\(\int \csc ^2(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\) [90]

Optimal result

Integrand size = 25, antiderivative size = 105 \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {3 b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f}-\frac {\cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{f} \]

[Out]

3/2*(a+b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f+3/2*b*(a+b+b*tan(f*x+e)^2)^(1/2)*ta
n(f*x+e)/f-cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/f

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4217, 283, 201, 223, 212} \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 f}+\frac {3 b \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 f}-\frac {\cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{f} \]

[In]

Int[Csc[e + f*x]^2*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(3*Sqrt[b]*(a + b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*f) + (3*b*Tan[e + f*x]*S
qrt[a + b + b*Tan[e + f*x]^2])/(2*f) - (Cot[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/f

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 4217

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1
+ ff^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b+b x^2\right )^{3/2}}{x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{f}+\frac {(3 b) \text {Subst}\left (\int \sqrt {a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {3 b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f}-\frac {\cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{f}+\frac {(3 b (a+b)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f} \\ & = \frac {3 b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f}-\frac {\cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{f}+\frac {(3 b (a+b)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f} \\ & = \frac {3 \sqrt {b} (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {3 b \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 f}-\frac {\cot (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{f} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.61 \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {(a+b) \cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {b \sin ^2(e+f x)}{a+b-a \sin ^2(e+f x)}\right ) \sqrt {a+b \sec ^2(e+f x)}}{f} \]

[In]

Integrate[Csc[e + f*x]^2*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-(((a + b)*Cot[e + f*x]*Hypergeometric2F1[-1/2, 2, 1/2, (b*Sin[e + f*x]^2)/(a + b - a*Sin[e + f*x]^2)]*Sqrt[a
+ b*Sec[e + f*x]^2])/f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(740\) vs. \(2(91)=182\).

Time = 10.17 (sec) , antiderivative size = 741, normalized size of antiderivative = 7.06

method	result	size
default	\(\frac {\left (3 b^{\frac {5}{2}} \ln \left (\frac {-4 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )-1}\right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+3 b^{\frac {5}{2}} \ln \left (-\frac {4 \left (-\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )+\sin \left (f x +e \right ) a -\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+a +b \right )}{\sin \left (f x +e \right )+1}\right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+3 b^{\frac {3}{2}} \ln \left (\frac {-4 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \sin \left (f x +e \right ) a -4 a -4 b}{\sin \left (f x +e \right )-1}\right ) a \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+3 b^{\frac {3}{2}} \ln \left (-\frac {4 \left (-\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \cos \left (f x +e \right )+\sin \left (f x +e \right ) a -\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+a +b \right )}{\sin \left (f x +e \right )+1}\right ) a \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )-4 \cos \left (f x +e \right )^{3} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b -6 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \cos \left (f x +e \right )^{3}-4 \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, a b -6 \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}+2 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2} \cos \left (f x +e \right )+2 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, b^{2}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \cot \left (f x +e \right )}{4 f b \left (b +a \cos \left (f x +e \right )^{2}\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (1+\cos \left (f x +e \right )\right )}\)	\(741\)

[In]

int(csc(f*x+e)^2*(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/4/f/b*(3*b^(5/2)*ln(4*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-a-b)/(sin(f*x+e)-1))*cos(f*x+e)^2*sin(f*x+e)+3*b^(5/2)*ln(-4*(-((b
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*
x+e))^2)^(1/2)+a+b)/(sin(f*x+e)+1))*cos(f*x+e)^2*sin(f*x+e)+3*b^(3/2)*ln(4*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e)
)^2)^(1/2)*b^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-a-b)/(sin(f*x+e
)-1))*a*cos(f*x+e)^2*sin(f*x+e)+3*b^(3/2)*ln(-4*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+
e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+a+b)/(sin(f*x+e)+1))*a*cos(f*x+e)^2*sin(f*
x+e)-4*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-6*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1
/2)*b^2*cos(f*x+e)^3-4*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-6*cos(f*x+e)^2*((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2+2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*cos(f*x+e)+2*((b+a*cos(
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2)*(a+b*sec(f*x+e)^2)^(3/2)/(b+a*cos(f*x+e)^2)/((b+a*cos(f*x+e)^2)/(1+cos(
f*x+e))^2)^(1/2)/(1+cos(f*x+e))*cot(f*x+e)

Fricas [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 370, normalized size of antiderivative = 3.52 \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [\frac {3 \, {\left (a + b\right )} \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (2 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, f \cos \left (f x + e\right ) \sin \left (f x + e\right )}, \frac {3 \, {\left (a + b\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left ({\left (2 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ] \]

[In]

integrate(csc(f*x+e)^2*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*(a + b)*sqrt(b)*cos(f*x + e)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 +
4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)
+ 8*b^2)/cos(f*x + e)^4)*sin(f*x + e) - 4*((2*a + 3*b)*cos(f*x + e)^2 - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*
x + e)^2))/(f*cos(f*x + e)*sin(f*x + e)), 1/4*(3*(a + b)*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*co
s(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*co
s(f*x + e)*sin(f*x + e) - 2*((2*a + 3*b)*cos(f*x + e)^2 - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*c
os(f*x + e)*sin(f*x + e))]

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(csc(f*x+e)**2*(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.93 \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 3 \, b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} b \tan \left (f x + e\right ) - \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}}}{\tan \left (f x + e\right )}}{2 \, f} \]

[In]

integrate(csc(f*x+e)^2*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*(3*a*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) + 3*b^(3/2)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) +
3*sqrt(b*tan(f*x + e)^2 + a + b)*b*tan(f*x + e) - 2*(b*tan(f*x + e)^2 + a + b)^(3/2)/tan(f*x + e))/f

Giac [F]

\[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(csc(f*x+e)^2*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^2} \,d x \]

[In]

int((a + b/cos(e + f*x)^2)^(3/2)/sin(e + f*x)^2,x)

[Out]

int((a + b/cos(e + f*x)^2)^(3/2)/sin(e + f*x)^2, x)