\(\int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\) [470]

Optimal result

Integrand size = 38, antiderivative size = 117 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {i B \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {i B \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {21, 3656, 924, 95, 211, 214} \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {i B \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {i B \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}} \]

[In]

Int[(Sqrt[Tan[c + d*x]]*(a*B + b*B*Tan[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[I*a - b]*d) - (I*B*ArcTanh[(Sq
rt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*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]

Rubi steps \begin{align*} \text {integral}& = B \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = \frac {B \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 {B \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 {B \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {B \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {B \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 {B \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 \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 {i B \text {arctanh}\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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt [4]{-1} B \left (-\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )}{d} \]

[In]

Integrate[(Sqrt[Tan[c + d*x]]*(a*B + b*B*Tan[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]))/d

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 0.57 (sec) , antiderivative size = 940499, normalized size of antiderivative = 8038.45

\[\text {output too large to display}\]

[In]

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

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4509 vs. \(2 (89) = 178\).

Time = 0.75 (sec) , antiderivative size = 4509, normalized size of antiderivative = 38.54 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/8*sqrt(((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2))*log((2*(2*
B^3*a^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) + (2*(B*a^5*b + 3*B*a^3*b^3 +
2*B*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*
b^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*
tan(d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a
^2*b^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3
*tan(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2
+ b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1)) - 1
/8*sqrt(((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2))*log(-(2*(2*B
^3*a^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) + (2*(B*a^5*b + 3*B*a^3*b^3 +
2*B*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b
^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*t
an(d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^
2*b^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*
tan(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 +
b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1)) + 1/
8*sqrt(((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2))*log((2*(2*B^3
*a^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) + (2*(B*a^5*b + 3*B*a^3*b^3 + 2*
B*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2
+ b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) - ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*tan
(d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*
b^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*ta
n(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b
^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1)) + 1/8*
sqrt(((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2))*log(-(2*(2*B^3*
a^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) + (2*(B*a^5*b + 3*B*a^3*b^3 + 2*B
*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2
+ b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) - ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*tan(
d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*b
^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*tan
(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b^
2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 - B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1)) - 1/8*s
qrt(-((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 + B^2*b)/((a^2 + b^2)*d^2))*log((2*(2*B^3*a
^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) - (2*(B*a^5*b + 3*B*a^3*b^3 + 2*B*
a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 +
b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*tan(d
*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*b^
4)*d + 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*tan(
d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(-((a^2 + b^
2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 + B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1)) - 1/8*s
qrt(-((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 + B^2*b)/((a^2 + b^2)*d^2))*log(-(2*(2*B^3*
a^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) - (2*(B*a^5*b + 3*B*a^3*b^3 + 2*B
*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2
+ b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*tan(
d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*b
^4)*d + 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*tan
(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(-((a^2 + b
^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 + B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1)) + 1/8*
sqrt(-((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 + B^2*b)/((a^2 + b^2)*d^2))*log((2*(2*B^3*
a^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) - (2*(B*a^5*b + 3*B*a^3*b^3 + 2*B
*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2
+ b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) - ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*tan(
d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*b
^4)*d + 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*tan
(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(-((a^2 + b
^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 + B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1)) + 1/8*
sqrt(-((a^2 + b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 + B^2*b)/((a^2 + b^2)*d^2))*log(-(2*(2*B^3
*a^4*b + 4*B^3*a^2*b^3 + (B^3*a^5 + 3*B^3*a^3*b^2 + 4*B^3*a*b^4)*tan(d*x + c) - (2*(B*a^5*b + 3*B*a^3*b^3 + 2*
B*a*b^5)*d^2*tan(d*x + c) - (B*a^6 + 4*B*a^4*b^2 + 7*B*a^2*b^4 + 4*B*b^6)*d^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2
+ b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) - ((B^2*a^6 + 7*B^2*a^4*b^2 + 12*B^2*a^2*b^4)*d*tan
(d*x + c)^2 + 2*(B^2*a^5*b + B^2*a^3*b^3 - 4*B^2*a*b^5)*d*tan(d*x + c) - (B^2*a^6 + 3*B^2*a^4*b^2 + 4*B^2*a^2*
b^4)*d + 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*ta
n(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(-((a^2 +
b^2)*sqrt(-B^4*a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*d^2 + B^2*b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1))

Sympy [F]

\[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=B \int \frac {\sqrt {\tan {\left (c + d x \right )}}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B b \tan \left (d x + c\right ) + B a\right )} \sqrt {\tan \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (B\,a+B\,b\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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