3.365 \(\int \frac{a B+b B \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx\)

Optimal. Leaf size=362 \[ \frac{b B \log \left (-\sqrt{2} \sqrt{\sqrt{a^2+b^2}+a} \sqrt{a+b \tan (c+d x)}+\sqrt{a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt{2} d \sqrt{\sqrt{a^2+b^2}+a}}-\frac{b B \log \left (\sqrt{2} \sqrt{\sqrt{a^2+b^2}+a} \sqrt{a+b \tan (c+d x)}+\sqrt{a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt{2} d \sqrt{\sqrt{a^2+b^2}+a}}+\frac{b B \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a^2+b^2}+a}-\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} d \sqrt{a-\sqrt{a^2+b^2}}}-\frac{b B \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a^2+b^2}+a}+\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} d \sqrt{a-\sqrt{a^2+b^2}}} \]

[Out]

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

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Rubi [A]  time = 0.328009, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {21, 3485, 700, 1129, 634, 618, 206, 628} \[ \frac{b B \log \left (-\sqrt{2} \sqrt{\sqrt{a^2+b^2}+a} \sqrt{a+b \tan (c+d x)}+\sqrt{a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt{2} d \sqrt{\sqrt{a^2+b^2}+a}}-\frac{b B \log \left (\sqrt{2} \sqrt{\sqrt{a^2+b^2}+a} \sqrt{a+b \tan (c+d x)}+\sqrt{a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt{2} d \sqrt{\sqrt{a^2+b^2}+a}}+\frac{b B \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a^2+b^2}+a}-\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} d \sqrt{a-\sqrt{a^2+b^2}}}-\frac{b B \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a^2+b^2}+a}+\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} d \sqrt{a-\sqrt{a^2+b^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*B*ArcTanh[(Sqrt[a + Sqrt[a^2 + b^2]] - Sqrt[2]*Sqrt[a + b*Tan[c + d*x]])/Sqrt[a - Sqrt[a^2 + b^2]]])/(Sqrt[
2]*Sqrt[a - Sqrt[a^2 + b^2]]*d) - (b*B*ArcTanh[(Sqrt[a + Sqrt[a^2 + b^2]] + Sqrt[2]*Sqrt[a + b*Tan[c + d*x]])/
Sqrt[a - Sqrt[a^2 + b^2]]])/(Sqrt[2]*Sqrt[a - Sqrt[a^2 + b^2]]*d) + (b*B*Log[a + Sqrt[a^2 + b^2] + b*Tan[c + d
*x] - Sqrt[2]*Sqrt[a + Sqrt[a^2 + b^2]]*Sqrt[a + b*Tan[c + d*x]]])/(2*Sqrt[2]*Sqrt[a + Sqrt[a^2 + b^2]]*d) - (
b*B*Log[a + Sqrt[a^2 + b^2] + b*Tan[c + d*x] + Sqrt[2]*Sqrt[a + Sqrt[a^2 + b^2]]*Sqrt[a + b*Tan[c + d*x]]])/(2
*Sqrt[2]*Sqrt[a + Sqrt[a^2 + b^2]]*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 3485

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rule 700

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(c*d^2 + a*e^2 - 2*c*d
*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1129

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/
c, 2]}, Dist[1/(2*c*r), Int[x^(m - 1)/(q - r*x + x^2), x], x] - Dist[1/(2*c*r), Int[x^(m - 1)/(q + r*x + x^2),
x], x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 1] && LtQ[m, 3] && NegQ[b^2 - 4*a*c]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{a B+b B \tan (c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx &=B \int \sqrt{a+b \tan (c+d x)} \, dx\\ &=\frac{(b B) \operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac{(2 b B) \operatorname{Subst}\left (\int \frac{x^2}{a^2+b^2-2 a x^2+x^4} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{d}\\ &=\frac{(b B) \operatorname{Subst}\left (\int \frac{x}{\sqrt{a^2+b^2}-\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} x+x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} d}-\frac{(b B) \operatorname{Subst}\left (\int \frac{x}{\sqrt{a^2+b^2}+\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} x+x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} d}\\ &=\frac{(b B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a^2+b^2}-\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} x+x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{2 d}+\frac{(b B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a^2+b^2}+\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} x+x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{2 d}+\frac{(b B) \operatorname{Subst}\left (\int \frac{-\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}}+2 x}{\sqrt{a^2+b^2}-\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} x+x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{2 \sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} d}-\frac{(b B) \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}}+2 x}{\sqrt{a^2+b^2}+\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} x+x^2} \, dx,x,\sqrt{a+b \tan (c+d x)}\right )}{2 \sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} d}\\ &=\frac{b B \log \left (a+\sqrt{a^2+b^2}+b \tan (c+d x)-\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} \sqrt{a+b \tan (c+d x)}\right )}{2 \sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} d}-\frac{b B \log \left (a+\sqrt{a^2+b^2}+b \tan (c+d x)+\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} \sqrt{a+b \tan (c+d x)}\right )}{2 \sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} d}-\frac{(b B) \operatorname{Subst}\left (\int \frac{1}{2 \left (a-\sqrt{a^2+b^2}\right )-x^2} \, dx,x,-\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}}+2 \sqrt{a+b \tan (c+d x)}\right )}{d}-\frac{(b B) \operatorname{Subst}\left (\int \frac{1}{2 \left (a-\sqrt{a^2+b^2}\right )-x^2} \, dx,x,\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}}+2 \sqrt{a+b \tan (c+d x)}\right )}{d}\\ &=\frac{b B \tanh ^{-1}\left (\frac{\sqrt{a+\sqrt{a^2+b^2}}-\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} \sqrt{a-\sqrt{a^2+b^2}} d}-\frac{b B \tanh ^{-1}\left (\frac{\sqrt{a+\sqrt{a^2+b^2}}+\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} \sqrt{a-\sqrt{a^2+b^2}} d}+\frac{b B \log \left (a+\sqrt{a^2+b^2}+b \tan (c+d x)-\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} \sqrt{a+b \tan (c+d x)}\right )}{2 \sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} d}-\frac{b B \log \left (a+\sqrt{a^2+b^2}+b \tan (c+d x)+\sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} \sqrt{a+b \tan (c+d x)}\right )}{2 \sqrt{2} \sqrt{a+\sqrt{a^2+b^2}} d}\\ \end{align*}

Mathematica [C]  time = 0.0830508, size = 88, normalized size = 0.24 \[ -\frac{i B \left (\sqrt{a-i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )-\sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.103, size = 662, normalized size = 1.8 \begin{align*}{\frac{aB}{4\,bd}\ln \left ( b\tan \left ( dx+c \right ) +a+\sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}+\sqrt{{a}^{2}+{b}^{2}} \right ) \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}}-{\frac{{a}^{2}B}{bd}\arctan \left ({ \left ( 2\,\sqrt{a+b\tan \left ( dx+c \right ) }+\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}-{\frac{B}{4\,bd}\ln \left ( b\tan \left ( dx+c \right ) +a+\sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}+\sqrt{{a}^{2}+{b}^{2}} \right ) \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\sqrt{{a}^{2}+{b}^{2}}}+{\frac{B \left ({a}^{2}+{b}^{2} \right ) }{bd}\arctan \left ({ \left ( 2\,\sqrt{a+b\tan \left ( dx+c \right ) }+\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}-{\frac{aB}{4\,bd}\ln \left ( \sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-b\tan \left ( dx+c \right ) -a-\sqrt{{a}^{2}+{b}^{2}} \right ) \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}}+{\frac{{a}^{2}B}{bd}\arctan \left ({ \left ( \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-2\,\sqrt{a+b\tan \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}+{\frac{B}{4\,bd}\ln \left ( \sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-b\tan \left ( dx+c \right ) -a-\sqrt{{a}^{2}+{b}^{2}} \right ) \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\sqrt{{a}^{2}+{b}^{2}}}-{\frac{B \left ({a}^{2}+{b}^{2} \right ) }{bd}\arctan \left ({ \left ( \sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}-2\,\sqrt{a+b\tan \left ( dx+c \right ) } \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)
^(1/2)+2*a)^(1/2)*a-1/d*B/b*a^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1
/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))-1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(
1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+1/d*B/b*(a^2+b^2)/(2*(a^2+b^2
)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/
2))-1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*B*(2*(a^2+
b^2)^(1/2)+2*a)^(1/2)*a+1/d*B/b*a^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b
*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(
1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)-1/d*B/b*(a^2+b^2)/(2*(a^2
+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)
^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.24889, size = 4263, normalized size = 11.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*sqrt(2)*sqrt(B^4*b^2/d^4)*d^4*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2)
)*((B^4*a^2 + B^4*b^2)/d^4)^(3/4)*arctan(-(sqrt(2)*sqrt(B^4*b^2/d^4)*B^3*b*d^5*sqrt((a*cos(d*x + c) + b*sin(d*
x + c))/cos(d*x + c))*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2))*((B^4*a^2 + B^
4*b^2)/d^4)^(3/4) - sqrt(2)*sqrt(B^4*b^2/d^4)*d^5*sqrt((sqrt(2)*B^3*b^3*d^3*sqrt((a*cos(d*x + c) + b*sin(d*x +
c))/cos(d*x + c))*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2))*((B^4*a^2 + B^4*b
^2)/d^4)^(3/4)*cos(d*x + c) + (B^4*a^2*b^2 + B^4*b^4)*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4)*cos(d*x + c) + (B^6*a^
3*b^2 + B^6*a*b^4)*cos(d*x + c) + (B^6*a^2*b^3 + B^6*b^5)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c)))*sqrt((B^2*
a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2))*((B^4*a^2 + B^4*b^2)/d^4)^(3/4) + (B^4*a^2 + B
^4*b^2)*sqrt(B^4*b^2/d^4)*d^4*sqrt((B^4*a^2 + B^4*b^2)/d^4) + (B^6*a^3 + B^6*a*b^2)*sqrt(B^4*b^2/d^4)*d^2)/(B^
8*a^2*b^2 + B^8*b^4)) + 4*sqrt(2)*sqrt(B^4*b^2/d^4)*d^4*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^
2)/d^4))/(B^2*b^2))*((B^4*a^2 + B^4*b^2)/d^4)^(3/4)*arctan(-(sqrt(2)*sqrt(B^4*b^2/d^4)*B^3*b*d^5*sqrt((a*cos(d
*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^
2))*((B^4*a^2 + B^4*b^2)/d^4)^(3/4) - sqrt(2)*sqrt(B^4*b^2/d^4)*d^5*sqrt(-(sqrt(2)*B^3*b^3*d^3*sqrt((a*cos(d*x
+ c) + b*sin(d*x + c))/cos(d*x + c))*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2)
)*((B^4*a^2 + B^4*b^2)/d^4)^(3/4)*cos(d*x + c) - (B^4*a^2*b^2 + B^4*b^4)*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4)*cos
(d*x + c) - (B^6*a^3*b^2 + B^6*a*b^4)*cos(d*x + c) - (B^6*a^2*b^3 + B^6*b^5)*sin(d*x + c))/((a^2 + b^2)*cos(d*
x + c)))*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2))*((B^4*a^2 + B^4*b^2)/d^4)^(
3/4) - (B^4*a^2 + B^4*b^2)*sqrt(B^4*b^2/d^4)*d^4*sqrt((B^4*a^2 + B^4*b^2)/d^4) - (B^6*a^3 + B^6*a*b^2)*sqrt(B^
4*b^2/d^4)*d^2)/(B^8*a^2*b^2 + B^8*b^4)) + sqrt(2)*(B^4*a^2 + B^4*b^2 - B^2*a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4
))*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2))*((B^4*a^2 + B^4*b^2)/d^4)^(1/4)*l
og((sqrt(2)*B^3*b^3*d^3*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*s
qrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2))*((B^4*a^2 + B^4*b^2)/d^4)^(3/4)*cos(d*x + c) + (B^4*a^2*b^2 + B^4*b^4
)*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4)*cos(d*x + c) + (B^6*a^3*b^2 + B^6*a*b^4)*cos(d*x + c) + (B^6*a^2*b^3 + B^6
*b^5)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c))) - sqrt(2)*(B^4*a^2 + B^4*b^2 - B^2*a*d^2*sqrt((B^4*a^2 + B^4*b
^2)/d^4))*sqrt((B^2*a^2 + B^2*b^2 + a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2))*((B^4*a^2 + B^4*b^2)/d^4)^
(1/4)*log(-(sqrt(2)*B^3*b^3*d^3*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt((B^2*a^2 + B^2*b^2 +
a*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4))/(B^2*b^2))*((B^4*a^2 + B^4*b^2)/d^4)^(3/4)*cos(d*x + c) - (B^4*a^2*b^2 +
B^4*b^4)*d^2*sqrt((B^4*a^2 + B^4*b^2)/d^4)*cos(d*x + c) - (B^6*a^3*b^2 + B^6*a*b^4)*cos(d*x + c) - (B^6*a^2*b
^3 + B^6*b^5)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c))))/(B^4*a^2 + B^4*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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