3.2.18 \(\int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx\) [118]

Optimal. Leaf size=224 \[ -\frac {2 a^2 \text {ArcTan}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3} \]

[Out]

-2*a^2*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)+2*a^2*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))/d/e^(3/2)+3*
a^2*sec(d*x+c)*(e*sin(d*x+c))^(3/2)/d/e^3-4*a^2/d/e/(e*sin(d*x+c))^(1/2)-2*a^2*cos(d*x+c)/d/e/(e*sin(d*x+c))^(
1/2)-2*a^2*sec(d*x+c)/d/e/(e*sin(d*x+c))^(1/2)+5*a^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*
d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/d/e^2/sin(d*x+c)^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3957, 2952, 2716, 2721, 2719, 2644, 331, 335, 304, 209, 212, 2650, 2651} \begin {gather*} -\frac {2 a^2 \text {ArcTan}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2/(e*Sin[c + d*x])^(3/2),x]

[Out]

(-2*a^2*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e^(3/2)) + (2*a^2*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/(d*e
^(3/2)) - (4*a^2)/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a^2*Sec[c
+ d*x])/(d*e*Sqrt[e*Sin[c + d*x]]) - (5*a^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*e^2*Sqrt
[Sin[c + d*x]]) + (3*a^2*Sec[c + d*x]*(e*Sin[c + d*x])^(3/2))/(d*e^3)

Rule 209

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

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 331

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

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2650

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*Cos[e + f*
x])^(n + 1)*((a*Sin[e + f*x])^(m + 1)/(a*b*f*(m + 1))), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Cos[e + f*
x])^n*(a*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2651

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*Sin[e +
f*x])^(n + 1))*((a*Cos[e + f*x])^(m + 1)/(a*b*f*(m + 1))), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Sin[e +
f*x])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{3/2}} \, dx &=\int \frac {(-a-a \cos (c+d x))^2 \sec ^2(c+d x)}{(e \sin (c+d x))^{3/2}} \, dx\\ &=\int \left (\frac {a^2}{(e \sin (c+d x))^{3/2}}+\frac {2 a^2 \sec (c+d x)}{(e \sin (c+d x))^{3/2}}+\frac {a^2 \sec ^2(c+d x)}{(e \sin (c+d x))^{3/2}}\right ) \, dx\\ &=a^2 \int \frac {1}{(e \sin (c+d x))^{3/2}} \, dx+a^2 \int \frac {\sec ^2(c+d x)}{(e \sin (c+d x))^{3/2}} \, dx+\left (2 a^2\right ) \int \frac {\sec (c+d x)}{(e \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {a^2 \int \sqrt {e \sin (c+d x)} \, dx}{e^2}+\frac {\left (3 a^2\right ) \int \sec ^2(c+d x) \sqrt {e \sin (c+d x)} \, dx}{e^2}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{x^{3/2} \left (1-\frac {x^2}{e^2}\right )} \, dx,x,e \sin (c+d x)\right )}{d e}\\ &=-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e^3}-\frac {\left (3 a^2\right ) \int \sqrt {e \sin (c+d x)} \, dx}{2 e^2}-\frac {\left (a^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{e^2 \sqrt {\sin (c+d x)}}\\ &=-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}+\frac {\left (4 a^2\right ) \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{e^2}} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e^3}-\frac {\left (3 a^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{2 e^2 \sqrt {\sin (c+d x)}}\\ &=-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}+\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{e-x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{e+x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e}\\ &=-\frac {2 a^2 \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}+\frac {2 a^2 \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2}}-\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \sec (c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {5 a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \sqrt {\sin (c+d x)}}+\frac {3 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e^3}\\ \end {align*}

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Mathematica [A]
time = 17.68, size = 260, normalized size = 1.16 \begin {gather*} -\frac {\left (1+\cos \left (2 \left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )^2 \cos (c+d x) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^2 \sin ^{\frac {3}{2}}(c+d x) \left (-1+\sin ^2(c+d x)\right ) \left (-2 \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right )-5 E\left (\left .\text {ArcSin}\left (\sqrt {\sin (c+d x)}\right )\right |-1\right )+5 F\left (\left .\text {ArcSin}\left (\sqrt {\sin (c+d x)}\right )\right |-1\right )-\log \left (1-\sqrt {\sin (c+d x)}\right )+\log \left (1+\sqrt {\sin (c+d x)}\right )-\frac {4}{\sqrt {\sin (c+d x)}}+\frac {-4+5 \sin ^2(c+d x)}{\sqrt {\sin (c+d x)} \sqrt {1-\sin ^2(c+d x)}}\right )}{4 d \left (1+\cos \left (2 \left (\frac {c}{2}+\frac {1}{2} (-c+\text {ArcSin}(\sin (c+d x)))\right )\right )\right )^2 (e \sin (c+d x))^{3/2} \sqrt {1-\sin ^2(c+d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^2/(e*Sin[c + d*x])^(3/2),x]

[Out]

-1/4*((1 + Cos[2*(c/2 + (d*x)/2)])^2*Cos[c + d*x]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[c + d*x]^(3/
2)*(-1 + Sin[c + d*x]^2)*(-2*ArcTan[Sqrt[Sin[c + d*x]]] - 5*EllipticE[ArcSin[Sqrt[Sin[c + d*x]]], -1] + 5*Elli
pticF[ArcSin[Sqrt[Sin[c + d*x]]], -1] - Log[1 - Sqrt[Sin[c + d*x]]] + Log[1 + Sqrt[Sin[c + d*x]]] - 4/Sqrt[Sin
[c + d*x]] + (-4 + 5*Sin[c + d*x]^2)/(Sqrt[Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])))/(d*(1 + Cos[2*(c/2 + (-c
+ ArcSin[Sin[c + d*x]])/2)])^2*(e*Sin[c + d*x])^(3/2)*Sqrt[1 - Sin[c + d*x]^2])

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Maple [A]
time = 0.24, size = 238, normalized size = 1.06

method	result	size
default	\(\frac {a^{2} \left (10 \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\sin \left (d x +c \right )+1}\, e^{\frac {3}{2}}-5 \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\sin \left (d x +c \right )+1}\, e^{\frac {3}{2}}-10 e^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right )-8 e^{\frac {3}{2}} \cos \left (d x +c \right )+4 \arctanh \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) e -4 \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) e +2 e^{\frac {3}{2}}\right )}{2 e^{\frac {5}{2}} \sqrt {e \sin \left (d x +c \right )}\, \cos \left (d x +c \right ) d}\)	\(238\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/2/e^(5/2)/(e*sin(d*x+c))^(1/2)/cos(d*x+c)*a^2*(10*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*
x+c)+1)^(1/2),1/2*2^(1/2))*(-sin(d*x+c)+1)^(1/2)*e^(3/2)-5*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((
-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*(-sin(d*x+c)+1)^(1/2)*e^(3/2)-10*e^(3/2)*cos(d*x+c)^2-8*e^(3/2)*cos(d*x+c)+4
*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))*(e*sin(d*x+c))^(1/2)*cos(d*x+c)*e-4*arctan((e*sin(d*x+c))^(1/2)/e^(1/2)
)*(e*sin(d*x+c))^(1/2)*cos(d*x+c)*e+2*e^(3/2))/d

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Maxima [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+a*sec(d*x+c))^2/(e*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.37, size = 262, normalized size = 1.17 \begin {gather*} \frac {{\left (-5 i \, \sqrt {2} \sqrt {-i} a^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 5 i \, \sqrt {2} \sqrt {i} a^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, a^{2} \arctan \left (\frac {\sin \left (d x + c\right ) - 1}{2 \, \sqrt {\sin \left (d x + c\right )}}\right ) \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{2} \cos \left (d x + c\right ) \log \left (\frac {\cos \left (d x + c\right )^{2} - 4 \, {\left (\sin \left (d x + c\right ) + 1\right )} \sqrt {\sin \left (d x + c\right )} - 6 \, \sin \left (d x + c\right ) - 2}{\cos \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) - 2}\right ) \sin \left (d x + c\right ) - 2 \, {\left (5 \, a^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sqrt {\sin \left (d x + c\right )}\right )} e^{\left (-\frac {3}{2}\right )}}{2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/2*(-5*I*sqrt(2)*sqrt(-I)*a^2*cos(d*x + c)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d
*x + c) + I*sin(d*x + c))) + 5*I*sqrt(2)*sqrt(I)*a^2*cos(d*x + c)*sin(d*x + c)*weierstrassZeta(4, 0, weierstra
ssPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*a^2*arctan(1/2*(sin(d*x + c) - 1)/sqrt(sin(d*x + c)))*cos
(d*x + c)*sin(d*x + c) + a^2*cos(d*x + c)*log((cos(d*x + c)^2 - 4*(sin(d*x + c) + 1)*sqrt(sin(d*x + c)) - 6*si
n(d*x + c) - 2)/(cos(d*x + c)^2 + 2*sin(d*x + c) - 2))*sin(d*x + c) - 2*(5*a^2*cos(d*x + c)^2 + 4*a^2*cos(d*x
+ c) - a^2)*sqrt(sin(d*x + c)))*e^(-3/2)/(d*cos(d*x + c)*sin(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \frac {1}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**2/(e*sin(d*x+c))**(3/2),x)

[Out]

a**2*(Integral((e*sin(c + d*x))**(-3/2), x) + Integral(2*sec(c + d*x)/(e*sin(c + d*x))**(3/2), x) + Integral(s
ec(c + d*x)**2/(e*sin(c + d*x))**(3/2), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^2/(e*sin(d*x + c))^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^2/(e*sin(c + d*x))^(3/2),x)

[Out]

int((a + a/cos(c + d*x))^2/(e*sin(c + d*x))^(3/2), x)

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