(c + d*x)^(1/2)*(-(a*1i - b)/(a^2*d^2 + b^2*d^2))^(5/2)*2i)/((a + b*tan(c + d*x))^(1/2) - a^(1/
2)) - (6*a*b^2*d^5*tan(c + d*x)^(1/2)*(-(a*1i - b)/(a^2*d^2 + b^2*d^2))^(5/2))/((a + b*tan(c + d*x))^(1/2) - a
^(1/2)) - (a^2*b*d^5*tan(c + d*x)^(1/2)*(-(a*1i - b)/(a^2*d^2 + b^2*d^2))^(5/2)*6i)/((a + b*tan(c + d*x))^(1/2
) - a^(1/2)))/((b*tan(c + d*x))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2 + 1))*(-(a*1i - b)/(a^2*d^2 + b^2*d^2
))^(1/2)*1i

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