Optimal. Leaf size=256 \[ \frac {(A+i B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}-\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}-\frac {2 b \left (a^2 A+2 A b^2-a b B\right )}{a^2 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 A \sqrt {\cot (c+d x)}}{a d \sqrt {a+b \tan (c+d x)}} \]
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Rubi [A]
time = 0.64, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4326, 3690,
3730, 3697, 3696, 95, 209, 212} \begin {gather*} -\frac {2 b \left (a^2 A-a b B+2 A b^2\right )}{a^2 d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {(A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac {(A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}}-\frac {2 A \sqrt {\cot (c+d x)}}{a d \sqrt {a+b \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 209
Rule 212
Rule 3690
Rule 3696
Rule 3697
Rule 3730
Rule 4326
Rubi steps
\begin {align*} \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx\\ &=-\frac {2 A \sqrt {\cot (c+d x)}}{a d \sqrt {a+b \tan (c+d x)}}-\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{2} (2 A b-a B)+\frac {1}{2} a A \tan (c+d x)+A b \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{a}\\ &=-\frac {2 b \left (a^2 A+2 A b^2-a b B\right )}{a^2 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 A \sqrt {\cot (c+d x)}}{a d \sqrt {a+b \tan (c+d x)}}-\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{4} a^2 (A b-a B)+\frac {1}{4} a^2 (a A+b B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=-\frac {2 b \left (a^2 A+2 A b^2-a b B\right )}{a^2 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 A \sqrt {\cot (c+d x)}}{a d \sqrt {a+b \tan (c+d x)}}-\frac {\left ((i a+b) (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 \left (a^2+b^2\right )}-\frac {\left (2 \left (\frac {1}{4} a^2 (A b-a B)-\frac {1}{4} i a^2 (a A+b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=-\frac {2 b \left (a^2 A+2 A b^2-a b B\right )}{a^2 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 A \sqrt {\cot (c+d x)}}{a d \sqrt {a+b \tan (c+d x)}}-\frac {\left ((i a+b) (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac {\left (2 \left (\frac {1}{4} a^2 (A b-a B)-\frac {1}{4} i a^2 (a A+b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{a^2 \left (a^2+b^2\right ) d}\\ &=-\frac {2 b \left (a^2 A+2 A b^2-a b B\right )}{a^2 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 A \sqrt {\cot (c+d x)}}{a d \sqrt {a+b \tan (c+d x)}}-\frac {\left ((i a+b) (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\left (a^2+b^2\right ) d}-\frac {\left (4 \left (\frac {1}{4} a^2 (A b-a B)-\frac {1}{4} i a^2 (a A+b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{a^2 \left (a^2+b^2\right ) d}\\ &=\frac {(A+i B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}-\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}-\frac {2 b \left (a^2 A+2 A b^2-a b B\right )}{a^2 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 A \sqrt {\cot (c+d x)}}{a d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 1.50, size = 255, normalized size = 1.00 \begin {gather*} \frac {\sqrt {\cot (c+d x)} \left (\frac {\sqrt [4]{-1} a \left (\frac {(a+i b) (A-i B) \text {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 {(a-i b) (A+i B) \text {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 ) \sqrt {\tan (c+d x)}}{a^2+b^2}-\frac {2 A}{\sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2 A+2 A b^2-a b B\right ) \tan (c+d x)}{a \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )}{a d} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 31.14, size = 18997, normalized size = 74.21
method | result | size |
default | \(\text {Expression too large to display}\) | \(18997\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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