Optimal. Leaf size=204 \[ -\frac {(i a-b)^{3/2} (A+i B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} (2 A b+3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \]
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Rubi [A]
time = 1.24, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3688, 3736,
6857, 65, 223, 212, 95, 211, 214} \begin {gather*} -\frac {(-b+i a)^{3/2} (A+i B) \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} (3 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(b+i a)^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 95
Rule 211
Rule 212
Rule 214
Rule 223
Rule 3688
Rule 3736
Rule 6857
Rubi steps
\begin {align*} \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx &=\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\int \frac {\frac {1}{2} a (2 a A-b B)+\left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {1}{2} b (2 A b+3 a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} a (2 a A-b B)+\left (2 a A b+a^2 B-b^2 B\right ) x+\frac {1}{2} b (2 A b+3 a B) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\text {Subst}\left (\int \left (\frac {b (2 A b+3 a B)}{2 \sqrt {x} \sqrt {a+b x}}+\frac {a^2 A-A b^2-2 a b B+\left (2 a A b+a^2 B-b^2 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\text {Subst}\left (\int \frac {a^2 A-A b^2-2 a b B+\left (2 a A b+a^2 B-b^2 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac {(b (2 A b+3 a B)) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\text {Subst}\left (\int \left (\frac {-2 a A b-a^2 B+b^2 B+i \left (a^2 A-A b^2-2 a b B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {2 a A b+a^2 B-b^2 B+i \left (a^2 A-A b^2-2 a b B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {(b (2 A b+3 a B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\left ((a+i b)^2 (i A-B)\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left ((a-i b)^2 (i A+B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(b (2 A b+3 a B)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {\sqrt {b} (2 A b+3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\left ((a+i b)^2 (i A-B)\right ) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left ((a-i b)^2 (i A+B)\right ) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=-\frac {(i a-b)^{3/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} (2 A b+3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\\ \end {align*}
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Mathematica [A]
time = 0.66, size = 243, normalized size = 1.19 \begin {gather*} \frac {-\sqrt [4]{-1} (-a+i b)^{3/2} (i A+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 )-(-1)^{3/4} (a+i b)^{3/2} (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 )+b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} \sqrt {b} (2 A b+3 a B) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}}{d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.54, size = 2398415, normalized size = 11756.94 \[\text {output too large to display}\]
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 ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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