3.10 \(\int (a+b \csc ^2(c+d x))^{3/2} \, dx\)

Optimal. Leaf size=119 \[ -\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{d}-\frac {b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}}{2 d}-\frac {\sqrt {b} (3 a+b) \tanh ^{-1}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{2 d} \]

[Out]

-a^(3/2)*arctan(cot(d*x+c)*a^(1/2)/(a+b+b*cot(d*x+c)^2)^(1/2))/d-1/2*(3*a+b)*arctanh(cot(d*x+c)*b^(1/2)/(a+b+b
*cot(d*x+c)^2)^(1/2))*b^(1/2)/d-1/2*b*cot(d*x+c)*(a+b+b*cot(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.10, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4128, 416, 523, 217, 206, 377, 203} \[ -\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{d}-\frac {b \cot (c+d x) \sqrt {a+b \cot ^2(c+d x)+b}}{2 d}-\frac {\sqrt {b} (3 a+b) \tanh ^{-1}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x]^2)^(3/2),x]

[Out]

-((a^(3/2)*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/d) - (Sqrt[b]*(3*a + b)*ArcTanh[(Sqr
t[b]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]])/(2*d) - (b*Cot[c + d*x]*Sqrt[a + b + b*Cot[c + d*x]^2])/(2
*d)

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])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \left (a+b \csc ^2(c+d x)\right )^{3/2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {b \cot (c+d x) \sqrt {a+b+b \cot ^2(c+d x)}}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {(a+b) (2 a+b)+b (3 a+b) x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-\frac {b \cot (c+d x) \sqrt {a+b+b \cot ^2(c+d x)}}{2 d}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\cot (c+d x)\right )}{d}-\frac {(b (3 a+b)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-\frac {b \cot (c+d x) \sqrt {a+b+b \cot ^2(c+d x)}}{2 d}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{d}-\frac {(b (3 a+b)) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{2 d}\\ &=-\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{d}-\frac {\sqrt {b} (3 a+b) \tanh ^{-1}\left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{2 d}-\frac {b \cot (c+d x) \sqrt {a+b+b \cot ^2(c+d x)}}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.91, size = 208, normalized size = 1.75 \[ \frac {b \sin ^3(c+d x) \left (a+b \csc ^2(c+d x)\right )^{3/2} \left (\sqrt {-b} \left (2 \sqrt {2} a^{3/2} \log \left (\sqrt {a \cos (2 (c+d x))-a-2 b}+\sqrt {2} \sqrt {a} \cos (c+d x)\right )-b \cot (c+d x) \csc (c+d x) \sqrt {a \cos (2 (c+d x))-a-2 b}\right )+\sqrt {2} b (3 a+b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {-b} \cos (c+d x)}{\sqrt {a \cos (2 (c+d x))-a-2 b}}\right )\right )}{(-b)^{3/2} d (a \cos (2 (c+d x))-a-2 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(a + b*Csc[c + d*x]^2)^(3/2)*(Sqrt[2]*b*(3*a + b)*ArcTanh[(Sqrt[2]*Sqrt[-b]*Cos[c + d*x])/Sqrt[-a - 2*b + a
*Cos[2*(c + d*x)]]] + Sqrt[-b]*(-(b*Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]*Cot[c + d*x]*Csc[c + d*x]) + 2*Sqrt[2]
*a^(3/2)*Log[Sqrt[2]*Sqrt[a]*Cos[c + d*x] + Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]]))*Sin[c + d*x]^3)/((-b)^(3/2)
*d*(-a - 2*b + a*Cos[2*(c + d*x)])^(3/2))

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fricas [B]  time = 1.02, size = 1607, normalized size = 13.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(-a)*a*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 + a^3*b)*cos(d*x + c)^6 + 160*(a^4 + 2*a^3*b + a^2*b^2)
*cos(d*x + c)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(d*x +
c)^2 - 8*(16*a^3*cos(d*x + c)^7 - 24*(a^3 + a^2*b)*cos(d*x + c)^5 + 10*(a^3 + 2*a^2*b + a*b^2)*cos(d*x + c)^3
- (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1)
)*sin(d*x + c))*sin(d*x + c) + (3*a + b)*sqrt(b)*log(2*((a^2 - 6*a*b + b^2)*cos(d*x + c)^4 - 2*(a^2 - 2*a*b -
3*b^2)*cos(d*x + c)^2 + 4*((a - b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(b)*sqrt((a*cos(d*x + c)^2 - a -
b)/(cos(d*x + c)^2 - 1))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1))*sin(d*x +
c) - 4*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x + c))/(d*sin(d*x + c)), -1/8*(2*(3*a +
b)*sqrt(-b)*arctan(-1/2*((a - b)*cos(d*x + c)^2 - a - b)*sqrt(-b)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x +
c)^2 - 1))*sin(d*x + c)/(a*b*cos(d*x + c)^3 - (a*b + b^2)*cos(d*x + c)))*sin(d*x + c) - sqrt(-a)*a*log(128*a^4
*cos(d*x + c)^8 - 256*(a^4 + a^3*b)*cos(d*x + c)^6 + 160*(a^4 + 2*a^3*b + a^2*b^2)*cos(d*x + c)^4 + a^4 + 4*a^
3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(d*x + c)^2 - 8*(16*a^3*cos(d*x +
c)^7 - 24*(a^3 + a^2*b)*cos(d*x + c)^5 + 10*(a^3 + 2*a^2*b + a*b^2)*cos(d*x + c)^3 - (a^3 + 3*a^2*b + 3*a*b^2
+ b^3)*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c))*sin(d*x + c)
+ 4*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x + c))/(d*sin(d*x + c)), 1/8*(2*a^(3/2)*ar
ctan(1/4*(8*a^2*cos(d*x + c)^4 - 8*(a^2 + a*b)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*sqrt(a)*sqrt((a*cos(d*x + c
)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c)/(2*a^3*cos(d*x + c)^5 - 3*(a^3 + a^2*b)*cos(d*x + c)^3 + (a^3
+ 2*a^2*b + a*b^2)*cos(d*x + c)))*sin(d*x + c) + (3*a + b)*sqrt(b)*log(2*((a^2 - 6*a*b + b^2)*cos(d*x + c)^4 -
2*(a^2 - 2*a*b - 3*b^2)*cos(d*x + c)^2 + 4*((a - b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(b)*sqrt((a*co
s(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(cos(d*x + c)^4 - 2*cos(d*x + c)
^2 + 1))*sin(d*x + c) - 4*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*cos(d*x + c))/(d*sin(d*x + c
)), 1/4*(a^(3/2)*arctan(1/4*(8*a^2*cos(d*x + c)^4 - 8*(a^2 + a*b)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*sqrt(a)*
sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c)/(2*a^3*cos(d*x + c)^5 - 3*(a^3 + a^2*b)*cos
(d*x + c)^3 + (a^3 + 2*a^2*b + a*b^2)*cos(d*x + c)))*sin(d*x + c) - (3*a + b)*sqrt(-b)*arctan(-1/2*((a - b)*co
s(d*x + c)^2 - a - b)*sqrt(-b)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c)/(a*b*cos(d*x
+ c)^3 - (a*b + b^2)*cos(d*x + c)))*sin(d*x + c) - 2*b*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*
cos(d*x + c))/(d*sin(d*x + c))]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(sin(d*x+c))]Error: Bad Argument Type

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maple [B]  time = 2.30, size = 1286, normalized size = 10.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*csc(d*x+c)^2)^(3/2),x)

[Out]

-1/4/d*((a*cos(d*x+c)^2-a-b)/(cos(d*x+c)^2-1))^(3/2)*(-1+cos(d*x+c))^2*(ln(-2*(-1+cos(d*x+c))*(b^(1/2)*cos(d*x
+c)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)+(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)*b^(1/2)+a*co
s(d*x+c)+a+b)/sin(d*x+c)^2/b^(1/2))*cos(d*x+c)*(-a)^(1/2)*b^(3/2)-ln(-4*(b^(1/2)*cos(d*x+c)*(-(a*cos(d*x+c)^2-
a-b)/(1+cos(d*x+c))^2)^(1/2)+(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)*b^(1/2)-a*cos(d*x+c)+a+b)/(-1+cos(
d*x+c)))*cos(d*x+c)*(-a)^(1/2)*b^(3/2)+3*ln(-2*(-1+cos(d*x+c))*(b^(1/2)*cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(1+c
os(d*x+c))^2)^(1/2)+(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)*b^(1/2)+a*cos(d*x+c)+a+b)/sin(d*x+c)^2/b^(1
/2))*cos(d*x+c)*(-a)^(1/2)*b^(1/2)*a-3*ln(-4*(b^(1/2)*cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2
)+(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)*b^(1/2)-a*cos(d*x+c)+a+b)/(-1+cos(d*x+c)))*cos(d*x+c)*(-a)^(1
/2)*b^(1/2)*a-b^(3/2)*ln(-2*(-1+cos(d*x+c))*(b^(1/2)*cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)
+(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)*b^(1/2)+a*cos(d*x+c)+a+b)/sin(d*x+c)^2/b^(1/2))*(-a)^(1/2)+b^(
3/2)*ln(-4*(b^(1/2)*cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)+(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*
x+c))^2)^(1/2)*b^(1/2)-a*cos(d*x+c)+a+b)/(-1+cos(d*x+c)))*(-a)^(1/2)+2*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2
)^(1/2)*cos(d*x+c)*(-a)^(1/2)*b-4*ln(4*(-a)^(1/2)*cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)+4*
(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)-4*a*cos(d*x+c))*cos(d*x+c)*a^2-3*a*b^(1/2)*ln(-2*(-1
+cos(d*x+c))*(b^(1/2)*cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)+(-(a*cos(d*x+c)^2-a-b)/(1+cos(
d*x+c))^2)^(1/2)*b^(1/2)+a*cos(d*x+c)+a+b)/sin(d*x+c)^2/b^(1/2))*(-a)^(1/2)+3*b^(1/2)*ln(-4*(b^(1/2)*cos(d*x+c
)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)+(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)*b^(1/2)-a*cos(
d*x+c)+a+b)/(-1+cos(d*x+c)))*a*(-a)^(1/2)+4*a^2*ln(4*(-a)^(1/2)*cos(d*x+c)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c
))^2)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(1/2)-4*a*cos(d*x+c)))/sin(d*x+c)^3/(-(a*cos
(d*x+c)^2-a-b)/(1+cos(d*x+c))^2)^(3/2)/(-a)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \csc \left (d x + c\right )^{2} + a\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*csc(d*x + c)^2 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {b}{{\sin \left (c+d\,x\right )}^2}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/sin(c + d*x)^2)^(3/2),x)

[Out]

int((a + b/sin(c + d*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*csc(c + d*x)**2)**(3/2), x)

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