3.7.54 \(\int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx\) [654]
Optimal. Leaf size=157 \[ \frac {i B \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a-b} d}-\frac {i B \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a+b} d} \]
[Out]
I*B*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d/(I*a-b)^
(1/2)-I*B*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d/(
I*a+b)^(1/2)
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Rubi [A]
time = 0.15, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {21, 4326, 3656,
924, 95, 211, 214} \begin {gather*} \frac {i B \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \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 {i B \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \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[(a*B + b*B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)),x]
[Out]
(I*B*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]
)/(Sqrt[I*a - b]*d) - (I*B*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d
*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[I*a + b]*d)
Rule 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 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 924
Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (c_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegr
and[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] &
& NeQ[c*d^2 + a*e^2, 0] && IGtQ[m + 1/2, 0]
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]
Rule 4326
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]
Rubi steps
\begin {align*} \int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx &=B \int \frac {1}{\sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=\left (B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {\left (B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \left (-\frac {1}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {1}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {\left (B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {\left (B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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 {\left (B \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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 B \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a-b} d}-\frac {i B \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a+b} d}\\ \end {align*}
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Mathematica [A]
time = 0.13, size = 144, normalized size = 0.92 \begin {gather*} \frac {\sqrt [4]{-1} B \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 ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a*B + b*B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)),x]
[Out]
((-1)^(1/4)*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
]) + ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqrt[a + I*b])*Sqrt[Cot[c
+ d*x]]*Sqrt[Tan[c + d*x]])/d
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 31.19, size = 1647, normalized size = 10.49
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a*B+b*B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
[Out]
B/d/(-I*a+(a^2+b^2)^(1/2)-b)/(I*a+(a^2+b^2)^(1/2)-b)*(I*EllipticPi((-(a*cos(d*x+c)-(a^2+b^2)^(1/2)*sin(d*x+c)+
b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(-I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/
2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*a*(a^2+b^2)^(1/2)-I*EllipticPi((-(a*cos(d*x+c)-(a^2+b^2)^(1/2
)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(I*a+(a^2+b^2)^(1/2)-
b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*a*(a^2+b^2)^(1/2)-I*EllipticPi((-(a*cos(d*x+c)-(a
^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(-I*a+(a^
2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*a*b+I*EllipticPi((-(a*cos(d*x+c)-(a^
2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(I*a+(a^2+
b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*a*b-2*EllipticPi((-(a*cos(d*x+c)-(a^2+
b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(-I*a+(a^2+b
^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*b*(a^2+b^2)^(1/2)-2*EllipticPi((-(a*cos
(d*x+c)-(a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))
/(I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*b*(a^2+b^2)^(1/2)+EllipticP
i((-(a*cos(d*x+c)-(a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b
^2)^(1/2))/(-I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*a^2+2*EllipticPi
((-(a*cos(d*x+c)-(a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^
2)^(1/2))/(-I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*b^2+EllipticPi((-
(a*cos(d*x+c)-(a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^
(1/2))/(I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*a^2+2*EllipticPi((-(a
*cos(d*x+c)-(a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1
/2))/(I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*b^2)*cos(d*x+c)*sin(d*x
+c)*((a*cos(d*x+c)+b*sin(d*x+c))/cos(d*x+c))^(1/2)*2^(1/2)*(a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))
^(1/2)*((a*cos(d*x+c)+(a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(a^2+b^2)^(1/2)/sin(d*x+c))^(1/2)*(-(a*cos(d*
x+c)-(a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)-a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)/(a*cos(d*x+c)+b*sin(d*x
+c))/(-1+cos(d*x+c))/(cos(d*x+c)/sin(d*x+c))^(1/2)
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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((a*B+b*B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
integrate((B*b*tan(d*x + c) + B*a)/((b*tan(d*x + c) + a)^(3/2)*sqrt(cot(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((a*B+b*B*tan(d*x+c))/cot(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} B \int \frac {1}{\sqrt {a + b \tan {\left (c + d x \right )}} \sqrt {\cot {\left (c + d x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a*B+b*B*tan(d*x+c))/cot(d*x+c)**(1/2)/(a+b*tan(d*x+c))**(3/2),x)
[Out]
B*Integral(1/(sqrt(a + b*tan(c + d*x))*sqrt(cot(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((a*B+b*B*tan(d*x+c))/cot(d*x+c)^(1/2)/(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 {B\,a+B\,b\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\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((B*a + B*b*tan(c + d*x))/(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(3/2)),x)
[Out]
int((B*a + B*b*tan(c + d*x))/(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))^(3/2)), x)
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